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Maxima and Minima

Define absolute extrema.

Absolute Extrema

Consider the function \(f(x)={x}^{2}+1\) over the interval \((\text{-}\infty ,\infty ).\) As \(x\to \text{\pm }\infty ,\) \(f(x)\to \infty .\) Therefore, the function does not have a largest value. However, since \({x}^{2}+1\ge 1\) for all real numbers \(x\) and \({x}^{2}+1=1\) when \(x=0,\) the function has a smallest value, 1, when \(x=0.\) We say that 1 is the absolute minimum of \(f(x)={x}^{2}+1\) and it occurs at \(x=0.\) We say that \(f(x)={x}^{2}+1\) does not have an absolute maximum (see the following figure).

Before proceeding, let’s note two important issues regarding this definition. First, the term absolute here does not refer to absolute value. An absolute extremum may be positive, negative, or zero. Second, if a function \(f\) has an absolute extremum over an interval \(I\) at \(c,\) the absolute extremum is \(f(c).\) The real number \(c\) is a point in the domain at which the absolute extremum occurs. For example, consider the function \(f(x)=1\text{/}({x}^{2}+1)\) over the interval \((\text{-}\infty ,\infty ).\) Since

\[f(0)=1\ge \frac{1}{{x}^{2}+1}=f(x)\]

for all real numbers \(x,\) we say \(f\) has an absolute maximum over \((\text{-}\infty ,\infty )\) at \(x=0.\) The absolute maximum is \(f(0)=1.\) It occurs at \(x=0,\) as shown in (b).

A function may have both an absolute maximum and an absolute minimum, just one extremum, or neither. shows several functions and some of the different possibilities regarding absolute extrema. However, the following theorem, called the Extreme Value Theorem, guarantees that a continuous function \(f\) over a closed, bounded interval \([a,b]\) has both an absolute maximum and an absolute minimum.

Before looking at how to find absolute extrema, let’s examine the related concept of local extrema. This idea is useful in determining where absolute extrema occur.

Condensed — the full section is in OpenStax Calculus Volume 1.

Local Extrema and Critical Points

Consider the function \(f\) shown in . The graph can be described as two mountains with a valley in the middle. The absolute maximum value of the function occurs at the higher peak, at \(x=2.\) However, \(x=0\) is also a point of interest. Although \(f(0)\) is not the largest value of \(f,\) the value \(f(0)\) is larger than \(f(x)\) for all \(x\) near 0. We say \(f\) has a local maximum at \(x=0.\) Similarly, the function \(f\) does not have an absolute minimum, but it does have a local minimum at \(x=1\) because \(f(1)\) is less than \(f(x)\) for \(x\) near 1.

Note that if \(f\) has an absolute extremum at \(c\) and \(f\) is defined over an interval containing \(c,\) then \(f(c)\) is also considered a local extremum. If an absolute extremum for a function \(f\) occurs at an endpoint, we do not consider that to be a local extremum, but instead refer to that as an endpoint extremum.

Given the graph of a function \(f,\) it is sometimes easy to see where a local maximum or local minimum occurs. However, it is not always easy to see, since the interesting features on the graph of a function may not be visible because they occur at a very small scale. Also, we may not have a graph of the function. In these cases, how can we use a formula for a function to determine where these extrema occur?

To answer this question, let’s look at again. The local extrema occur at \(x=0,\) \(x=1,\) and \(x=2.\) Notice that at \(x=0\) and \(x=1,\) the derivative \(f'(x)=0.\) At \(x=2,\) the derivative \(f'(x)\) does not exist, since the function \(f\) has a corner there. In fact, if \(f\) has a local extremum at a point \(x=c,\) the derivative \(f'(c)\) must satisfy one of the following conditions: either \(f'(c)=0\) or \(f'(c)\) is undefined. Such a value \(c\) is known as a critical number and it is important in finding extreme values for functions.

As mentioned earlier, if \(f\) has a local extremum at a point \(x=c,\) then \(c\) must be a critical number of \(f.\) This fact is known as Fermat’s theorem.

Condensed — the full section is in OpenStax Calculus Volume 1.

Locating Absolute Extrema

The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. As shown in , one or both of these absolute extrema could occur at an endpoint. If an absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute extremum is a local extremum. Therefore, by , the point \(c\) at which the local extremum occurs must be a critical point. We summarize this result in the following theorem.

With this idea in mind, let’s examine a procedure for locating absolute extrema.

Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions.

At this point, we know how to locate absolute extrema for continuous functions over closed intervals. We have also defined local extrema and determined that if a function \(f\) has a local extremum at a point \(c,\) then \(c\) must be a critical number of \(f.\) However, \(c\) being a critical point is not a sufficient condition for \(f\) to have a local extremum at \(c.\) Later in this chapter, we show how to determine whether a function actually has a local extremum at a critical point. First, however, we need to introduce the Mean Value Theorem, which will help as we analyze the behavior of the graph of a function.

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

A function may have both an absolute maximum and an absolute minimum, have just one absolute extremum, or have no absolute maximum or absolute minimum.
If a function has a local extremum, the point at which it occurs must be a critical point. However, a function need not have a local extremum at a critical point.
A continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. Each extremum occurs at a critical point or an endpoint.

Maxima and Minima

For the following exercises, determine where the local and absolute maxima and minima occur on the graph given. Assume the graph represents the entirety of each function. For any extrema located at an endpoint, approximate the x-value.

For the following problems, draw graphs of \(f(x),\) which is continuous, over the interval \([-4,4]\) with the following properties:

For the following exercises, find the critical numbers in the domains of the following functions.

For the following exercises, find the local and/or absolute extrema for the functions over the specified domain.

For the following exercises, find the local and absolute minima and maxima for the functions over \((\text{-}\infty ,\infty ).\)

For the following functions, use a calculator to graph the function and to estimate the absolute and local maxima and minima. Then, solve for them explicitly.

For the following exercises, consider the production of gold during the California gold rush (1848–1888). The production of gold can be modeled by \(G(t)=\frac{(25t)}{({t}^{2}+16)},\) where \(t\) is the number of years since the rush began \((0\le t\le 40)\) and \(G\) is ounces of gold produced (in millions). A summary of the data is shown in the following figure.

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

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

For each of the following functions, find all critical points. Use a graphing utility to determine whether the function has a local extremum at each of the critical points.
1. \(f(x)=\frac{1}{3}{x}^{3}-\frac{5}{2}{x}^{2}+4x\)
2. \(f(x)={({x}^{2}-1)}^{3}\)
3. \(f(x)=\frac{4x}{1+{x}^{2}}\)
Die Antwort aufzeigen
1. The derivative \(f'(x)={x}^{2}-5x+4\) is defined for all real numbers \(x.\) Therefore, we only need to find the values for \(x\) where \(f'(x)=0.\) Since \(f'(x)={x}^{2}-5x+4=(x-4)(x-1),\) the critical numbers are \(x=1\) and \(x=4.\) From the graph of \(f\) in , we see that \(f\) has a local maximum at \(x=1\) and a local minimum at \(x=4.\)
2. Using the chain rule, we see the derivative is
  \[f'(x)=3{({x}^{2}-1)}^{2}(2x)=6x{({x}^{2}-1)}^{2}.\]
  Therefore, \(f\) has critical points when \(x=0\) and when \({x}^{2}-1=0.\) We conclude that the critical numbers are \(x=0,\text{\pm }1.\) From the graph of \(f\) in , we see that \(f\) has a local (and absolute) minimum at \(x=0,\) but does not have a local extremum at \(x=1\) or \(x=-1.\)
3. By the quotient rule, we see that the derivative is
  \[f'(x)=\frac{(1+{x}^{2})(4)-4x(2x)}{{(1+{x}^{2})}^{2}}=\frac{4-4{x}^{2}}{{(1+{x}^{2})}^{2}}.\]
  The derivative is defined everywhere. Therefore, we only need to find values for \(x\) where \(f'(x)=0.\) Solving \(f'(x)=0,\) we see that \(4-4{x}^{2}=0,\) which implies \(x=\text{\pm }1.\) Therefore, the critical numbers are \(x=\text{\pm }1.\) From the graph of \(f\) in , we see that \(f\) has an absolute maximum at \(x=1\) and an absolute minimum at \(x=-1.\) Hence, \(f\) has a local maximum at \(x=1\) and a local minimum at \(x=-1.\) (Note that if \(f\) has an absolute extremum over an interval \(I\) at a point \(c\) that is not an endpoint of \(I,\) then \(f\) has a local extremum at \(c.)\)
Find all critical points for \(f(x)={x}^{3}-\frac{1}{2}{x}^{2}-2x+1.\)

Die Antwort aufzeigen

\(x=-\frac{2}{3},\) \(x=1\)
For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.
1. \(f(x)=\text{-}{x}^{2}+3x-2\) over \([1,3].\)
2. \(f(x)={x}^{2}-3{x}^{2\text{/}3}\) over \([0,2].\)
Die Antwort aufzeigen
1. Step 1. Evaluate \(f\) at the endpoints \(x=1\) and \(x=3.\)
  \[f(1)=0\ \text{and}\ f(3)=-2\]
  Step 2. Since \(f'(x)=-2x+3,\) \(f'\) is defined for all real numbers \(x.\) Therefore, there are no critical points where the derivative is undefined. It remains to check where \(f'(x)=0.\) Since \(f'(x)=-2x+3=0\) at \(x=\frac{3}{2}\) and \(\frac{3}{2}\) is in the interval \([1,3],\) \(f(\frac{3}{2})\) is a candidate for an absolute extremum of \(f\) over \([1,3].\) We evaluate \(f(\frac{3}{2})\) and find
  \[f(\frac{3}{2})=\frac{1}{4}.\]
  Step 3. We set up the following table to compare the values found in steps 1 and 2.
  
  \(x\) \(f(x)\) Conclusion
  \(1\) \(0\)
  \(\frac{3}{2}\) \(\frac{1}{4}\) Absolute maximum
  \(3\) \(-2\) Absolute minimum
  
  From the table, we find that the absolute maximum of \(f\) over the interval [1, 3] is \(\frac{1}{4},\) and it occurs at \(x=\frac{3}{2}.\) The absolute minimum of \(f\) over the interval [1, 3] is \(-2,\) and it occurs at \(x=3\) as shown in the following graph.
2. Step 1. Evaluate \(f\) at the endpoints \(x=0\) and \(x=2.\)
  \[f(0)=0\ \text{and}\ f(2)=4-3\sqrt[3]{4}\approx -0.762\]
  Step 2. The derivative of \(f\) is given by
  \[f'(x)=2x-\frac{2}{{x}^{1\text{/}3}}=\frac{2{x}^{4\text{/}3}-2}{{x}^{1\text{/}3}}\]
  for \(x\ne 0.\) The derivative is zero when \(2{x}^{4\text{/}3}-2=0,\) which implies \(x=\text{\pm }1.\) The derivative is undefined at \(x=0.\) Therefore, the critical numbers of \(f\) are \(x=0,1,-1.\) The point \(x=0\) is an endpoint, so we already evaluated \(f(0)\) in step 1. The point \(x=-1\) is not in the interval of interest, so we need only evaluate \(f(1).\) We find that
  \[f(1)=-2.\]
  Step 3. We compare the values found in steps 1 and 2, in the following table.
  
  \(x\) \(f(x)\) Conclusion
  \(0\) \(0\) Absolute maximum
  \(1\) \(-2\) Absolute minimum
  \(2\) \(-0.762\)
  
  We conclude that the absolute maximum of \(f\) over the interval [0, 2] is zero, and it occurs at \(x=0.\) The absolute minimum is −2, and it occurs at \(x=1\) as shown in the following graph.
Find the absolute maximum and absolute minimum of \(f(x)={x}^{2}-4x+3\) over the interval \([1,4].\)

Die Antwort aufzeigen

The absolute maximum is \(3\) and it occurs at \(x=4.\) The absolute minimum is \(-1\) and it occurs at \(x=2.\)
In precalculus, you learned a formula for the position of the maximum or minimum of a quadratic equation \(y=a{x}^{2}+bx+c,\) which was \(h=-\frac{b}{(2a)}.\) Prove this formula using calculus.
If you are finding an absolute minimum over an interval \([a,b],\) why do you need to check the endpoints? Draw a graph that supports your hypothesis.

Die Antwort aufzeigen

Answers may vary
If you are examining a function over an interval \((a,b),\) for \(a\) and \(b\) finite, is it possible not to have an absolute maximum or absolute minimum?
When you are checking for critical points, explain why you also need to determine points where \(f'(x)\) is undefined. Draw a graph to support your explanation.

Die Antwort aufzeigen

Answers will vary
Can you have a finite absolute maximum for \(y=a{x}^{2}+bx+c\) over \((\text{-}\infty ,\infty )?\) Explain why or why not using graphical arguments.
Can you have a finite absolute maximum for \(y=a{x}^{3}+b{x}^{2}+cx+d\) over \((\text{-}\infty ,\infty )\) assuming a is non-zero? Explain why or why not using graphical arguments.

Die Antwort aufzeigen

No; answers will vary
Let \(m\) be the number of local minima and \(M\) be the number of local maxima. Can you create a function where \(M>m+2?\) Draw a graph to support your explanation.
Is it possible to have more than one absolute maximum? Use a graphical argument to prove your hypothesis.

Die Antwort aufzeigen

Since the absolute maximum is the function (output) value rather than the x value, the answer is no; answers will vary
Is it possible to have no absolute minimum or maximum for a function? If so, construct such a function. If not, explain why this is not possible.
[T] Graph the function \(y={e}^{ax}.\) For which values of \(a,\) on any infinite domain, will you have an absolute minimum and absolute maximum?

Die Antwort aufzeigen

When \(a=0\)
Absolute maximum at \(x=2\) and absolute minima at \(x=\text{\pm }3\)
Absolute minimum at \(x=1\) and absolute maximum at \(x=2\)

Die Antwort aufzeigen

Answers may vary.
Absolute maximum at \(x=4,\) absolute minimum at \(x=-1,\) local maximum at \(x=-2,\) and a critical point that is not a maximum or minimum at \(x=2\)
Absolute maxima at \(x=2\) and \(x=-3,\) local minimum at \(x=1,\) and absolute minimum at \(x=4\)

Die Antwort aufzeigen

Answers may vary.
\(y=4{x}^{3}-3x\)
\(y=4\sqrt{x}-{x}^{2}\)

Die Antwort aufzeigen

\(x=1\)
\(y=\frac{1}{x-1}\)
\(y=\text{ln}(x-2)\)

Die Antwort aufzeigen

None
\(y=\text{tan}(x)\)
\(y=\sqrt{4-{x}^{2}}\)

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\(x=0\text{;}x=\pm 2\)
\(y={x}^{3\text{/}2}-3{x}^{5\text{/}2}\)
\(y=\frac{{x}^{2}-1}{{x}^{2}+2x-3}\)

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None
\(y={\text{sin}}^{2}(x)\)
\(y=x+\frac{1}{x}\)

Die Antwort aufzeigen

\(x=-1,1\)
\(f(x)={x}^{2}+3\) over \([-1,4]\)
\(y={x}^{2}+\frac{2}{x}\) over \([1,4]\)

Die Antwort aufzeigen

Absolute maximum: \(x=4,\) \(y=\frac{33}{2};\) absolute minimum: \(x=1,\) \(y=3\)
\(y={(x-{x}^{2})}^{2}\) over \([-1,1]\)
\(y=\frac{1}{(x-{x}^{2})}\) over \((0,1)\)

Die Antwort aufzeigen

Absolute minimum: \(x=\frac{1}{2},\) \(y=4\)
\(y=\sqrt{9-x}\) over \([1,9]\)
\(y=x+\text{sin}(x)\) over \([0,2\pi ]\)

Die Antwort aufzeigen

Absolute maximum: \(x=2\pi ,\) \(y=2\pi ;\) absolute minimum: \(x=0,\) \(y=0\)
\(y=\frac{x}{1+x}\) over \([0,100]\)
\(y=|x+1|+|x-1|\) over \([-3,2]\)

Die Antwort aufzeigen

Absolute maximum: \(x=-3\) and \(y=6;;\) absolute minimum: \(-1\le x\le 1,\) \(y=2\)
\(y=\sqrt{x}-\sqrt{{x}^{3}}\) over \([0,4]\)
\(y=\text{sin}\ x+\text{cos}\ x\) over \([0,2\pi ]\)

Die Antwort aufzeigen

Absolute maximum: \(x=\frac{\pi }{4},\) \(y=\sqrt{2};\) absolute minimum: \(x=\frac{5\pi }{4},\) \(y=\text{-}\sqrt{2}\)
\(y=4\ \text{sin}\ \theta -3\ \text{cos}\ \theta\) over \([0,2\pi ]\)
\(y={x}^{2}+4x+5\)

Die Antwort aufzeigen

Absolute minimum: \(x=-2,\) \(y=1\)

Symbols used here

\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.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

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

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

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

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

\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.

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

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

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: Maxima and Minima

Define absolute extrema.
Define local extrema.
Explain how to find the critical points of a function over a closed interval.
Describe how to use critical points to locate absolute extrema over a closed interval.
The derivative
Using the chain rule, we see the derivative is
By the quotient rule, we see that the derivative is
Evaluate

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.

Versuch es selbst.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mehr in 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

\(x\)	\(f(x)\)	Conclusion
\(1\)	\(0\)
\(\frac{3}{2}\)	\(\frac{1}{4}\)	Absolute maximum
\(3\)	\(-2\)	Absolute minimum