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Review of Functions

Use functional notation to evaluate a function.

Functions

Given two sets \(A\) and \(B,\) a set with elements that are ordered pairs \((x,y),\) where \(x\) is an element of \(A\) and \(y\) is an element of \(B,\) is a relation from \(A\) to \(B.\) A relation from \(A\) to \(B\) defines a relationship between those two sets. A function is a special type of relation in which each element of the first set is related to exactly one element of the second set. The element of the first set is called the input; the element of the second set is called the output. Functions are used all the time in mathematics to describe relationships between two sets. For any function, when we know the input, the output is determined, so we say that the output is a function of the input. For example, the area of a square is determined by its side length, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in the air can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function of the weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology to study them.

For example, consider the function \(f,\) where the domain is the set of all real numbers and the rule is to square the input. Then, the input \(x=3\) is assigned to the output \({3}^{2}=9.\) Since every nonnegative real number has a real-value square root, every nonnegative number is an element of the range of this function. Since there is no real number with a square that is negative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative real numbers.

For a general function \(f\) with domain \(D,\) we often use \(x\) to denote the input and \(y\) to denote the output associated with \(x.\) When doing so, we refer to \(x\) as the independent variable and \(y\) as the dependent variable, because it depends on \(x.\) Using function notation, we write \(y=f(x),\) and we read this equation as \(\text{“}y\) equals \(f\) of \(x.\text{”}\) For the squaring function described earlier, we write \(f(x)={x}^{2}.\)

The concept of a function can be visualized using , , and .

\[\{x|x\ \text{has some property}\}.\]\[\{x|1Condensed — the full section is in OpenStax Calculus Volume 1.

Representing Functions

Typically, a function is represented using one or more of the following tools:

  • A table
  • A graph
  • A formula

We can identify a function in each form, but we can also use them together. For instance, we can plot on a graph the values from a table or create a table from a formula.

Functions described using a table of values arise frequently in real-world applications. Consider the following simple example. We can describe temperature on a given day as a function of time of day. Suppose we record the temperature every hour for a 24-hour period starting at midnight. We let our input variable \(x\) be the time after midnight, measured in hours, and the output variable \(y\) be the temperature \(x\) hours after midnight, measured in degrees Fahrenheit. We record our data in .

Hours after MidnightTemperature \((\text{^{\circ}}F)\)Hours after MidnightTemperature \((\text{^{\circ}}F)\)
0581284
1541385
2531485
3521583
4521682
5551780
6601877
7641974
8722069
9752165
10782260
11802358

We can see from the table that temperature is a function of time, and the temperature decreases, then increases, and then decreases again. However, we cannot get a clear picture of the behavior of the function without graphing it.

Given a function \(f\) described by a table, we can provide a visual picture of the function in the form of a graph. Graphing the temperatures listed in can give us a better idea of their fluctuation throughout the day. shows the plot of the temperature function.

From the points plotted on the graph in , we can visualize the general shape of the graph. It is often useful to connect the dots in the graph, which represent the data from the table. In this example, although we cannot make any definitive conclusion regarding what the temperature was at any time for which the temperature was not recorded, given the number of data points collected and the pattern in these points, it is reasonable to suspect that the temperatures at other times followed a similar pattern, as we can see in .

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

Combining Functions

Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when we combine functions in different ways, using basic mathematical operations to create new functions. For example, if the cost for a company to manufacture \(x\) items is described by the function \(C(x)\) and the revenue created by the sale of \(x\) items is described by the function \(R(x),\) then the profit on the manufacture and sale of \(x\) items is defined as \(P(x)=R(x)-C(x).\) Using the difference between two functions, we created a new function.

Alternatively, we can create a new function by composing two functions. For example, given the functions \(f(x)={x}^{2}\) and \(g(x)=3x+1,\) the composite function \(f∘g\) is defined such that

\[(f∘g)(x)=f(g(x))={(g(x))}^{2}={(3x+1)}^{2}.\]

The composite function \(g∘f\) is defined such that

\[(g∘f)(x)=g(f(x))=3f(x)+1=3{x}^{2}+1.\]

Note that these two new functions are different from each other.

To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Given two functions \(f\) and \(g,\) we can define four new functions:

\[\begin{array}{llll}(f+g)(x)=f(x)+g(x) & & & Sum \\ (f-g)(x)=f(x)-g(x) & & & \text{Difference} \\ (f\cdot g)(x)=f(x)g(x) & & & Product \\ (\frac{f}{g})(x)=\frac{f(x)}{g(x)}\ \text{for}\ g(x)\ne 0 & & & Quotient\end{array}\]
Example

Try it.

Given the functions \(f(x)=2x-3\) and \(g(x)={x}^{2}-1,\) find each of the following functions and state its domain.

  1. \((f+g)(x)\)
  2. \((f-g)(x)\)
  3. \((f\cdot g)(x)\)
  4. \((\frac{f}{g})(x)\)
Solution
  1. \((f+g)(x)=(2x-3)+({x}^{2}-1)={x}^{2}+2x-4.\) The domain of this function is the interval \((\text{-}\infty ,\infty ).\)
  2. \((f-g)(x)=(2x-3)-({x}^{2}-1)=\text{-}{x}^{2}+2x-2.\) The domain of this function is the interval \((\text{-}\infty ,\infty ).\)
  3. \((f\cdot g)(x)=(2x-3)({x}^{2}-1)=2{x}^{3}-3{x}^{2}-2x+3.\) The domain of this function is the interval \((\text{-}\infty ,\infty ).\)
  4. \((\frac{f}{g})(x)=\frac{2x-3}{{x}^{2}-1}.\) The domain of this function is \(\{x|x\ne \text{\pm }1\}.\)

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

Symmetry of Functions

The graphs of certain functions have symmetry properties that help us understand the function and the shape of its graph. For example, consider the function \(f(x)={x}^{4}-2{x}^{2}-3\) shown in (a). If we take the part of the curve that lies to the right of the y-axis and flip it over the y-axis, it lays exactly on top of the curve to the left of the y-axis. In this case, we say the function has symmetry about the y-axis. On the other hand, consider the function \(f(x)={x}^{3}-4x\) shown in (b). If we take the graph and rotate it \(180\text{^{\circ}}\) about the origin, the new graph will look exactly the same. In this case, we say the function has symmetry about the origin.

If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But without a graph, how can we determine algebraically whether a function \(f\) has symmetry? Looking at again, we see that since \(f\) is symmetric about the \(y\)-axis, if the point \((x,y)\) is on the graph, the point \((\text{-}x,y)\) is on the graph. In other words, \(f(\text{-}x)=f(x).\) If a function \(f\) has this property, we say \(f\) is an even function, which has symmetry about the y-axis. For example, \(f(x)={x}^{2}\) is even because

\[f(\text{-}x)={(\text{-}x)}^{2}={x}^{2}=f(x).\]

In contrast, looking at again, if a function \(f\) is symmetric about the origin, then whenever the point \((x,y)\) is on the graph, the point \((\text{-}x,\text{-}y)\) is also on the graph. In other words, \(f(\text{-}x)=\text{-}f(x).\) If \(f\) has this property, we say \(f\) is an odd function, which has symmetry about the origin. For example, \(f(x)={x}^{3}\) is odd because

\[f(\text{-}x)={(\text{-}x)}^{3}=\text{-}{x}^{3}=\text{-}f(x).\]

One symmetric function that arises frequently is the absolute value function, written as \(|x|.\) The absolute value function is defined as

\[f(x)=\{\begin{array}{l}\text{-}x,x<0 \\ x,x\ge 0\end{array}.\]

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

Key Concepts

  • A function is a mapping from a set of inputs to a set of outputs with exactly one output for each input.
  • If no domain is stated for a function \(y=f(x),\) the domain is considered to be the set of all real numbers \(x\) for which the function is defined.
  • When sketching the graph of a function \(f,\) each vertical line may intersect the graph, at most, once.
  • A function may have any number of zeros, but it has, at most, one y-intercept.
  • To define the composition \(g∘f,\) the range of \(f\) must be contained in the domain of \(g.\)
  • Even functions are symmetric about the \(y\)-axis whereas odd functions are symmetric about the origin.

Practice (40)

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

  1. For the function \(f(x)=3{x}^{2}+2x-1,\) evaluate

    1. \(f(-2)\)
    2. \(f(\sqrt{2})\)
    3. \(f(a+h)\)
    જવાબ બતાવો

    Substitute the given value for x in the formula for \(f(x).\)

    1. \(f(-2)=3{(-2)}^{2}+2(-2)-1=12-4-1=7\)
    2. \(f(\sqrt{2})=3{(\sqrt{2})}^{2}+2\sqrt{2}-1=6+2\sqrt{2}-1=5+2\sqrt{2}\)
    3. \(\begin{array}{ll}f(a+h)=3{(a+h)}^{2}+2(a+h)-1 & =3({a}^{2}+2ah+{h}^{2})+2a+2h-1 \\ & =3{a}^{2}+6ah+3{h}^{2}+2a+2h-1\end{array}\)

  2. For \(f(x)={x}^{2}-3x+5,\) evaluate \(f(1)\) and \(f(a+h).\)

    જવાબ બતાવો

    \(f(1)=3\) and \(f(a+h)={a}^{2}+2ah+{h}^{2}-3a-3h+5\)

  3. For each of the following functions, determine the i. domain and ii. range.

    1. \(f(x)={(x-4)}^{2}+5\)
    2. \(f(x)=\sqrt{3x+2}-1\)
    3. \(f(x)=\frac{3}{x-2}\)
    જવાબ બતાવો
    1. Consider \(f(x)={(x-4)}^{2}+5.\)
      1. Since \(f(x)={(x-4)}^{2}+5\) is a real number for any real number \(x,\) the domain of \(f\) is the interval \((\text{-}\infty ,\infty ).\)
      2. Since \({(x-4)}^{2}\ge 0,\) we know \(f(x)={(x-4)}^{2}+5\ge 5.\) Therefore, the range must be a subset of \(\{y|y\ge 5\}.\) To show that every element in this set is in the range, we need to show that for a given \(y\) in that set, there is a real number \(x\) such that \(f(x)={(x-4)}^{2}+5=y.\) Solving this equation for \(x,\) we see that we need \(x\) such that \[{(x-4)}^{2}=y-5.\]
        This equation is satisfied as long as there exists a real number \(x\) such that \[x-4=\pm \sqrt{y-5}.\]
        Since \(y\ge 5,\) the square root is well-defined. We conclude that for \(x=4\pm \sqrt{y-5},f(x)=y,\) and therefore the range is \(\{y|y\ge 5\}.\)
    2. Consider \(f(x)=\sqrt{3x+2}-1.\)
      1. To find the domain of \(f,\) we need the expression \(3x+2\ge 0.\) Solving this inequality, we conclude that the domain is \(\{x|x\ge -2\text{/}3\}.\)
      2. To find the range of \(f,\) we note that since \(\sqrt{3x+2}\ge 0,f(x)=\sqrt{3x+2}-1\ge -1.\) Therefore, the range of \(f\) must be a subset of the set \(\{y|y\ge -1\}.\) To show that every element in this set is in the range of \(f,\) we need to show that for all \(y\) in this set, there exists a real number \(x\) in the domain such that \(f(x)=y.\) Let \(y\ge -1.\) Then, \(f(x)=y\) if and only if \[\sqrt{3x+2}-1=y.\]
        Solving this equation for \(x,\) we see that \(x\) must solve the equation \[\sqrt{3x+2}=y+1.\]
        Since \(y\ge -1,\) such an \(x\) could exist. Squaring both sides of this equation, we have \(3x+2={(y+1)}^{2}.\)
        Therefore, we need \[3x={(y+1)}^{2}-2,\]
        which implies \[x=\frac{1}{3}{(y+1)}^{2}-\frac{2}{3}.\]
        We just need to verify that \(x\) is in the domain of \(f.\) Since the domain of \(f\) consists of all real numbers greater than or equal to \(-2\text{/}3,\) and \[\frac{1}{3}{(y+1)}^{2}-\frac{2}{3}\ge -\frac{2}{3},\]
        there does exist an \(x\) in the domain of \(f.\) We conclude that the range of \(f\) is \(\{y|y\ge -1\}.\)
    3. Consider \(f(x)=3\text{/}(x-2).\)
      1. Since \(3\text{/}(x-2)\) is defined when the denominator is nonzero, the domain is \(\{x|x\ne 2\}.\)
      2. To find the range of \(f,\) we need to find the values of \(y\) such that there exists a real number \(x\) in the domain with the property that \[\frac{3}{x-2}=y.\]
        Solving this equation for \(x,\) we find that \[x=\frac{3}{y}+2.\]
        Therefore, as long as \(y\ne 0,\) there exists a real number \(x\) in the domain such that \(f(x)=y.\) Thus, the range is \(\{y|y\ne 0\}.\)

  4. Find the domain and range for \(f(x)=\sqrt{4-2x}+5.\)

    જવાબ બતાવો

    Domain = \(\{x|x\le 2\},\) range = \(\{y|y\ge 5\}\)

  5. Consider the function \(f(x)=-4x+2.\)

    1. Find all zeros of \(f.\)
    2. Find the \(y\)-intercept (if any).
    3. Sketch a graph of \(f.\)
    જવાબ બતાવો
    1. To find the zeros, solve \(f(x)=-4x+2=0.\) We discover that \(f\) has one zero at \(x=1\text{/}2.\)
    2. The \(y\)-intercept is given by \((0,f(0))=(0,2).\)
    3. Given that \(f\) is a linear function of the form \(f(x)=mx+b\) that passes through the points \((1\text{/}2,0)\) and \((0,2),\) we can sketch the graph of \(f\) ().

  6. Consider the function \(f(x)=\sqrt{x+3}+1.\)

    1. Find all zeros of \(f.\)
    2. Find the \(y\)-intercept (if any).
    3. Sketch a graph of \(f.\)
    જવાબ બતાવો
    1. To find the zeros, solve \(\sqrt{x+3}+1=0.\) This equation implies \(\sqrt{x+3}=-1.\) Since \(\sqrt{x+3}\ge 0\) for all \(x,\) this equation has no solutions, and therefore \(f\) has no zeros.
    2. The \(y\)-intercept is given by \((0,f(0))=(0,\sqrt{3}+1).\)
    3. To graph this function, we make a table of values. Since we need \(x+3\ge 0,\) we need to choose values of \(x\ge -3.\) We choose values that make the square-root function easy to evaluate.
      \(\text{x}\)\(-3\)\(-2\)\(1\)
      \(\text{f}(\text{x})\)\(1\)\(2\)\(3\)

    Making use of the table and knowing that, since the function is a square root, the graph of \(f\) should be similar to the graph of \(y=\sqrt{x},\) we sketch the graph ().

  7. Find the zeros of \(f(x)={x}^{3}-5{x}^{2}+6x.\)

    જવાબ બતાવો

    \(x=0,2,3\)

  8. If a ball is dropped from a height of \(100\) ft, its height \(s\) at time \(t\) is given by the function \(s(t)=-16{t}^{2}+100,\) where \(s\) is measured in feet and \(t\) is measured in seconds. The domain is restricted to the interval \([0,c],\) where \(t=0\) is the time when the ball is dropped and \(t=c\) is the time when the ball hits the ground.

    1. Create a table showing the height \(s(t)\) when \(t=0,0.5,1,1.5,2,\text{and}\ 2.5.\) Using the data from the table, determine the domain for this function. That is, find the time \(c\) when the ball hits the ground.
    2. Sketch a graph of \(s.\)
    જવાબ બતાવો

    1. \(\text{t}\)\(0\)\(0.5\)\(1\)\(1.5\)\(2\)\(2.5\)
      \(\text{s}(\text{t})\)\(100\)\(96\)\(84\)\(64\)\(36\)\(0\)

      Since the ball hits the ground when \(t=2.5,\) the domain of this function is the interval \([0,2.5].\)


  9. Given the functions \(f(x)=2x-3\) and \(g(x)={x}^{2}-1,\) find each of the following functions and state its domain.

    1. \((f+g)(x)\)
    2. \((f-g)(x)\)
    3. \((f\cdot g)(x)\)
    4. \((\frac{f}{g})(x)\)
    જવાબ બતાવો
    1. \((f+g)(x)=(2x-3)+({x}^{2}-1)={x}^{2}+2x-4.\) The domain of this function is the interval \((\text{-}\infty ,\infty ).\)
    2. \((f-g)(x)=(2x-3)-({x}^{2}-1)=\text{-}{x}^{2}+2x-2.\) The domain of this function is the interval \((\text{-}\infty ,\infty ).\)
    3. \((f\cdot g)(x)=(2x-3)({x}^{2}-1)=2{x}^{3}-3{x}^{2}-2x+3.\) The domain of this function is the interval \((\text{-}\infty ,\infty ).\)
    4. \((\frac{f}{g})(x)=\frac{2x-3}{{x}^{2}-1}.\) The domain of this function is \(\{x|x\ne \text{\pm }1\}.\)

  10. For \(f(x)={x}^{2}+3\) and \(g(x)=2x-5,\) find \((f\text{/}g)(x)\) and state its domain.

    જવાબ બતાવો

    \((\frac{f}{g})(x)=\frac{{x}^{2}+3}{2x-5}.\) The domain is \(\{x|x\ne \frac{5}{2}\}.\)

  11. Consider the functions \(f(x)={x}^{2}+1\) and \(g(x)=1\text{/}x.\)

    1. Find \((g∘f)(x)\) and state its domain and range.
    2. Evaluate \((g∘f)(4),(g∘f)(-1\text{/}2).\)
    3. Find \((f∘g)(x)\) and state its domain and range.
    4. Evaluate \((f∘g)(4),(f∘g)(-1\text{/}2).\)
    જવાબ બતાવો
    1. We can find the formula for \((g∘f)(x)\) in two different ways. We could write \[(g∘f)(x)=g(f(x))=g({x}^{2}+1)=\frac{1}{{x}^{2}+1}.\]
      Alternatively, we could write \[(g∘f)(x)=g(f(x))=\frac{1}{f(x)}=\frac{1}{{x}^{2}+1}.\]
      Since \({x}^{2}+1\ne 0\) for all real numbers \(x,\) the domain of \((g∘f)(x)\) is the set of all real numbers. Since \(0<1\text{/}({x}^{2}+1)\le 1,\) the range is, at most, the interval \((0,1].\) To show that the range is this entire interval, we let \(y=1\text{/}({x}^{2}+1)\) and solve this equation for \(x\) to show that for all \(y\) in the interval \((0,1],\) there exists a real number \(x\) such that \(y=1\text{/}({x}^{2}+1).\) Solving this equation for \(x,\) we see that \({x}^{2}+1=1\text{/}y,\) which implies that \[x=\pm \sqrt{\frac{1}{y}-1}.\]
      If \(y\) is in the interval \((0,1],\) the expression under the radical is nonnegative, and therefore there exists a real number \(x\) such that \(1\text{/}({x}^{2}+1)=y.\) We conclude that the range of \(g∘f\) is the interval \((0,1].\)
    2. \((g∘f)(4)=g(f(4))=g({4}^{2}+1)=g(17)=\frac{1}{17}\)
      \((g∘f)(-\frac{1}{2})=g(f(-\frac{1}{2}))=g({(-\frac{1}{2})}^{2}+1)=g(\frac{5}{4})=\frac{4}{5}\)
    3. We can find a formula for \((f∘g)(x)\) in two ways. First, we could write \[(f∘g)(x)=f(g(x))=f(\frac{1}{x})={(\frac{1}{x})}^{2}+1.\]
      Alternatively, we could write \[(f∘g)(x)=f(g(x))={(g(x))}^{2}+1={(\frac{1}{x})}^{2}+1.\]
      The domain of \(f∘g\) is the set of all real numbers \(x\) such that \(x\ne 0.\) To find the range of \(f,\) we need to find all values \(y\) for which there exists a real number \(x\ne 0\) such that \[{(\frac{1}{x})}^{2}+1=y.\]
      Solving this equation for \(x,\) we see that we need \(x\) to satisfy \[{(\frac{1}{x})}^{2}=y-1,\]
      which simplifies to \[\frac{1}{x}=\pm \sqrt{y-1}.\]
      Finally, we obtain \[x=\pm \frac{1}{\sqrt{y-1}}.\]
      Since \(1\text{/}\sqrt{y-1}\) is a real number if and only if \(y>1,\) the range of \(f◦g\) is the set \(\{y|y>1\}.\)
    4. \((f∘g)(4)=f(g(4))=f(\frac{1}{4})={(\frac{1}{4})}^{2}+1=\frac{17}{16}\)
      \((f∘g)(-\frac{1}{2})=f(g(-\frac{1}{2}))=f(-2)={(-2)}^{2}+1=5\)

  12. Let \(f(x)=2-5x.\) Let \(g(x)=\sqrt{x}.\) Find \((f∘g)(x).\)

    જવાબ બતાવો

    \((f∘g)(x)=2-5\sqrt{x}.\)

  13. Consider the functions \(f\) and \(g\) described by and .

    \(\text{x}\)\(-3\)\(-2\)\(-1\)01234
    \(\text{f}(\text{x})\)0424\(-2\)0\(-2\)4
    \(\text{x}\)\(-4\)\(-2\)024
    \(\text{g}(\text{x})\)10305
    1. Evaluate \((g∘f)(3),(g∘f)(0).\)
    2. State the domain and range of \((g∘f)(x).\)
    3. Evaluate \((f∘f)(3),(f∘f)(1).\)
    4. State the domain and range of \((f∘f)(x).\)
    જવાબ બતાવો
    1. \((g∘f)(3)=g(f(3))=g(-2)=0\)
      \(\ (g∘f)(0)=g(4)=5\)
    2. The domain of \(g∘f\) is the set \(\{-3,-2,-1,0,1,2,3,4\}.\) Since the range of \(f\) is the set \(\{-2,0,2,4\},\) the range of \(g∘f\) is the set \(\{0,3,5\}.\)
    3. \((f∘f)(3)=f(f(3))=f(-2)=4\)
      \((f∘f)(1)=f(f(1))=f(-2)=4\)
    4. The domain of \(f∘f\) is the set \(\{-3,-2,-1,0,1,2,3,4\}.\) Since the range of \(f\) is the set \(\{-2,0,2,4\},\) the range of \(f∘f\) is the set \(\{0,4\}.\)

  14. A store is advertising a sale of \(20\%\) off all merchandise. Caroline has a coupon that entitles her to an additional \(15\%\) off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of \(x\) dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by using a composite function.

    જવાબ બતાવો

    Since the sale price is \(20\%\) off the original price, if an item is \(x\) dollars, its sale price is given by \(f(x)=0.80x.\) Since the coupon entitles an individual to \(15\%\) off the price of any item, if an item is \(y\) dollars, the price, after applying the coupon, is given by \(g(y)=0.85y.\) Therefore, if the price is originally \(x\) dollars, its sale price will be \(f(x)=0.80x\) and then its final price after the coupon will be \(g(f(x))=0.85(0.80x)=0.68x.\)

  15. If items are on sale for \(10\%\) off their original price, and a customer has a coupon for an additional \(30\%\) off, what will be the final price for an item that is originally \(x\) dollars, after applying the coupon to the sale price?

    જવાબ બતાવો

    \((g∘f)(x)=0.63x\)

  16. Determine whether each of the following functions is even, odd, or neither.

    1. \(f(x)=-5{x}^{4}+7{x}^{2}-2\)
    2. \(f(x)=2{x}^{5}-4x+5\)
    3. \(f(x)=\frac{3x}{{x}^{2}+1}\)
    જવાબ બતાવો

    To determine whether a function is even or odd, we evaluate \(f(\text{-}x)\) and compare it to f(x) and \(\text{-}f(x).\)

    1. \(f(\text{-}x)=-5{(\text{-}x)}^{4}+7{(\text{-}x)}^{2}-2=-5{x}^{4}+7{x}^{2}-2=f(x).\) Therefore, \(f\) is even.
    2. \(f(\text{-}x)=2{(\text{-}x)}^{5}-4(\text{-}x)+5=-2{x}^{5}+4x+5.\) Now, \(f(\text{-}x)\ne f(x).\) Furthermore, noting that \(\text{-}f(x)=-2{x}^{5}+4x-5,\) we see that \(f(\text{-}x)\ne \text{-}f(x).\) Therefore, \(f\) is neither even nor odd.
    3. \(f(\text{-}x)=3(\text{-}x)\text{/}({(\text{-}x)}^{2}+1)=-3x\text{/}({x}^{2}+1)=\text{-}[3x\text{/}({x}^{2}+1)]=\text{-}f(x).\) Therefore, \(f\) is odd.

  17. Determine whether \(f(x)=4{x}^{3}-5x\) is even, odd, or neither.

    જવાબ બતાવો

    \(f(x)\) is odd.

  18. Find the domain and range of the function \(f(x)=2|x-3|+4.\)

    જવાબ બતાવો

    Since the absolute value function is defined for all real numbers, the domain of this function is \((\text{-}\infty ,\infty ).\) Since \(|x-3|\ge 0\) for all \(x,\) the function \(f(x)=2|x-3|+4\ge 4.\) Therefore, the range is, at most, the set \(\{y|y\ge 4\}.\) To see that the range is, in fact, this whole set, we need to show that for \(y\ge 4\) there exists a real number \(x\) such that

    \[2|x-3|+4=y.\]

    A real number \(x\) satisfies this equation as long as

    \[|x-3|=\frac{1}{2}(y-4).\]

    Since \(y\ge 4,\) we know \(y-4\ge 0,\) and thus the right-hand side of the equation is nonnegative, so it is possible that there is a solution. Furthermore,

    \[|x-3|=\{\begin{array}{l}-(x-3)\ \text{if}\ x<3 \\ x-3\ \text{if}\ x\ge 3\end{array}.\]

    Therefore, we see there are two solutions:

    \[x=\pm \frac{1}{2}(y-4)+3.\]

    The range of this function is \(\{y|y\ge 4\}.\)

  19. For the function \(f(x)=|x+2|-4,\) find the domain and range.

    જવાબ બતાવો

    Domain = \((\text{-}\infty ,\infty ),\) range = \(\{y|y\ge -4\}.\)

  20. \(x\)\(y\)\(x\)\(y\)
    −3911
    −2424
    −1139
    00
    જવાબ બતાવો

    a. Domain = \(\{-3,-2,-1,0,1,2,3\},\) range = \(\{0,1,4,9\}\) b. Yes, a function

  21. \(x\)\(y\)\(x\)\(y\)
    −3−211
    −2−828
    −1−13−2
    00
  22. \(x\)\(y\)\(x\)\(y\)
    1−311
    2−222
    3−133
    00
    જવાબ બતાવો

    a. Domain = \(\{0,1,2,3\},\) range = \(\{-3,-2,-1,0,1,2,3\}\) b. No, not a function

  23. \(x\)\(y\)\(x\)\(y\)
    1151
    2161
    3171
    41
  24. \(x\)\(y\)\(x\)\(y\)
    33151
    52212
    81333
    100
    જવાબ બતાવો

    a. Domain = \(\{3,5,8,10,15,21,33\},\) range = \(\{0,1,2,3\}\) b. Yes, a function

  25. \(x\)\(y\)\(x\)\(y\)
    −7111−2
    −2534
    −21611
    0−1

  26. \(f(x)=5x-2\)

    જવાબ બતાવો

    a. \(-2\) b. 3 c. 13 d. \(-5x-2\) e. \(5a-2\) f. \(5a+5h-2\)

  27. \(f(x)=4{x}^{2}-3x+1\)

  28. \(f(x)=\frac{2}{x}\)

    જવાબ બતાવો

    a. Undefined b. 2 c. \(\frac{2}{3}\) d. \(-\frac{2}{x}\) e \(\frac{2}{a}\) f. \(\frac{2}{a+h}\)

  29. \(f(x)=|x-7|+8\)

  30. \(f(x)=\sqrt{6x+5}\)

    જવાબ બતાવો

    a. \(\sqrt{5}\) b. \(\sqrt{11}\) c. \(\sqrt{23}\) d. \(\sqrt{-6x+5}\) e. \(\sqrt{6a+5}\) f. \(\sqrt{6a+6h+5}\)

  31. \(f(x)=\frac{x-2}{3x+7}\)

  32. \(f(x)=\frac{x}{{x}^{2}-16}\)

  33. \(g(x)=\sqrt{8x-1}\)

    જવાબ બતાવો

    \(x\ge \frac{1}{8};y\ge 0;x=\frac{1}{8};\) no y-intercept

  34. \(h(x)=\frac{3}{{x}^{2}+4}\)

  35. \(f(x)=-1+\sqrt{x+2}\)

    જવાબ બતાવો

    \(x\ge -2;y\ge -1;x=-1;y=-1+\sqrt{2}\)

  36. \(f(x)=\frac{1}{\sqrt{x-9}}\)

  37. \(g(x)=\frac{3}{x-4}\)

    જવાબ બતાવો

    \(x\ne 4;y\ne 0;\) no x-intercept; \(y=-\frac{3}{4}\)

  38. \(f(x)=4|x+5|\)

  39. \(g(x)=\sqrt{\frac{7}{x-5}}\)

    જવાબ બતાવો

    \(x>5;y>0;\) no intercepts

  40. \(f(x)={x}^{2}+1\)

    \(x\)\(y\)\(x\)\(y\)
    −31012
    −2525
    −12310
    01

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.
P(A),\ P(A \mid B)
probability, conditional probability
Chance of A; chance of A given that B happened.
i
imaginary unit
i² = −1.
^\circ
degrees
1/360 of a full turn. 180° = π radians.
\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.
\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…
e
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.

How to: Review of Functions

  1. Use functional notation to evaluate a function.
  2. Determine the domain and range of a function.
  3. Draw the graph of a function.
  4. Find the zeros of a function.
  5. Recognize a function from a table of values.
  6. Make new functions from two or more given functions.
  7. Describe the symmetry properties of a function.
  8. Consider

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.

તમારા પોતાના પ્રયત્ન કરો

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.

આમાં વધુ Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests