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Using derivatives to identify extreme values

In many different settings, we are interested in knowing where a function achieves its least and greatest values.

Introduction

In many different settings, we are interested in knowing where a function achieves its least and greatest values. These can be important in applications say to identify a point at which maximum profit or minimum cost occurs or in theory to characterize the behavior of a function or a family of related functions.

Consider the simple and familiar example of a parabolic function such as \(s(t) = -16t^2 + 24t + 32\) (shown at left in Figure) that represents the height of an object tossed vertically: its maximum value occurs at the vertex of the parabola and represents the greatest height the object reaches. This maximum value is an especially important point on the graph, the point at which the curve changes from increasing to decreasing.

For instance, on the right in Figure, \(g\) has a global maximum of \(g(c)\), but \(g\) does not appear to have a global minimum, as the graph of \(g\) seems to decrease indefinitely. Note that the point \((c,g(c))\) marks a fundamental change in the behavior of \(g\), where \(g\) changes from increasing to decreasing; similar things happen at both \((a,g(a))\) and \((b,g(b))\), although these points are not global minima or maxima.

For example, on the right in Figure, \(g\) has a relative minimum of \(g(b)\) at the point \((b,g(b))\) and a relative maximum of \(g(a)\) at \((a,g(a))\). We have already identified the global maximum of \(g\) as \(g(c)\); it can also be considered a relative maximum. Any maximum or minimum may also be called an extreme value of \(f\).

Exploration

Condensed — the full section is in Boelkins, Active Calculus.

Critical numbers and the first derivative test

As seen in the first two functions on the left in Figure, when a continuous function defined on \((a,b)\) changes from being always increasing on interval \((a,c)\) to being always decreasing on interval \((c, b)\) (where \(a \lt c \lt b\)), the function has a relative maximum at \(c\). Similarly, when a continuous function defined on \((a,b)\) changes from being always decreasing on interval \((a,c)\) to being always increasing on interval \((c, b)\) (as in the two rightmost functions in the figure), the function has a relative minimum at \(c\). The middle option in Figure demonstrates that it's possible for a function to have neither a maximum nor minimum at a critical number, as it can be the case that the function doesn't change from increasing to decreasing or vice versa.

Because the sign of the derivative changes at every location \(c\) where a continuous function changes from increasing to decreasing or from decreasing to increasing, there are only two possible ways for these changes in behavior to occur: either \(f'(c) = 0\) or \(f'(c)\) is undefined. Because these values of \(c\) are so important, we call them critical numbers.

Critical numbers are the only possible locations where the function \(f\) may have relative extremes. Again, note that not every critical number produces a maximum or minimum; in the middle graph of Figure, the function pictured there has a horizontal tangent line at the noted point, but the function is increasing before and increasing after, so the critical number does not yield a maximum or minimum.

When \(c\) is a critical number, we say that \((c,f(c))\) is a critical point of the function, or that \(f(c)\) is a critical value. The first derivative test summarizes how sign changes in the first derivative (which can only occur at critical numbers) indicate the presence of a local maximum or minimum for a given function.

Let \(p\) be a critical number of a continuous function \(f\) that is differentiable near \(p\) (except possibly at \(x = p\)). If \(f'\) changes sign from positive to negative at \(p\), then \(f\) has a relative maximum at \(p\). If \(f'\) changes sign from negative to positive at \(p\), then \(f\) has a relative minimum at \(p\).

Condensed — the full section is in Boelkins, Active Calculus.

The second derivative test

Recall that the second derivative of a function tells us several important things about the behavior of the function itself. For instance, if \(f''\) is positive on an interval, then we know that \(f'\) is increasing on that interval and, consequently, that \(f\) is concave up, so throughout that interval the tangent line to \(y = f(x)\) lies below the curve at every point. At a point where \(f'(p) = 0\), the sign of the second derivative determines whether \(f\) has a local minimum or local maximum at the critical number \(p\).

In Figure, we see the four possibilities for a function \(f\) that has a critical number \(p\) at which \(f'(p) = 0\), provided \(f''(p)\) is not zero on an interval including \(p\) (except possibly at \(p\)). On either side of the critical number, \(f''\) can be either positive or negative, and hence \(f\) can be either concave up or concave down. In the first two graphs, \(f\) does not change concavity at \(p\), and in those situations, \(f\) has either a local minimum or local maximum. In particular, if \(f'(p) = 0\) and \(f''(p) \lt 0\), then \(f\) is concave down at \(p\) with a horizontal tangent line, so \(f\) has a local maximum there. This fact, along with the corresponding statement for when \(f''(p)\) is positive, is the substance of the second derivative test.

If \(p\) is a critical number of a continuous function \(f\) such that \(f'(p) = 0\) and \(f''(p) \ne 0\), then \(f\) has a relative maximum at \(p\) if and only if \(f''(p) \lt 0\), and \(f\) has a relative minimum at \(p\) if and only if \(f''(p) \gt 0\).

In the event that \(f''(p) = 0\), the second derivative test is inconclusive. That is, the test doesn't provide us any information. This is because if \(f''(p) = 0\), it is possible that \(f\) has a local minimum, local maximum, or neither. Consider the functions \(f(x) = x^4\), \(g(x) = -x^4\), and \(h(x) = x^3\) at the critical point \(p = 0\).

Just as a first derivative sign chart reveals all of the increasing and decreasing behavior of a function, we can construct a second derivative sign chart that demonstrates all of the important information involving concavity.

Points \(B\), \(C\), and \(D\) in Figure are locations at which the concavity of \(f\) changes. We give a special name to any such point.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

The critical numbers of a continuous function \(f\) are the values of \(p\) for which \(f'(p) = 0\) or \(f'(p)\) does not exist. These values are important because they identify horizontal tangent lines or corner points on the graph, which are the only possible locations at which a local maximum or local minimum can occur.
Given a differentiable function \(f\), whenever \(f'\) is positive, \(f\) is increasing; whenever \(f'\) is negative, \(f\) is decreasing. The first derivative test tells us that at any point where \(f\) changes from increasing to decreasing, \(f\) has a local maximum, while conversely at any point where \(f\) changes from decreasing to increasing \(f\) has a local minimum.
Given a twice differentiable function \(f\), if we have a horizontal tangent line at \(x = p\) and \(f''(p)\) is nonzero, the sign of \(f''\) tells us the concavity of \(f\) and hence whether \(f\) has a maximum or minimum at \(x = p\). In particular, if \(f'(p) = 0\) and \(f''(p) \lt 0\), then \(f\) is concave down at \(p\) and \(f\) has a local maximum there, while if \(f'(p) = 0\) and \(f''(p) \gt 0\), then \(f\) has a local minimum at \(p\). If \(f'(p) = 0\) and \(f''(p) = 0\), then the second derivative does not tell us whether \(f\) has a local extreme at \(p\) or not.

Practice (11)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

In this section we'll learn about things called the First Derivative Test and the Second Derivative Test. It will be helpful before we begin to remind ourselves of a few facts about the relationship between the behavior of a function and the signs of its first and second derivatives.
Drag each of the statements on the left onto the equivalent statement on the right. Each statement on the left corresponds to exactly one statement on the right.
This problem concerns a function about which the following information is known:
- \(f\) is a differentiable function defined at every real number \(x\)
- \(f(0) = -1/2\)
- \(y = f'(x)\) has its graph given at center in Figure
1. Construct a first derivative sign chart for \(f\). Clearly identify all critical numbers of \(f\), where \(f\) is increasing and decreasing, and where \(f\) has local extrema.
2. On the right-hand axes, sketch an approximate graph of \(y = f''(x)\).
3. Construct a second derivative sign chart for \(f\). Clearly identify where \(f\) is concave up and concave down, as well as all inflection points.
4. On the left-hand axes, sketch a possible graph of \(y = f(x)\).
גלה את התשובה
1. From the given graph of \(f'\), the first derivative sign chart for \(f\) shows that \(f\) has critical numbers at \(x = -1, 1\) and that \(f'\) is positive for \(-1 \lt x \lt 1\) and for \(x \gt 1\), while being negative for all \(x \lt -1\). Hence \(f\) is increasing for \(-1 \lt x \lt 1\) and for \(x \gt 1\), and decreasing for \(x \lt -1\). It follows by the first derivative test that \(f\) has a local minimum at \(x = -1\) because \(f\) changes from decreasing to increasing there.
2. Based on where \(f'(x)\) has horizontal tangent lines and where its slope is positive and negative, we can sketch an approximate graph of \(y = f''(x)\) in the usual way that we sketch the derivative of a given function. From the given graph of \(y = f'(x)\), we can see that \(f''(x) = 0\) at \(x \approx -0.35\) and at \(x = 1\), and that between these points, \(f''(x)\) is negative because \(f'(x)\) is decreasing, while \(f''(x)\) is positive everywhere else since \(f'(x)\) is increasing everywhere else. A possible graph of \(y = f''(x)\) is shown at right in the following figure.
3. From our sketch of the second derivative in (b), we can develop a second derivative sign chart for \(f\). In particular, we saw that \(f''(x) = 0\) at \(x \approx -0.35\) and at \(x = 1\), and that between these points, \(f''(x)\) is negative, while \(f''(x)\) is positive everywhere else. This shows that \(f\) is concave down for \(-0.35 \lt x \lt 1\) while being concave up everywhere else. We thus see that \(f\) has points of inflection at both \(x \approx -0.35\) and \(x = 1\).
4. Now that we know where \(f\) is increasing and decreasing as well as where \(f\) is concave up and concave down, along with the fact that \(f(0) = -0.5\), we can sketch an approximate graph of \(y = f(x)\). Note that we also know \(f\) has a horizontal tangent line at \(x = 1\) while being increasing on either side of \(x = 1\), and earlier we established that \(f\) has a local minimum at \(x = -1\). After making a few rough sketches and shifting the graph vertically to pass through \((0,-0.5)\), we arrive at the graph shown at left below.
Suppose that \(g\) is a differentiable function and \(g'(2) = 0\). In addition, suppose that on \(1 \lt x\lt 2\) and \(2 \lt x \lt 3\) it is known that \(g'(x)\) is positive.
1. Does \(g\) have a local maximum, local minimum, or neither at \(x = 2\)? Why?
2. Suppose that \(g''(x)\) exists for every \(x\) such that \(1 \lt x \lt 3\). Reasoning graphically, describe the behavior of \(g''(x)\) for \(x\)-values near \(2\).
3. Besides being a critical number of \(g\), what is special about the value \(x = 2\) in terms of the behavior of the graph of \(g\)?
גלה את התשובה
1. \(g\) has neither a local maximum nor local minimum at \(x = 2\) because \(f'\) does not change sign at \(x = 2\).
2. Because \(g'(2) = 0\) and \(g'\) is positive for \(1 \lt x \lt 2\) and for \(2 \lt x \lt 3\), it must be that \(g'\) is decreasing for \(1 \lt x \lt 2\) and \(g'\) is increasing for \(2 \lt x \lt 3\). Since \(g'\) changes from decreasing to increasing at \(x = 2\), it follows that \(g'\) has a local minimum at \(x = 2\) and thus \(g''(2) = 0\). In addition, \(g''(x)\) must change sign from negative to positive at \(x = 2\). In particular, \(g''\) is negative for \(1 \lt x \lt 2\) and positive for \(2 \lt x \lt 3\).
3. \(g\) has a horizontal tangent line at \(x = 2\), is increasing before \(x = 2\) and decreasing after \(x = 2\), and concave down before \(x = 2\) and concave up after \(x = 2\), and thus \(g\) has a point of inflection at \(x = 2\).
Suppose that \(h\) is a differentiable function whose first derivative is given by the graph in Figure.
1. How many real number solutions can the equation \(h(x) = 0\) have? Why?
2. If \(h(x) = 0\) has two distinct real solutions, what can you say about the signs of the two solutions? Why?
3. Assume that \(\lim_{x \to \infty} h'(x) = 3\), as appears to be indicated in Figure. How will the graph of \(y = h(x)\) appear as \(x \to \infty\)? Why?
4. Describe the concavity of \(y = h(x)\) as fully as you can from the provided information.
גלה את התשובה
1. Since \(h'\) is negative to the left of \(0\) and positive to the right of \(0\), the function \(h\) is decreasing for \(x \lt 0\) and increasing for \(x \gt 0\). So \(h\) can have at most \(2\) real roots (solutions to \(h(x)=0\). As the graphs in the following figure show, it is possible that \(h\) has no real zeros, one real zero, or two real zeros.
2. Given that \(h\) is decreasing for \(x \lt 0\) and increasing for \(x \gt 0\), the function \(h\) has an absolute minimum value at \(x=0\). So one root is to the left of this minimum and the other to the right. Thus, one root is negative and the other positive. See, for instance, the lowest of the three possibilities for \(h\) shown in the figure in (a).
3. The slopes of the tangent lines to \(h\) will all be near \(3\), but less than \(3\), as \(x\) gets large. So the graph of \(h\) will look very much like a line with slope \(3\) for large values of \(x\).
4. We note that \(h'\) is increasing everywhere, \(h\) is concave up everywhere. Since the graph of \(h'\) levels off for large values of \(|x|\), the concavity of \(h\) is small when \(|x|\) is large. Also, the largest value of the slope of the tangent line to \(h'\) occurs when \(x=0\), so \(h\) has its largest second dervitive when \(x=0\). So \(h\) is almost linear as \(x\) decreases without bound, \(h\) has more bend as \(x\) approaches \(0\), has its greatest bend at \(x=0\), then the second derivative of \(h\) as \(x\) increases, ultimately having \(h\) become almost linear as \(x\) increases without bound.
Suppose that \(g(x)\) is a function continuous for every value of \(x \ne 2\) whose first derivative is \(g'(x) = \frac{(x+4)(x-1)^2}{x-2}\). Further, assume that it is known that \(g\) has a vertical asymptote at \(x = 2\).
1. Determine all critical numbers of \(g\).
2. By developing a carefully labeled first derivative sign chart, decide whether \(g\) has as a local maximum, local minimum, or neither at each critical number. Note: observe that \(g'(x)\) can change sign at the vertical asymptote of \(g(x)\).
3. Does \(g\) have a global maximum? global minimum? Justify your claims.
4. What is the value of \(\lim_{x \to \infty} g'(x)\)? What does the value of this limit tell you about the long-term behavior of \(g\)?
5. Sketch a possible graph of \(y = g(x)\).
גלה את התשובה
1. Since \(g'(x) = \frac{(x+4)(x-1)^2}{x-2}\), we see that \(g'(x) = 0\) implies that \(x = -4\) or \(x = 1\). While \(x = 2\) makes \(g'\) undefined, we are told that \(g\) has a vertical asymptote at \(x = 2\), so \(x = 2\) is not in the domain of \(g\), and hence is technically not a critical number of \(g\). Nonetheless, we place \(x = 2\) on our first derivative sign chart since the vertical asymptote is a location at which \(g'\) may change sign.
2. The first derivative sign chart shows that \(g'(x) \gt 0\) for \(x \lt -4\), \(g'(x) \lt 0\) for \(-4 \lt x \lt 1\), \(g'(x) \lt 0\) for \(1 \lt x \lt 2\), and \(g'(x) \gt 0\) for \(x \gt 2\). By the first derivative test, \(g\) has a local maximum at \(x = -4\) and neither a max nor min at \(x = 1\). As these are the only two critical numbers, these are the only two locations for possible extremes. (Note: although \(g\) changes from decreasing to increasing at \(x = 2\), this is due to a vertical asymptote, and \(g\) does not have a minimum there.)
3. Because \(g\) is decreasing as \(x \to 2^-\) (where \(g\) has a vertical asymptote), \(g\) does not have a global minimum. For \(x \gt 2\), \(g\) is always increasing, which suggests that \(g\) does not have a global maximum (though we do not know for sure that \(g\) increases without bound).
4. We observe that \[\begin{aligned}\lim_{x \to \infty} g'(x) =\mathstrut \amp \lim_{x \to \infty} \frac{(x+4)(x-1)^2}{x-2} \\ =\mathstrut \amp \lim_{x \to \infty} \frac{x^3 + 2x^2 - 7x + 4}{x-2} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} \\ =\mathstrut \amp \lim_{x \to \infty} \frac{x^2 + 2x - 7 + \frac{4}{x}}{1 - \frac{2}{x}} \\ =\mathstrut \amp \infty\end{aligned}\] Since \(g'(x) \to \infty\) as \(x \to \infty\), this tells us that \(g\) increases without bound as \(x \to \infty\).
5. From all of our work above, we know that \(g\) has a local maximum at \(x = -4\), a horizontal tangent line with neither a max nor min at \(x = 1\), and a vertical asymptote at \(x = 2\), plus \(g\) and \(g'\) both increase without bound as \(x \to \infty\). Thus, a possible graph of \(g\) is the following.
Let \(p\) be a function whose second derivative is \(p''(x) = (x+1)(x-2)e^{-x}\).
1. Construct a second derivative sign chart for \(p\) and determine all inflection points of \(p\).
2. Suppose you also know that \(x = \frac{\sqrt{5}-1}{2}\) is a critical number of \(p\). Does \(p\) have a local minimum, local maximum, or neither at \(x = \frac{\sqrt{5}-1}{2}\)? Why?
3. If the point \((2, \frac{12}{e^2})\) lies on the graph of \(y = p(x)\) and \(p'(2) = -\frac{5}{e^2}\), find the equation of the tangent line to \(y = p(x)\) at the point where \(x = 2\). Does the tangent line lie above the curve, below the curve, or neither at this value? Why?
גלה את התשובה
1. Note that \(p''(x)=0\) at \(x=-1\) and \(x=2\). The second derivative sign chart for \(p\) shows that \(p''(x)\) is negative for \(-1 \lt x \lt 2\) and positive for all other values of \(x\). Since \(p''\) changes sign at \(x = -1\) and \(x = 2\), \(p\) has points of inflection at \(x = -1\) and \(x = 2\). In particular, \(p\) is concave down for \(-1 \lt x \lt 2\) and concave up for all other values of \(x\).
2. If we compute \(p''\left(\frac{\sqrt{5}-1}{2} \right)\), we find that \(p''\left(\frac{\sqrt{5}-1}{2} \right) \approx -1.205\), which means that \(p\) is concave down at \(x = \frac{\sqrt{5}-1}{2}\). Since \(p\) has a horizontal tangent line at this point, it follows that \(p\) has a local maximum at \(x = \frac{\sqrt{5}-1}{2}\) by the second derivative test.
3. Since \(p(2) = \frac{12}{e^2}\) and \(p'(2) = -\frac{5}{e^2}\), the equation of the tangent line to \(p\) at \(x = 2\) is given by \(y = -\frac{5}{e^2}(x-2) + \frac{12}{e^2}\). To determine whether the tangent line lies above or below the graph of \(y=p(x)\) at \(x = 2\), we consider the concavity of \(p\). Recall we know that \(p''(2) = 0\) and that \(p\) is concave down to the left of \(x = 2\) and concave up to the right. Thus, the tangent line lies neither solely below nor solely above the graph of \(y = p(x)\) at \(x = 2\); it lies below \(y = p(x)\) to the right of \(x = 2\) and above \(y = p(x)\) to the left of \(x = 2\).
This problem concerns a function about which the following information is known:
- \(f\) is a differentiable function defined at every real number \(x\)
- \(f(0) = -1/2\)
- \(y = f'(x)\) has its graph given at center in Figure
1. Construct a first derivative sign chart for \(f\). Clearly identify all critical numbers of \(f\), where \(f\) is increasing and decreasing, and where \(f\) has local extrema.
2. On the right-hand axes, sketch an approximate graph of \(y = f''(x)\).
3. Construct a second derivative sign chart for \(f\). Clearly identify where \(f\) is concave up and concave down, as well as all inflection points.
4. On the left-hand axes, sketch a possible graph of \(y = f(x)\).
גלה את התשובה
1. From the given graph of \(f'\), the first derivative sign chart for \(f\) shows that \(f\) has critical numbers at \(x = -1, 1\) and that \(f'\) is positive for \(-1 \lt x \lt 1\) and for \(x \gt 1\), while being negative for all \(x \lt -1\). Hence \(f\) is increasing for \(-1 \lt x \lt 1\) and for \(x \gt 1\), and decreasing for \(x \lt -1\). It follows by the first derivative test that \(f\) has a local minimum at \(x = -1\) because \(f\) changes from decreasing to increasing there.
2. Based on where \(f'(x)\) has horizontal tangent lines and where its slope is positive and negative, we can sketch an approximate graph of \(y = f''(x)\) in the usual way that we sketch the derivative of a given function. From the given graph of \(y = f'(x)\), we can see that \(f''(x) = 0\) at \(x \approx -0.35\) and at \(x = 1\), and that between these points, \(f''(x)\) is negative because \(f'(x)\) is decreasing, while \(f''(x)\) is positive everywhere else since \(f'(x)\) is increasing everywhere else. A possible graph of \(y = f''(x)\) is shown at right in the following figure.
3. From our sketch of the second derivative in (b), we can develop a second derivative sign chart for \(f\). In particular, we saw that \(f''(x) = 0\) at \(x \approx -0.35\) and at \(x = 1\), and that between these points, \(f''(x)\) is negative, while \(f''(x)\) is positive everywhere else. This shows that \(f\) is concave down for \(-0.35 \lt x \lt 1\) while being concave up everywhere else. We thus see that \(f\) has points of inflection at both \(x \approx -0.35\) and \(x = 1\).
4. Now that we know where \(f\) is increasing and decreasing as well as where \(f\) is concave up and concave down, along with the fact that \(f(0) = -0.5\), we can sketch an approximate graph of \(y = f(x)\). Note that we also know \(f\) has a horizontal tangent line at \(x = 1\) while being increasing on either side of \(x = 1\), and earlier we established that \(f\) has a local minimum at \(x = -1\). After making a few rough sketches and shifting the graph vertically to pass through \((0,-0.5)\), we arrive at the graph shown at left below.
Suppose that \(g\) is a differentiable function and \(g'(2) = 0\). In addition, suppose that on \(1 \lt x\lt 2\) and \(2 \lt x \lt 3\) it is known that \(g'(x)\) is positive.
1. Does \(g\) have a local maximum, local minimum, or neither at \(x = 2\)? Why?
2. Suppose that \(g''(x)\) exists for every \(x\) such that \(1 \lt x \lt 3\). Reasoning graphically, describe the behavior of \(g''(x)\) for \(x\)-values near \(2\).
3. Besides being a critical number of \(g\), what is special about the value \(x = 2\) in terms of the behavior of the graph of \(g\)?
גלה את התשובה
1. \(g\) has neither a local maximum nor local minimum at \(x = 2\) because \(f'\) does not change sign at \(x = 2\).
2. Because \(g'(2) = 0\) and \(g'\) is positive for \(1 \lt x \lt 2\) and for \(2 \lt x \lt 3\), it must be that \(g'\) is decreasing for \(1 \lt x \lt 2\) and \(g'\) is increasing for \(2 \lt x \lt 3\). Since \(g'\) changes from decreasing to increasing at \(x = 2\), it follows that \(g'\) has a local minimum at \(x = 2\) and thus \(g''(2) = 0\). In addition, \(g''(x)\) must change sign from negative to positive at \(x = 2\). In particular, \(g''\) is negative for \(1 \lt x \lt 2\) and positive for \(2 \lt x \lt 3\).
3. \(g\) has a horizontal tangent line at \(x = 2\), is increasing before \(x = 2\) and decreasing after \(x = 2\), and concave down before \(x = 2\) and concave up after \(x = 2\), and thus \(g\) has a point of inflection at \(x = 2\).
Suppose that \(h\) is a differentiable function whose first derivative is given by the graph in Figure.
1. How many real number solutions can the equation \(h(x) = 0\) have? Why?
2. If \(h(x) = 0\) has two distinct real solutions, what can you say about the signs of the two solutions? Why?
3. Assume that \(\lim_{x \to \infty} h'(x) = 3\), as appears to be indicated in Figure. How will the graph of \(y = h(x)\) appear as \(x \to \infty\)? Why?
4. Describe the concavity of \(y = h(x)\) as fully as you can from the provided information.
גלה את התשובה
1. Since \(h'\) is negative to the left of \(0\) and positive to the right of \(0\), the function \(h\) is decreasing for \(x \lt 0\) and increasing for \(x \gt 0\). So \(h\) can have at most \(2\) real roots (solutions to \(h(x)=0\). As the graphs in the following figure show, it is possible that \(h\) has no real zeros, one real zero, or two real zeros.
2. Given that \(h\) is decreasing for \(x \lt 0\) and increasing for \(x \gt 0\), the function \(h\) has an absolute minimum value at \(x=0\). So one root is to the left of this minimum and the other to the right. Thus, one root is negative and the other positive. See, for instance, the lowest of the three possibilities for \(h\) shown in the figure in (a).
3. The slopes of the tangent lines to \(h\) will all be near \(3\), but less than \(3\), as \(x\) gets large. So the graph of \(h\) will look very much like a line with slope \(3\) for large values of \(x\).
4. We note that \(h'\) is increasing everywhere, \(h\) is concave up everywhere. Since the graph of \(h'\) levels off for large values of \(|x|\), the concavity of \(h\) is small when \(|x|\) is large. Also, the largest value of the slope of the tangent line to \(h'\) occurs when \(x=0\), so \(h\) has its largest second dervitive when \(x=0\). So \(h\) is almost linear as \(x\) decreases without bound, \(h\) has more bend as \(x\) approaches \(0\), has its greatest bend at \(x=0\), then the second derivative of \(h\) as \(x\) increases, ultimately having \(h\) become almost linear as \(x\) increases without bound.
Suppose that \(g(x)\) is a function continuous for every value of \(x \ne 2\) whose first derivative is \(g'(x) = \frac{(x+4)(x-1)^2}{x-2}\). Further, assume that it is known that \(g\) has a vertical asymptote at \(x = 2\).
1. Determine all critical numbers of \(g\).
2. By developing a carefully labeled first derivative sign chart, decide whether \(g\) has as a local maximum, local minimum, or neither at each critical number. Note: observe that \(g'(x)\) can change sign at the vertical asymptote of \(g(x)\).
3. Does \(g\) have a global maximum? global minimum? Justify your claims.
4. What is the value of \(\lim_{x \to \infty} g'(x)\)? What does the value of this limit tell you about the long-term behavior of \(g\)?
5. Sketch a possible graph of \(y = g(x)\).
גלה את התשובה
1. Since \(g'(x) = \frac{(x+4)(x-1)^2}{x-2}\), we see that \(g'(x) = 0\) implies that \(x = -4\) or \(x = 1\). While \(x = 2\) makes \(g'\) undefined, we are told that \(g\) has a vertical asymptote at \(x = 2\), so \(x = 2\) is not in the domain of \(g\), and hence is technically not a critical number of \(g\). Nonetheless, we place \(x = 2\) on our first derivative sign chart since the vertical asymptote is a location at which \(g'\) may change sign.
2. The first derivative sign chart shows that \(g'(x) \gt 0\) for \(x \lt -4\), \(g'(x) \lt 0\) for \(-4 \lt x \lt 1\), \(g'(x) \lt 0\) for \(1 \lt x \lt 2\), and \(g'(x) \gt 0\) for \(x \gt 2\). By the first derivative test, \(g\) has a local maximum at \(x = -4\) and neither a max nor min at \(x = 1\). As these are the only two critical numbers, these are the only two locations for possible extremes. (Note: although \(g\) changes from decreasing to increasing at \(x = 2\), this is due to a vertical asymptote, and \(g\) does not have a minimum there.)
3. Because \(g\) is decreasing as \(x \to 2^-\) (where \(g\) has a vertical asymptote), \(g\) does not have a global minimum. For \(x \gt 2\), \(g\) is always increasing, which suggests that \(g\) does not have a global maximum (though we do not know for sure that \(g\) increases without bound).
4. We observe that \[\begin{aligned}\lim_{x \to \infty} g'(x) =\mathstrut \amp \lim_{x \to \infty} \frac{(x+4)(x-1)^2}{x-2} \\ =\mathstrut \amp \lim_{x \to \infty} \frac{x^3 + 2x^2 - 7x + 4}{x-2} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} \\ =\mathstrut \amp \lim_{x \to \infty} \frac{x^2 + 2x - 7 + \frac{4}{x}}{1 - \frac{2}{x}} \\ =\mathstrut \amp \infty\end{aligned}\] Since \(g'(x) \to \infty\) as \(x \to \infty\), this tells us that \(g\) increases without bound as \(x \to \infty\).
5. From all of our work above, we know that \(g\) has a local maximum at \(x = -4\), a horizontal tangent line with neither a max nor min at \(x = 1\), and a vertical asymptote at \(x = 2\), plus \(g\) and \(g'\) both increase without bound as \(x \to \infty\). Thus, a possible graph of \(g\) is the following.
Let \(p\) be a function whose second derivative is \(p''(x) = (x+1)(x-2)e^{-x}\).
1. Construct a second derivative sign chart for \(p\) and determine all inflection points of \(p\).
2. Suppose you also know that \(x = \frac{\sqrt{5}-1}{2}\) is a critical number of \(p\). Does \(p\) have a local minimum, local maximum, or neither at \(x = \frac{\sqrt{5}-1}{2}\)? Why?
3. If the point \((2, \frac{12}{e^2})\) lies on the graph of \(y = p(x)\) and \(p'(2) = -\frac{5}{e^2}\), find the equation of the tangent line to \(y = p(x)\) at the point where \(x = 2\). Does the tangent line lie above the curve, below the curve, or neither at this value? Why?
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1. Note that \(p''(x)=0\) at \(x=-1\) and \(x=2\). The second derivative sign chart for \(p\) shows that \(p''(x)\) is negative for \(-1 \lt x \lt 2\) and positive for all other values of \(x\). Since \(p''\) changes sign at \(x = -1\) and \(x = 2\), \(p\) has points of inflection at \(x = -1\) and \(x = 2\). In particular, \(p\) is concave down for \(-1 \lt x \lt 2\) and concave up for all other values of \(x\).
2. If we compute \(p''\left(\frac{\sqrt{5}-1}{2} \right)\), we find that \(p''\left(\frac{\sqrt{5}-1}{2} \right) \approx -1.205\), which means that \(p\) is concave down at \(x = \frac{\sqrt{5}-1}{2}\). Since \(p\) has a horizontal tangent line at this point, it follows that \(p\) has a local maximum at \(x = \frac{\sqrt{5}-1}{2}\) by the second derivative test.
3. Since \(p(2) = \frac{12}{e^2}\) and \(p'(2) = -\frac{5}{e^2}\), the equation of the tangent line to \(p\) at \(x = 2\) is given by \(y = -\frac{5}{e^2}(x-2) + \frac{12}{e^2}\). To determine whether the tangent line lies above or below the graph of \(y=p(x)\) at \(x = 2\), we consider the concavity of \(p\). Recall we know that \(p''(2) = 0\) and that \(p\) is concave down to the left of \(x = 2\) and concave up to the right. Thus, the tangent line lies neither solely below nor solely above the graph of \(y = p(x)\) at \(x = 2\); it lies below \(y = p(x)\) to the right of \(x = 2\) and above \(y = p(x)\) to the left of \(x = 2\).

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Using derivatives to identify extreme values

What are the critical numbers of a function f and how are they connected to identifying the most extreme values the function achieves?
How does the first derivative of a function reveal important information about the behavior of the function, including the function's extreme values?
How can the second derivative of a function be used to help identify extreme values of the function?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

נסה את שלך.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

יותר בפנים. Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests