Functions

Definition of Function

A function f from set D to set E is a rule or correspondence that assigns to each element x of set D exactly one element y of set E. The set D is called the domain of the function. The element y of E is called the image of x under f, or the value of f at x, and is written $ f(x) $ . The subset R of E consisting of all images of elements of D is called the range of the function. The members of the domain D and range R are referred to as the input and output values, respectively.

EXAMPLE 9.1 Let D be the set of all words in English having fewer than 20 letters. Let f be the rule that assigns to each word the number of letters in the word. Then E can be the set of all integers (or some larger set); R is the set $ {x N x < 20} $ . f assigns to the word “truth” the number 5; this would be written $ f() = 5 $ . Moreover, $ f(a) = 1 $ and $ f() = 9 $ .

Note that a function assigns a unique function value to each element in its domain; however, more than one element may be assigned the same function value.

EXAMPLE 9.2 Let D be the set of real numbers and g be the rule given by $ g(x) = x^{2} + 3 $ . Find: $ g(4) $ , $ g(-4) $ , $ g(a) + g(b) $ , $ g(a + b) $ . What is the range of g?

Find values of g by substituting for x in the rule $ g(x) = x^{2} + 3 $ :

\[ \begin{aligned}&g(4)=4^{2}+3=19\quad g(-4)=(-4)^{2}+3=19\\ &\quad g(a)+g(b)=a^{2}+3+b^{2}+3=a^{2}+b^{2}+6\\ &\quad g(a+b)=(a+b)^{2}+3=a^{2}+2ab+b^{2}+3\\ \end{aligned} \]

The range of g is found by noting that the square of a number, $ x^{2} $ , is always greater than or equal to zero. Hence $ g(x) = x^{2} + 3 $ . Thus, the range of g is $ {y R | y } $ .

Function Notation

A function is indicated by the notation $ f:D E $ . The effect of a function on an element of D is then written $ f:x f(x) $ . A picture of the type shown in Fig. 9-1 is often used to visualize the function relationship.

Figure 9-1

Domain and Range

The domain and range of a function are normally sets of real numbers. If a function is defined by an expression and the domain is not stated, the domain is assumed to be the set of all real numbers for which the expression is defined. This set is called the implied domain, or the largest possible domain, of the function.

EXAMPLE 9.3 Find the (largest possible) domain for (a) $ f(x) = $ ; (b) $ g(x) = $ ; (c) $ h(x) = x^{2} - 4 $

  1. The expression $ $ is defined for all real numbers x except when $ x+6=0 $ , that is, when x=-6. Thus the domain of f is $ {xx} $ .

  2. The expression $ $ is defined when $ x-5 $ , that is, when $ x $ . Thus the domain of g is $ {x R x } $

  3. The expression $ x^{2}-4 $ is defined for all real numbers. Thus the domain of h is R.

Graph of a Function

The graph of a function f is the graph of all points $ (x,y) $ such that x is in the domain of f, and $ y = f(x) $ .

V ertical Line Test

Since for each value of x in the domain of f there is exactly one value of y such that $ y = f(x) $ , a vertical line x = c can cross the graph of a function at most once. Thus, if a vertical line crosses a graph more than once, the graph is not the graph of a function.

I ncreasing, Decreasing, and Constant Functions

  1. If, for all x in an interval, as x increases, the value of $ f(x) $ increases, thus, the graph of the function rises from left to right, then the function f is called an increasing function on the interval. A function that is increasing throughout its domain is referred to as an increasing function. Algebraically, then, f is increasing on $ (a, b) $ if for all $ x_{1}, x_{2} $ in $ (a, b) $ , when $ x_{1} < x_{2}, f(x_{1}) < f(x_{2}) $ .

  2. If, for all x in an interval, as x increases, the value of $ f(x) $ decreases, thus, the graph of the function falls from left to right, then the function f is called a decreasing function on the interval. A function that is decreasing throughout its domain is referred to as a decreasing function. Algebraically, then, f is decreasing on $ (a, b) $ if for all $ x_{1}, x_{2} $ in $ (a, b) $ , when $ x_{1} < x_{2}, f(x_{1}) > f(x_{2}) $ .

  3. If the value of a function does not change on an interval, thus, the graph of the function is a horizontal line segment, then the function is called a constant function on the interval. A function that is constant throughout its domain is referred to as a constant function. Algebraically, then, f is constant on $ (a, b) $ if for all $ x_{1}, x_{2} $ in $ (a, b) $ , $ f(x_{1}) = f(x_{2}) $ .

EXAMPLE 9.4 Given the graph of $ f(x) $ shown in Fig. 9-2, assuming the domain of f is R, identify the intervals on which f is increasing or decreasing:

As x increases through the domain of f, y decreases until x = 2, then increases. Thus the function is decreasing on $ (-∞, 2) $ and increasing on $ (2, ∞) $ .

Even and Odd Functions

  1. If, for all x in the domain of a function $ f, f(-x) = f(x) $ , the function is called an even function. Since, for an even function, the equation $ y = f(x) $ is not changed when -x is substituted for x, the graph of an even function has y-axis symmetry.

  2. If, for all x in the domain of a function $ f, f(-x) = -f(x) $ , the function is called an odd function. Since, for an odd function, the equation $ y = f(x) $ is not changed when -x is substituted for x and -y is substituted for y, the graph of an odd function has origin symmetry.

  3. Most functions are neither even nor odd.

EXAMPLE 9.5 Determine whether the following functions are even, odd, or neither:

  1. $ f(x) = 7x^{2} $ (b) $ g(x) = 4x + 6 $ (c) $ h(x) = 6x - $ (d) $ F(x) = $

  2. Consider $ f(-x) $ . $ f(-x) = 7(-x)^{2} = 7x^{2} $ . Since $ f(-x) = f(x) $ , f is an even function.

  3. Consider $ g(-x) $ . $ g(-x) = 4(-x) + 6 = -4x + 6 $ . Also, $ -g(x) = -(4x + 6) = -4x - 6 $ . Since neither $ g(-x) = g(x) $ nor $ g(-x) = -g(x) $ is the case, the function g is neither even nor odd.

  4. Consider $ h(-x) $ . $ h(-x) = 6(-x) - = -6x + $ . Thus, $ h(-x) = -h(x) $ and h is an odd function.

  5. Consider $ F(-x) $ . $ F(-x) = = - $ . Since neither $ F(-x) = F(x) $ nor $ F(-x) = -F(x) $ is the case, the function F is neither even nor odd.

Average Rate of Change of a Function

Let f be a function. The average rate of change $ (x) $ with respect to x over the interval $ [a, b] $ is defined as

\[ \frac{Change in f(x)}{Change in x}=\frac{f(b)-f(a)}{b-a} \]

Over an interval from x to $ x + h $ this quantity becomes

\[ \frac{f(x+h)-f(x)}{h} \]

which is referred to as the difference quotient.

EXAMPLE 9.6 Find the average rate of change of $ f(x) = x^{2} $ on the interval [1,4].

Calculate: $ ==5. $

EXAMPLE 9.7 Find the difference quotient for $ f(x) = x^{2} $

\[ \frac{f(x+h)-f(x)}{h}=\frac{(x+h)^{2}-x^{2}}{h}=\frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\frac{2xh+h^{2}}{h}=2x+h,for h\neq0. \]

I ndependent and Dependent Variables

In applications, if $ y = f(x) $ , the language “y is a function of x” is used. x is referred to as the independent variable, and y as the dependent variable.

EXAMPLE 9.8 In the formula $ A = r^{2} $ , the area A of a circle is written as a function of the radius r. To write the radius as a function of the area, solve this equation for r in terms of A, thus: $ r^{2} = $ , $ r = $ . Since the radius is a positive quantity, $ r = $ gives r as a function of A.

SOLVED PROBLEMS

9.1. Which of the following equations defines y as a function of x?

  1. Since for each value of x there is exactly one corresponding value of y, this defines y as a function of x.

  2. Let x = 6. Then $ 6 = y^{2} + 5 $ ; thus $ y^{2} = 1 $ and $ y = $ . Since for at least one value of x there correspond two values of y, this equation does not define y as a function of x.

  3. Since for each value of x there is exactly one corresponding value of y, this defines y as a function of x. Note that the radical symbol defines y as the positive square root only.

  4. Since for each value of x there is exactly one corresponding value of y, namely 5, this defines y as a function of x.

  5. Let x = 10. Then $ 10^{2} - y^{2} = 36 $ , thus $ y^{2} = 64 $ and $ y = $ . Since for at least one value of x there correspond two values of y, this equation does not define y as a function of x.

9.2. Given $ f(x) = x^{2} - 4x + 2 $ , find (a) $ f(5) $ ; (b) $ f(-3) $ ; (c) $ f(a) $ ; (d) $ f(a + b) $ ; (e) $ f(a) + f(b) $ .

Replace x by the various input values provided:

\[ \begin{aligned}(a)f(5)=5^{2}-4\cdot5+2=7;(b)f(-3)=(-3)^{2}-4(-3)+2=23;(c)f(a)=a^{2}-4a+2\end{aligned} \]

  1. Here x is replaced by the entire quantity $ a + b $ .

\[ f(a+b)=(a+b)^{2}-4(a+b)+2=a^{2}+2ab+b^{2}-4a-4b+2 \]

  1. Here \(x\) is replaced by \(a\) and by \(b\), then the results are added. \(f(a)=a^{2}-4a+2\); \(f(b)=b^{2}-4b+2\); hence \(f(a)+f(b)=a^{2}-4a+2+b^{2}-4b+2=a^{2}+b^{2}-4a-4b+4\)

9.3. Given $ g(x) = -2x^{2} + 3x $ , find and simplify (a) $ g(h) $ ; (b) $ g(x + h) $ ; (c) $ $

  1. Replace x by h. $ g(h) = -2h^{2} + 3h $ .

  2. Replace x by the entire quantity $ x + h $ .

\[ g(x+h)=-2(x+h)^{2}+3(x+h)=-2x^{2}-4xh-2h^{2}+3x+3h \]

  1. Use the result of part (b).

\[ \begin{aligned}\frac{g(x+h)-g(x)}{h}&=\frac{[-2(x+h)^{2}+3(x+h)]-(-2x^{2}+3x)}{h}\\&=\frac{-2x^{2}-4xh-2h^{2}+3x+3h+2x^{2}-3x}{h}\\&=\frac{-4xh-2h^{2}+3h}{h}=-4x-2h+3\end{aligned} \]

9.4. Given $ f(x) = $ and $ g(x) = 4 - x^{2} $ , find (a) $ f(a)g(b) $ ; (b) $ f(g(a)) $ ; (c) $ g(f(b)) $ .

  1. To find $ f(a)g(b) $ , substitute, then multiply: $ f(a)= $ ; $ g(b)=4-b^{2} $ ; hence $ f(a)g(b)=()(4-b{2})=\frac{4-b{2}}{a^{2}} $ .

  2. To find $ f(g(a)) $ , first substitute a into the rule for g to obtain $ g(a) = 4 - a^{2} $ , then substitute this into the rule for f to obtain $ f(g(a)) = f(4 - a^{2}) = $ .

  3. To find $ g(f(b)) $ , first substitute b into the rule for f to obtain $ f(b) = $ , then substitute this into the rule for g to obtain $ g(f(b)) = g() = 4 - ()^{2} $ .

9.5. Find the domain for each of the following functions: (a) $ f(x) = 3x - x^{3} $ ; (b) $ f(x) = $ ;

\[ \left(\mathrm{c}\right)f(x)=\frac{x^{2}-3x+2}{x^{3}+2x^{2}-24x};\left(\mathrm{d}\right)f(x)=\sqrt{x+5};\left(\mathrm{e}\right)f(x)=\sqrt{x^{2}-8x+12};\left(\mathrm{f}\right)f(x)=\sqrt[3]{\frac{x+1}{x^{3}-8}}. \]

  1. This is an example of a polynomial function. Since the polynomial is defined for all real x, the domain of the function is all real numbers, R.

  2. The expression $ $ is defined for all real numbers except if the denominator is 0. This occurs when $ x^{2}-9=0 $ ; thus $ x= $ . The domain is therefore $ {xR|x} $ .

  3. The expression on the right is defined for all real numbers except if the denominator is 0. This occurs when $ x^{3} + 2x^{2} - 24x = 0 $ , or $ x(x - 4)(x + 6) = 0 $ , thus x = 0, 4, -6. The domain is therefore $ {x R | x , 4, -6} $ .

  4. The expression $ $ is defined as long as the expression under the radical is nonnegative. This occurs when $ x + 5 $ or $ x $ . The domain is therefore $ {x R | x } $ , or the interval $ [-5, ) $ .

  5. The expression on the right is defined as long as the expression under the radical is nonnegative. Solving $ x^{2}-8x+12 $ by the methods of Chapter 6, $ x $ or $ x $ is obtained. The domain is therefore $ {xR|x $ or $ x} $ .

  6. The cube root is defined for all real numbers. Thus the expression on the right is defined for all real numbers except if the denominator is 0. This occurs when $ x^{3}-8=0 $ or $ (x-2)(x^{2}+2x+4)=0 $ , thus only when x=2. The domain is therefore $ {x|x} $ .

9.6. Write the circumference C of a circle as a function of its area A.

In Example 9.8 the radius r of a circle was expressed as a function of its area A: $ r = $ . Since $ C = 2r $ , it follows that $ C = 2 $ expresses C as a function of A.

9.7. A theater operator estimates that 500 tickets can be sold if they are priced at 7 per ticket, and that for each .25 increase in the price of a seat, two fewer seats will be sold. Express the revenue R as a function of the number n of .25 price increases of a ticket.

The price of a ticket is $ 7 + 0.25n $ and the number of tickets sold is 500 - 2n. Since revenue = (number of seats sold) $ $ (price per seat), $ R = (7 + 0.25n)(500 - 2n) $ .

9.8. A field is to be marked off in the shape of a rectangle, with one side formed by a straight river. If 100 feet is available for fencing, express the area A of the rectangle as a function of the length of one of the two equal sides x:

Figure 9-3

Since there are two sides of length x, the remaining side has length 100 - 2x.

Since Area = length × width for a rectangle,

\[ A = x(100 - 2x). \]

9.9. A rectangle is inscribed in a circle of radius r (see Fig. 9-4). Express the area A of the rectangle as a function of one side x of the rectangle.

Figure 9-4

From the Pythagorean theorem, it is clear that the sides of the rectangle are related by $ x^{2} + y^{2} = (2r)^{2} $ . Thus $ y = $ and $ A = x $ .

9.10. A right circular cylinder is inscribed in a right circular cone of height H and base radius R (Fig. 9-5). Express the volume V of the cylinder as a function of its base radius r.

Figure 9-5

In the figure, a cross-section through the axis of the cone and cylinder is shown. Triangle ADC is similar to triangle EFC, hence ratios of corresponding sides are equal. In particular, $ = $ thus, $ = $ . Solving for h, $ h = (R - r) $ . Since for a right circular cylinder, $ V = r^{2}h $ , the volume of this cylinder is $ V = (R - r)r^{2} $ .

9.11. Let $ F(x) = mx $ , $ G(x) = x^{2} $ .

  1. Show that $ F(kx) = kF(x) $ . (b) Show that $ F(a + b) = F(a) + F(b) $ . (c) Show that neither of these relations holds in general for the function G.

  2. $ F(kx) = m(kx) = mkx = kmx = kF(x) $

  3. $ F(a + b) = m(a + b) = ma + mb = F(a) + F(b) $

  4. For the function G, compare $ G(kx) = (kx)^{2} = k{2}x{2} $ with $ kG(x) = kx^{2} $ . These are only equal for the special cases k = 0 or k = 1. Similarly, compare $ G(a + b) = (a + b)^{2} = a^{2} + 2ab + b^{2} $ with $ G(a) + G(b) = a^{2} + b^{2} $ . These are only equal in case a = 0 or b = 0.

9.12. Make a table of values and draw graphs of the following functions: (a) $ f(x) = 4 $ ;

\[ \begin{aligned}(b)f(x)=\frac{4x+3}{5};(c)f(x)=4x-x^{2};(d)f(x)=\left\{\begin{aligned}&\quad4\quad&if x\geq0\\&-4\quad&if x<0\end{aligned}\right.;(e)f(x)=\left\{\begin{aligned}&4\quad&if x\geq2\\&-x\quad&if-1<x<2\\&x+2\quad&if x\leq-1\end{aligned}\right.\end{aligned} \]

  1. Form a table of values; then plot the points and connect them. The graph (Fig. 9-6) is a horizontal straight line.
x -2 0 2 4
y 4 4 4 4

Figure 9-6
  1. Form a table of values; then plot the points and connect them. The graph (Fig. 9-7) is a straight line.
x -2 0 2 4
y -1 3/5 11/5 19/5

Figure 9-7
  1. Form a more extensive table of values; then plot the points and connect them. The graph (Fig. 9-8) is a smooth curve.
x -4 -2 0 2 4 6 8
y -32 -12 0 4 0 -12 -32

Figure 9-8
  1. Form a table of values. The graph (Fig. 9-9) is discontinuous at the point where x = 0.
x -4 -2 0 2 4
y -4 -4 4 4 4

Figure 9-9
  1. Form a table of values. The graph (Fig. 9-10) is discontinuous at the point where x = 2. Note that the graph consists of three separate “pieces,” since the rule defining the function does so also.
x -3 -2 -1 0 1 2 3
y -1 -2 1 0 -1 4 4

Figure 9-10

9.13. Find the range for each of the functions defined in the previous problem.

  1. $ f(x) = 4 $ . The only possible function value is 4, hence the range is $ {4} $ .

  2. $ f(x) = $ . Set $ k = $ and solve for x in terms of k to obtain $ x = $ . There are no restrictions on k, hence the range is R.

  3. $ f(x) = 4x - x^{2} $ . Set k = 4x - x^{2} and solve for x in terms of k to obtain $ x = 2 $ . This expression represents a real number only if $ k $ ; hence the range is $ (- , 4] $ .

  4. $ f(x) = {

    \[\begin{array}{ll} 4 & \text{if } x \geq 0 \\ -4 & \text{if } x < 0 \end{array}\]

    . $ . The only possible function values are 4 and -4; hence the range is the set $ {4, -4} $ .

\[ \left(\mathrm{e}\right)f(x)=\begin{cases}4&\text{if}x\geq2\\-x&\text{if}-1<x<2.\\x+2&\text{if}x\leq-1\end{cases} \]

If $ x $ , $ x + 2 $ ; thus f can take on any value in $ (-, 1] $ . If -1 < x < 2, -2 < -x < 1; this adds nothing to the range. If $ x $ , $ f(x) = 4 $ ; hence the range consists of the set union $ (-, 1] {4} $ .

9.14. Find the average rate of change for (a) $ f(x) = 7x + 12 $ on $ [2,8] $ ; (b) $ f(x) = $ on $ [-5,0] $

\[ \left(a\right)\frac{f(8)-f(2)}{8-2}=\frac{(7\cdot8+12)-(7\cdot2+12)}{6}=\frac{68-26}{6}=7 \]

\[ \left(b\right)\frac{f(0)-f(-5)}{0-(-5)}=\frac{\left(\frac{3-5\cdot0}{9}\right)-\left(\frac{3-5(-5)}{9}\right)}{5}=\frac{\frac{3}{9}-\frac{28}{9}}{5}=\frac{-25/9}{5}=-\frac{5}{9} \]

9.15. Find the difference quotient for (a) $ f(x) = x^{3} $ ; (b) $ f(x) = $

\[ \begin{aligned}(a)\frac{f(x+h)-f(x)}{h}&=\frac{(x+h)^{3}-x^{3}}{h}=\frac{x^{3}+3x^{2}h+3xh^{2}+h^{3}-x^{3}}{h}\\&=\frac{3x^{2}h+3xh^{2}+h^{3}}{h}=3x^{2}+3xh+h^{2}\end{aligned} \]

\[ \begin{aligned}(b)\frac{f(x+h)-f(x)}{h}&=\frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\frac{x^{2}-(x+h)^{2}}{hx^{2}(x+h)^{2}}=\frac{x^{2}-x^{2}-2xh-h^{2}}{hx^{2}(x+h)^{2}}\\&=\frac{-2xh-h^{2}}{hx^{2}(x+h)^{2}}=\frac{-2x-h}{x^{2}(x+h)^{2}}\end{aligned} \]

SUPPLEMENTARY PROBLEMS

9.16. Let F be any function whose domain contains -x whenever it contains x. Define:

\[ g(x)=\frac{F(x)\;+\;F(-x)}{2}\;and\;h(x)=\frac{F(x)\;-\;F(-x)}{2}. \]

  1. Show that g is an even function and h is an odd function.

  2. Show that $ F(x) = g(x) + h(x) $ . Thus, any function can be written as the sum of an odd function and an even function.

  3. Show that the only function which is both even and odd is $ f(x)=0 $ .

9.17. Are the following functions even, odd, or neither?

\[ \begin{aligned}(a)f(x)=\frac{x^{3}}{x^{4}+1};(b)f(x)=\frac{x^{4}}{x^{3}+1};(c)f(x)=\left|x\right|-\frac{1}{x^{2}};(d)f(x)=(x-1)^{3}+(x+1)^{3}\end{aligned} \]

Ans. (a) odd; (b) neither; (c) even; (d) odd

9.18. Find the domain for the following functions:

\[ \begin{aligned}(a)f(x)=\sqrt{x-3};(b)f(x)=\sqrt{3-x};(c)f(x)=\frac{1}{\sqrt{x-3}};(d)f(x)=\frac{1}{\sqrt{3-x}}\end{aligned} \]

\[ Ans.\quad(a)[3,\infty);(b)(-\infty,3];(c)(3,\infty);(d)(-\infty,3) \]

9.19. Find the domain for the following functions:

\[ \begin{aligned}(a)\ g(x)&=|x-3|;(\mathbf{b})\ g(x)=\frac{x^{2}+9}{x-3};(\mathbf{c})\ g(x)=\sqrt{\frac{x-3}{x^{2}-3x+2}};(\mathbf{d})\ g(x)=\sqrt{x^{3}-9x^{2}}.\end{aligned} \]

\[ Ans.\quad(a)\ \boldsymbol{R};(b)\ \{x\in\boldsymbol{R}|x\neq3\};(c)\ (1,2)\cup[3,\infty);(d)\ [9,\infty) \]

9.20. The income tax rate in a certain state is 4% on taxable income up to 30,000, 5% on taxable income between 30,000 and 50,000, and 6% on taxable income over 50,000. Express the income tax $ T(x) $ as a function of taxable income x.

\[ Ans.\quad T(x)=\left\{\begin{aligned}&0.04x\quad&\quad&if0<x\leq30,000\\&1200+0.05(x-30,000)\quad&\quad&if30,000<x\leq50,000\\&2200+0.06(x-50,000)\quad&\quad&if50,000<x\end{aligned}\right. \]

9.21. (a) Express the length of a diagonal d of a square as a function of the length of one side s. (b) Express d as a function of the area A of the square. (c) Express d as a function of the perimeter P of the square.

\[ Ans.\quad(a)d(s)=s\sqrt{2};(b)d(A)=\sqrt{2A};(c)d(P)=\frac{P\sqrt{2}}{4} \]

9.22. (a) Express the area A of an equilateral triangle as a function of one side s. (b) Express the perimeter of the triangle P as a function of the area A.

\[ Ans.\quad(a)A(s)=s^{2}\sqrt{3}/4;(b)P(A)=(6\sqrt{A})/\sqrt[4]{3} \]

9.23. An equilateral triangle of side s is inscribed in a circle of radius r.

  1. Express s as a function of r. (b) Express the area A of the triangle as a function of r. (c) Express the area A of the triangle as a function of a, the area of the circle.

\[ Ans.\quad(a)s(r)=r\sqrt{3};(b)A(r)=\frac{3r^{2}\sqrt{3}}{4};(c)A(a)=\frac{3a\sqrt{3}}{4\pi} \]

9.24. (a) Express the volume V of a sphere as a function of its radius r. (b) Express the surface area S of the sphere as a function of r. (c) Express r as a function of S. (d) Express V as a function of S.

\[ Ans.\quad(a)V(r)=\frac{4}{3}\pi r^{3};(b)S(r)=4\pi r^{2};(c)r(S)=\sqrt{\frac{S}{4\pi}};(d)V(S)=\frac{1}{3}\sqrt{\frac{S^{3}}{4\pi}} \]

9.25. A right circular cylinder is inscribed in a sphere of radius R. (R is a constant.)

  1. Express the height h of the cylinder as a function of the radius r of the cylinder.

  2. Express the total surface area S of the cylinder as a function of r.

  3. Express the volume V of the cylinder as a function of r.

\[ Ans.\quad(a)h(r)=2\sqrt{R^{2}-r^{2}};(b)S(r)=4\pi r\sqrt{R^{2}-r^{2}}+2\pi r^{2};(c)V(r)=2\pi r^{2}\sqrt{R^{2}-r^{2}} \]

9.26. Which of Figs. 9-11 to 9-14 are graphs of functions?

  1. Figure 9-11

Figure 9-11
  1. Figure 9-12

Figure 9-12
  1. Figure 9-13

Figure 9-13
  1. Figure 9-14

Figure 9-14

Ans. (a) and (c) are graphs of functions; (b) and (d) fail the vertical line test and are not graphs of functions.

9.27. Given $ f(x) = x^{2} - 3x + 1 $ , find (a) $ f(2) $ ; (b) $ f(-3) $ ; (c) $ $ .

Ans. (a) -1; (b) 19; (c) $ 1 + h $

9.28. Given $ f(x) = - x $ , find (a) $ f(2) $ ; (b) $ f(-3) $ ; (c) $ $ .

Ans. (a) $ - $ ; (b) $ $ ; (c) $ $

9.29. The distance s an object falls from rest in time t seconds is given in feet by $ s(t) = 16t^{2} $ . Find (a) $ s(2) $ ; (b) $ s(3) $ ; (c) $ $

Ans. (a) 64 feet; (b) 144 feet; (c) 96.16 feet

9.30. Given $ f(x) = $ , find and write in simplest form: (a) $ f(f(b)) $ ; (b) $ $

Ans. (a) b; (b) $ $

9.31. Given $ f(x) = x^{2} $ , find and write in simplest form: (a) $ f(f(b)) $ ; (b) $ $ ; (c) $ $

Ans. (a) $ b^{4} $ ; (b) $ x + a $ ; (c) $ 2x + h $

9.32. Given $ f(x) = $ , find and write in simplest form: (a) $ f(f(b)) $ ; (b) $ $ ; (c) $ $

Ans. (a) b; (b) $ $ ; (c) $ $

9.33. Given $ f(x)= $ , find and write in simplest form: (a) $ f(f(b)) $ ; (b) $ $

Ans. (a) $ $ ; (b) $ $

9.34. Find the average rate of change for $ f(x) = 9x - 7 $ on the interval [0,5].

Ans. 9

9.35. (a) Find the average rate of change for $ f(x) = $ on the interval [4,9].

  1. Find the difference quotient for $ f(x) = $ . Rationalize the numerator in the answer.

Ans. (a) $ $ ; (b) $ = $

9.36. Find the average rate of change for $ f(x) = x^{2} - 6x + 9 $ (a) on the interval $ [0, 6] $ ; (b) on the interval $ [1, 7] $ .

Ans. (a) 0; (b) 2

9.37. Find the average rate of change for $ f(x) = $ on the interval $ [0, 5] $ .

Ans. $ - $

9.38. Find the difference quotient for (a) $ f(x) = $ ; (b) $ f(x) = $ . Rationalize the numerator in the answer.

\[ \frac{1}{(x+1)(x+h+1)};(b)\ \frac{\sqrt{2(x+h)-1}-\sqrt{2x-1}}{h}=\frac{2}{\sqrt{2(x+h)-1}+\sqrt{2x-1}} \]