Algebra of Functions; Inverse Functions

Algebraic Combinations of Functions

Algebraic combinations of functions can be obtained in several ways: Given two functions \(f\) and \(g\), the sum, difference, product, and quotient functions can be defined as follows:

NAME DEFINITION DOMAIN
Sum $ (f + g)(x) = f(x) + g(x) $ The set of all $ x $ that are in the domain of both $ f $ and $ g \(</td></tr><tr><td>Difference</td><td>\) (f - g)(x) = f(x) - g(x) $ The set of all $ x $ that are in the domain of both $ f $ and $ g \(</td></tr><tr><td>Difference</td><td>\) (g - f)(x) = g(x) - f(x) $ The set of all $ x $ that are in the domain of both $ f $ and $ g \(</td></tr><tr><td>Product</td><td>\) (fg)(x) = f(x)g(x) $ The set of all $ x $ that are in the domain of both $ f $ and $ g \(</td></tr><tr><td>Quotient</td><td>\) ()(x) = $ The set of all $ x $ that are in the domain of both $ f $ and $ g $ , with $ g(x) \(</td></tr><tr><td>Quotient</td><td>\) ()(x) = $ The set of all $ x $ that are in the domain of both $ f $ and $ g $ , with $ f(x) $

EXAMPLE 13.1 Given $ f(x) = x^{2} $ and $ g(x) = $ , find $ (f + g)x $ and $ (f/g)(x) $ and state the domains of the functions.

$ (f + g)(x) = f(x) + g(x) = x^{2} + $ . Since the domain of f is R and the domain of g is $ {x R x } $ the domain of this function is also $ {x R x } $ .

$ ()(x)= $ . The domain of this function is the same as the domain of $ f+g $ , with the further restriction that $ g(x) $ , that is, $ {xx>2} $ .

Definition of Composite Function

The composite function $ f g $ of two functions f and g is defined by:

\[ f\circ g(x)=f(g(x)) \]

The domain of $ f g $ is the set of all x in the domain of g such that $ g(x) $ is in the domain of f.

EXAMPLE 13.2 Given $ f(x) = 3x - 8 $ and $ g(x) = 1 - x^{2} $ , find $ f g $ and state its domain.

$ f g(x) = f(g(x)) = f(1 - x^{2}) = 3(1 - x^{2}) - 8 = -5 - 3x^{2} $ . Since the domains of f and g are both R, the domain of $ f g $ is also R.

EXAMPLE 13.3 Given $ f(x) = x^{2} $ and $ g(x) = $ , find $ f g $ and state its domain.

$ f g(x) = f(g(x)) = f() = ()^{2} = x - 5 $ . The domain of $ f g $ is not all of R. Since the domain of $ g $ is $ {x R x } $ , the domain of $ f g $ is the set of all $ x $ in the domain of f, that is, all of $ {x R x } $ .

Fig. 13-1 shows the relationships among f, g, and $ f g $ .

Figure 13-1

One-to-One Functions

A function with domain D and range R is called a one-to-one function if exactly one element of set D corresponds to each element of set R.

EXAMPLE 13.4 Let $ f(x) = x^{2} $ and $ g(x) = 2x $ . Show that f is not a one-to-one function and that g is a one-to-one function.

The domain of f is R. Since $ f(3) = f(-3) = 9 $ , that is, the elements 3 and -3 in the domain of f correspond to 9 in the range, f is not one-to-one.

The domain and range of g are both R. Let k be an arbitrary real number. If 2x = k, then the only x that corresponds to k is x = k/2. Thus g is one-to-one.

A function with domain D and range R is one-to-one if either of the following equivalent conditions is satisfied.

  1. Whenever $ f(u) = f(v) $ in R, then u = v in D.

  2. Whenever $ u v $ in D, then $ f(u) f(v) $ in R.

Horizontal Line Test

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

Definition of Inverse Function

Let f be a one-to-one function with domain D and range R. Since for each y in R there is exactly one x in D such that $ y = f(x) $ , define a function g with domain R and range D such that $ g(y) = x $ . Then g reverses the correspondence defined by f. The function g is called the inverse function of f.

Function-Inverse Function Relationship

If g is the inverse function of f, then, by the above definition,

  1. $ g(f(x)) = x $ for every x in D.

  2. $ f(g(y)) = y $ for every y in R.

Notation for Inverse Functions

If f is a one-to-one function with domain D and range R, then the inverse function of f with domain R and range D is often denoted by $ f^{-1} $ . Then $ f^{-1} $ is also a one-to-one function and $ x = f^{-1}(y) $ if and only if $ y = f(x) $ . With this notation, the function-inverse function relationship becomes:

  1. $ f^{-1}(f(x)) = x $ for every x in D.

  2. $ f(f^{-1}(y)) = y $ for every y in R.

Figure 13-2 shows the relationship between f and $ f^{-1} $

Figure 13-2

To Find the Inverse Function for a Given Function f:

  1. Verify that f is one-to-one.

  2. Solve the equation $ y = f(x) $ for x in terms of y, if possible. This gives an equation of form $ x = f^{-1}(y) $ .

  3. Interchange x and y in the equation found in step 2. This gives an equation of the form $ y = f^{-1}(x) $ .

EXAMPLE 13.5 Find the inverse function for $ f(x) = 3x - 1 $

First, show that \(f\) is one-to-one. Assume \(f(u) = f(v)\). Then it follows that

\[ \begin{aligned}3u-1&=3\nu-1\\3u&=3\nu\\u&=\nu\end{aligned} \]

Thus, \(f\) is one-to-one. Now solve \(y = 3x - 1\) for \(x\) to obtain

\[ y=3x-1 \]

\[ y+1=3x \]

\[ x=\frac{y+1}{3} \]

Now interchange x and y to obtain $ y = f^{-1}(x) = $ .

Graph of an Inverse Function

The graphs of $ y = f(x) $ and $ y = f^{-1}(x) $ are symmetric with respect to the line y = x.

SOLVED PROBLEMS

13.1. Given $ f(x) = ax + b $ and $ g(x) = cx + d $ , $ a, c $ , find $ f + g $ , f - g, fg, f/g and state their domains.

\[ \begin{align*}(f+g)(x)=f(x)+g(x)=ax+b+cx+d=(a+c)x+(b+d)\ $ f-g)(x)=f(x)-g(x)=(ax+b)-(cx+d)=ax+b-cx-d=(a-c)x+(b-d)\ $ fg)(x)=f(x)g(x)=(ax+b)(cx+d)=acx^{2}+(ad+bc)x+bd\end{align*} \]

Since R is the domain of both f and g, the domain of each of these functions is R.

\[ (f/g)(x)=f(x)/g(x)=(ax+b)/(cx+d) \]

The domain of this function is \(\{x\in\mathbb{R}\mid x\neq-d/c\}\).

13.2. Given $ f(x) = $ and $ g(x) = $ , find $ f + g $ , f - g, fg, f/g and state their domains.

\[ \begin{aligned}(f+g)(x)&=f(x)+g(x)=\frac{x+1}{x^{2}-4}+\frac{2}{x}=\frac{x(x+1)+2(x^{2}-4)}{x(x^{2}-4)}=\frac{3x^{2}+x-8}{x(x^{2}-4)}\ $ f-g)(x)&=f(x)-g(x)=\frac{x+1}{x^{2}-4}-\frac{2}{x}=\frac{x(x+1)-2(x^{2}-4)}{x(x^{2}-4)}=\frac{-x^{2}+x+8}{x(x^{2}-4)}\\ (fg)(x)&=f(x)g(x)=\frac{x+1}{x^{2}-4}\cdot\frac{2}{x}=\frac{2x+2}{x(x^{2}-4)}\end{aligned} \]

Since the domain of \(f\) is \(\{x\in\mathbb{R}\mid x\neq-2,2\}\) and the domain of \(g\) is \(\{x\in\mathbb{R}\mid x\neq0\}\), the domain of each of these functions is \(\{x\in\mathbb{R}\mid x\neq-2,2,0\}\).

\[ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{x^{2}-4}\div\frac{2}{x}=\frac{x+1}{x^{2}-4}\cdot\frac{x}{2}=\frac{x^{2}+x}{2x^{2}-8} \]

The domain of this function may not be apparent from its final form. From the definition of the quotient function, the domain of this function must be those elements of $ {xx,2,0} $ for which $ g(x) $ is not 0. Since $ g(x) $ is never 0, the domain of this quotient function is $ {xx,2,0} $ .

13.3. If f and g are even functions, show that $ f + g $ and fg are even functions.

$ (f + g)(-x) = f(-x) + g(-x) $ and $ (fg)(-x) = f(-x)g(-x) $ by definition. Since f and g are even functions,

\[ f(-x) + g(-x) = f(x) + g(x) and f(-x)g(-x) = f(x)g(x). Therefore, \]

\[ (f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x)and \]

\[ (f g)(-x)=f(-x)g(-x)=f(x)g(x)=(f g)(x), \]

that is, $ f + g $ and fg are even functions.

13.4. Given $ f(x) = $ and $ g(x) = $ , find $ g + f $ , g - f, gf, f/g and state their domains.

\[ (g+f)(x)=g(x)+f(x)=\sqrt{x^{2}-4}+\sqrt{1-x} \]

\[ (g-f)(x)=g(x)-f(x)=\sqrt{x^{2}-4}-\sqrt{1-x} \]

\[ (gf)(x)=g(x)f(x)=\sqrt{x^{2}-4}\cdot\sqrt{1-x} \]

Since the domain of \(f\) is \(\{x\in\mathbb{R}\mid x\leq1\}\) and the domain of \(g\) is \(\{x\in\mathbb{R}\mid x\leq-2\) or \(x\geq2\}\), the domain of each of these functions is the intersection of these two sets, that is, \(\{x\in\mathbb{R}\mid x\leq-2\}\).

\[ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{1-x}}{\sqrt{x^{2}-4}} \]

The domain of this function is those elements of $ {x R x } $ for which $ g(x) $ , that is, $ {x R x < -2} $ .

13.5. Given $ f(x) = x^{4} $ and $ g(x) = 3x + 5 $ , find $ f g $ and $ g f $ and state their domains.

\[ \begin{aligned}f\circ g(x)&=f(g(x))=f(3x+5)=(3x+5)^{4}\\g\circ f(x)&=g(f(x))=g(x^{4})=3x^{4}+5\end{aligned} \]

Since the domains of \(f\) and \(g\) are both \(R\), the domains of \(f\circ g\) and \(g\circ f\) are also \(R\).

13.6 Given $ f(x) = |x| $ and $ g(x) = -5 $ , find $ f g $ and $ g f $ and state their domains.

\[ \begin{aligned}f\circ g(x)&=f(g(x))=f(-5)=|-5|=5\\g\circ f(x)&=g(f(x))=g(|x|)=-5\end{aligned} \]

Since the domains of \(f\) and \(g\) are both \(R\), the domains of \(f \circ g\) and \(g \circ f\) are also \(R\).

13.7. Given $ f(x) = $ and $ g(x) = x^{2} + 5x $ , find $ f g $ and $ g f $ and state their domains.

\[ f\circ g(x)=f(g(x))=f(x^{2}+5x)=\sqrt{x^{2}+5x-6} \]

\[ g\circ f(x)=g(f(x))=g(\sqrt{x-6})=(\sqrt{x-6})^{2}+5\sqrt{x-6}=x-6+5\sqrt{x-6} \]

Since the domain of g is R, the domain of f. g is the set of all real numbers with $ g(x) $ in the domain of f, that is, $ g(x) $ , or $ x^{2} + 5x $ , or $ {x R x or x } $ .

Since the domain of f is $ {x R x } $ , the domain of $ g f $ is the set of all numbers in this set with $ f(x) $ in the domain of g; this is all of $ {x R x } $ .

13.8. Given $ f(x) = | x - 1 | $ and $ g(x) = 1/x $ , find $ f g $ and $ g f $ and state their domains.

\[ f\circ g(x)=f(g(x))=f\bigg(\frac{1}{x}\bigg)=\left|\frac{1}{x}-1\right|\mathrm{o r}\left|\frac{1-x}{x}\right| \]

\[ g\circ f(x)=g(f(x))=g(|x-1|)=\frac{1}{|x-1|} \]

Since the domain of g is $ {x R x } $ , the domain of $ f g $ is the set of all nonzero real numbers with $ g(x) $ in the domain of f, that is, $ {x R x } $ .

Since the domain of f is R, the domain of g ∘ f is the set of all real numbers with $ f(x) $ in the domain of g, that is, $ {x ∈ R x ≠ 1} $ .

13.9. Given $ f(x) = $ and $ g(x) = $ , find $ f g $ and $ g f $ and state their domains.

\[ f\circ g(x)=f(g(x))=f(\sqrt{4-x^{2}})=\sqrt{(\sqrt{4-x^{2}})^{2}+5}=\sqrt{4-x^{2}+5}=\sqrt{9-x^{2}} \]

\[ g\circ f(x)=g(f(x))=g(\sqrt{x^{2}+5})=\sqrt{4-(\sqrt{x^{2}+5})^{2}}=\sqrt{4-(x^{2}+5)}=\sqrt{-1-x^{2}} \]

Since the domain of g is $ {xx} $ , the domain of $ fg $ is the set of all numbers in this set with $ f(x) $ in the domain of g; this is all of $ {xx} $ . Note that the domain of $ fg $ cannot be determined from its final form.

Since $ -1 - x^{2} $ is negative for all real x, the domain of $ g f $ is empty.

13.10. Find a composite function form for each of the following:

\[ \begin{aligned}(a)\quad y&=(5x-3)^{4}\quad&(b)\quad y&=\sqrt{1-x^{2}}\quad&(c)\quad y&=\frac{1}{(x^{2}-5x+6)^{2/3}}\end{aligned} \]

  1. Let $ y = u^{4} $ and u = 5x - 3. Then $ y = f(u) $ and $ u = g(x) $ ; hence $ y = f(g(x)) $ .

  2. Let $ y = $ and $ u = 1 - x^{2} $ . Then $ y = f(u) $ and $ u = g(x) $ ; hence $ y = f(g(x)) $ .

  3. Let $ y = u^{-2/3} $ and $ u = x^{2} - 5x + 6 $ . Then $ y = f(u) $ and $ u = g(x) $ ; hence $ y = f(g(x)) $ .

13.11. A spherical balloon is being inflated at the constant rate of $ 6, ft^{3}/min $ . Express its radius r as a function of time t (in minutes), assuming that r = 0 when t = 0.

Express the radius r as a function of the volume V and V as a function of the time t.

Since $ V = r^{3} $ for a sphere, solve for r to obtain $ r = f(V) = $ . V is a linear function of t with slope $ 6$ ;

since V=0 when t=0, $ V=g(t)=6t $ . Hence, $ r=f(g(t))== $ feet.

13.12. The revenue (in dollars) from the sale of x units of a certain product is given by the function $ R(x) = 20x - x^{2}/200 $ . The cost (in dollars) of producing x units is given by the function $ C(x) = 4x + 8000 $ . Find the profit on sales of x units.

The profit function $ P(x) $ is given by $ P(x) = (R - C)(x) $ . Hence

\[ \begin{aligned}P(x)&=(R-C)(x)\\&=R(x)-C(x)\\&=(20x-x^{2}/200)-(4x+8000)\\&=20x-x^{2}/200-4x-8000\\&=-x^{2}/200+16x-8000\\ \end{aligned} \]

13.13. In the previous problem, if the demand x and the price p (in dollars) for the product are related by the function $ x = f(p) = 4000 - 200p $ , $ 0 p $ , write the profit as a function of the demand p.

\[ \begin{aligned}F(p)&=P\circ f(p)=P(f(p))=P(4000-200p)\\&=-(4000-200p)^{2}/200+16(4000-200p)-8000\end{aligned} \]

13.14. In the previous problem, find the price which would yield the maximum profit and also find this maximum profit.

Simplifying, obtain

\[ \begin{aligned}F(p)&=-(4000-200p)^{2}/200+16(4000-200p)-8000\\&-(16,000,000-1,600,000p+40,000p^{2})/200+64,000-3200p-8000\\&=-80,000+8000p-200p^{2}+64,000-3200p-8000\\&=-200p^{2}+4800p-24,000\\ \end{aligned} \]

This is a quadratic function. Completing the square yields $ F(p) = -200(p - 12)^{2} + 4800 $ . The function attains a maximum value (maximum profit) of 4800 when the price $ p = \12 $ .

13.15. Show that every increasing function is one-to-one on its domain.

Let f be an increasing function, that is, for every a,b in the domain of f, if a < b, then $ f(a) < f(b) $ . Now, if $ u v $ , then either u < v or u > v. Thus either $ f(u) < f(v) $ or $ f(u) > f(v) $ ; in either case, $ f(u) f(v) $ and f is a one-to-one function.

13.16. Determine whether or not each of the following functions is one-to-one.

  1. $ f(x) = 5 $ ; (b) $ f(x) = 5x $ ; (c) $ f(x) = x^{2} + 5 $ ; (d) $ f(x) = $

  2. Since $ f(2) = 5 $ and $ f(3) = 5 $ , this function is not one-to-one.

  3. Assume $ f(u) = f(v) $ . Then it follows that 5u = 5v; hence u = v. Therefore, f is one-to-one.

  4. Since $ f(2) = 9 $ and $ f(-2) = 9 $ , this function is not one-to-one.

  5. Assume $ f(u) = f(v) $ . Then it follows that

\[ \begin{aligned}\sqrt{u-5}&=\sqrt{v-5}\\u-5&=v-5\\u&=v\end{aligned} \]

Therefore, \(f\) is one-to-one.

13.17. Use the function-inverse function relationship to show that f and g are inverses of each other, and sketch the graphs of f and g and the line y = x on the same Cartesian coordinate system.

\[ \begin{aligned}&(a)f(x)=2x-3\quad&g(x)=\frac{x+3}{2}\\&(b)f(x)=x^{2}+3,x\geq0\quad&g(x)=\sqrt{x-3},x\geq3\\&(c)f(x)=-\sqrt{4-x},x\leq4\quad&g(x)=4-x^{2},x\leq0\end{aligned} \]

  1. Note first that Dom f = Range g = R.

Also Dom g = Range f = R.

\[ \begin{aligned}g(f(x))&=g(2x-3)=\frac{2x-3+3}{2}\\&=x\\f(g(y))&=f\Big(\frac{y+3}{2}\Big)=2\Big(\frac{y+3}{2}\Big)-3\\&=y\end{aligned} \]

The line y = x is shown dashed in Fig. 13-3.

Figure 13-3
  1. Note first that Dom \(f = \operatorname{Range} g = [0, \infty)\). Also Dom \(g = \operatorname{Range} f = [3, \infty)\).

\[ \begin{aligned}g(f(x))&=g(x^{2}+3)=\sqrt{x^{2}+3-3}\\&=\sqrt{x^{2}}=x\quad on[0,\infty).\\f(g(y))&=f(\sqrt{y-3})=(\sqrt{y-3})^{2}+3)\\&=y-3+3=y\end{aligned} \]

The line y = x is shown dashed in Fig. 13-4.

Figure 13-4
  1. Note first that Dom \(f = \operatorname{Range} g = (-\infty, 4]\).

Also Dom \(g = \mathrm{Range} f = (-\infty, 0].\)

\[ \begin{aligned}g(f(x))&=g(-\sqrt{4-x})=4-(-\sqrt{4-x})^{2}\\&=4-(4-x)=x\end{aligned} \]

\[ \begin{aligned}f(g(y))&=f(4-y^{2})=-\sqrt{4-(4-y^{2})}\\&=-\sqrt{y^{2}}=y\quad on(-\infty,0]\end{aligned} \]

The line y = x is shown dashed in Fig. 13-5.

Figure 13-5

13.18. The following functions are one-to-one. Find the inverse functions for each.

  1. $ f(x) = 4x - 1 $

  2. $ f(x)= $

  3. $ f(x) = x^{2} - 9, x $

  4. $ f(x) = 4 + (x + 3)^{2}, x $

  5. Set y = 4x - 1.

Solve for x in terms of y.

  1. Set $ y = 2/(x + 3) $ .

Solve for x in terms of y.

\[ \begin{aligned}4x-1&=y\\4x&=y+1\\x&=\frac{y+1}{4}\end{aligned} \]

\[ y=\frac{2}{x+3} \]

\[ x+3=\frac{2}{y} \]

\[ x=\frac{2}{y}-3 \]

Interchange x and y.

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

Interchange x and y.

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

Note:

Note:

Dom f = Range $ f^{-1} = R $

Dom f = Range $ f^{-1} = (-, -3) (-3, ) $

Dom $ f^{-1}= $ Range f = R

\[ Dom f^{-1} = Range f = (-\infty, 0) \cup (0, \infty) \]

  1. Set $ y = x^{2} - 9 $ , $ x $ .
  1. Set y = 4 + (x + 3)2, x ≤ −3.

Solve for x in terms of y.

\[ \begin{aligned}x^{2}-9&=y\\x^{2}&=y+9\\x&=\sqrt{y+9}\end{aligned} \]

Solve for x in terms of y.

\[ 4+(x+3)^{2}=y \]

(since \(x\) must be nonnegative)

\[ (x+3)^{2}=y-4 \]

\[ x+3=-\sqrt{y-4} \]

(since $ x + 3 $ must be nonpositive)

\[ x=-3-\sqrt{y-4} \]

Interchange x and y.

\[ y=f^{-1}(x)=\sqrt{x+9} \]

Interchange x and y.

\[ y=f^{-1}(x)=-3-\sqrt{x-4} \]

Note:

Note:

\[ Dom f=Range f^{-1}=[0,\infty) \]

\[ Dom f=Range f^{-1}=(-\infty,-3] \]

\[ Dom f^{-1}=Range f=[-9,\infty) \]

\[ Dom f^{-1}=Range f=[4,\infty) \]

13.19. The function $ F(x) = (x - 4)^{2} $ is not one-to-one. Find the inverse of the function defined by restricting the domain of F to (a) $ x $ ; (b) $ x $ .

  1. First, show that the function $ f(x) = (x - 4)^{2} $ , $ x $ , is one-to-one.

Assume $ f(u) = f(v) $ . Then it follows that

\[ (u-4)^{2}=(v-4)^{2},\quad u,v\geq4 \]

\[ u-4=\pm\sqrt{(v-4)^{2}} \]

\[ u=4\pm(v-4)\qquad since v\geq4 \]

Now, since u must be greater than or equal to 4, the positive sign must be chosen.

\[ u=4+\nu-4=\nu \]

Therefore, f is one-to-one.

Now set $ y = f(x) = (x - 4)^{2} $ , $ x $ , and solve for x in terms of y.

\[ \begin{array}{l}(x-4)^{2}=y\\x-4=\sqrt{y}\text{since}x\geq4\\x=4+\sqrt{y}\end{array} \]

Interchange x and y to obtain $ y = f^{-1}(x) = 4 + $ .

Note: Dom \(f = \mathrm{Range} f^{-1} = [4, \infty)\). Dom \(f^{-1} = \mathrm{Range} f = [0, \infty)\)

  1. First, show that the function $ f(x) = (x - 4)^{2} $ , $ x $ , is one-to-one.

Assume $ f(u) = f(v) $ . Then it follows that

\[ \begin{aligned}(u-4)^{2}&=(v-4)^{2},\quad&u,v\leq4\\u-4&=\pm\sqrt{(v-4)^{2}}\\u&=4\pm(4-v)\quad&since v\leq4\end{aligned} \]

Now, since u must be less than or equal to 4, the negative sign must be chosen.

\[ u=4-(4-v)=v \]

Therefore, f is one-to-one.

Now set $ y = f(x) = (x - 4)^{2} $ , $ x $ , and solve for x in terms of y.

\[ \begin{array}{l}(x-4)^{2}=y\\x-4=-\sqrt{y}\\x=4-\sqrt{y}\end{array}\quad since x\leq4 \]

Interchange x and y to obtain $ y = f^{-1}(x) = 4 - $ .

Note: Dom f = Range $ f^{-1} = (-, 4] $ . Dom $ f^{-1} = $ Range $ f = [0, ) $ .

SUPPLEMENTARY PROBLEMS

13.20. Show that if f and g are odd functions, then $ f + g $ and f - g are odd functions, but fg and f/g are even functions.

13.21. Given $ f(x) = $ and $ g(x) = $ , find $ f g $ and $ g f $ and state their domains.

Ans. $ f g(x) = g f(x) = x $ for all $ x R $ .

13.22. The revenue (in dollars) from the sale of x units of a certain product is given by the function $ R(x) = 60x - x^{2}/100 $ . The cost (in dollars) of producing x units is given by the function $ C(x) = 15x + 40,000 $ . Find the profit on sales of x units.

\[ Ans.\quad P(x)=-x^{2}/100+45x-40,000 \]

13.23. In the previous problem, suppose that the demand x and the price p (in dollars) for the product are related by the function $ x = f(p) = 5000 - 50p $ $ 0 p $ . Write the profit as a function of the demand p.

\[ Ans.\quad F(p)=-(5000-50p)^{2}/100+45(5000-50p)-40,000 \]

13.24. In the previous problem, find the price which would yield the maximum profit and also find this maximum profit.

Ans. Price of $55 yields a maximum profit of $10,625.

13.25. A 300-foot-long cable, originally of diameter 5 inches, is submerged in seawater. Because of corrosion, the surface area of the cable diminishes at the rate of $ 1250 , in^{2}/year $ . Express the diameter d of the cable as a function of time t (in years).

Ans. $ d = 5 - $ inches

13.26. Show that every decreasing function is one-to-one on its domain.

13.27. A function is periodic if there exists some nonzero real number p, called a period, such that $ f(x + p) = f(x) $ for all x in the domain of the function. Show that no periodic function is one-to-one.

13.28. Show that the graphs of $ f^{-1} $ and f are reflections of each other in the line y = x by verifying the following: (a) If $ P(u,v) $ is on the graph of f, then $ Q(v,u) $ is on the graph of $ f^{-1} $ . (b) The midpoint of line segment PQ is on the line y = x. (c) The line PQ is perpendicular to the line y = x.

13.29. The following functions are one-to-one. Find the inverse functions for each.

\[ f(x)=5-10x \]

  1. $ f(x) = $

  2. $ f(x) = $

\[ \left(\mathbf{d}\right)f(x)=2-x^{3} \]

\[ (\mathrm{e})f(x)=\sqrt{9-x^{2}},0\leq x\leq3 \]

\[ (f)f(x)=3-\sqrt{x-2} \]

Ans. (a) $ f^{-1}(x) = $ ; (b) $ f^{-1}(x) = $ ; (c) $ f^{-1}(x) = $ ; (d) $ f^{-1}(x) = $ ;

\[ \left(\mathrm{e}\right)f^{-1}(x)=\sqrt{9-x^{2}},0\leq x\leq3;\\\left(\mathrm{f}\right)f^{-1}(x)=(3-x)^{2}+2,x\leq3 \]

13.30. The following functions are one-to-one. Find the inverse function for each.

\[ f(x)=2+\sqrt{4-x^{2}}\qquad0\leq x\leq2 \]

\[ f(x)=2+\sqrt{4-x^{2}}\qquad-2\leq x\leq0 \]

\[ (c)f(x)=2-\sqrt{4-x^{2}}\qquad0\leq x\leq2 \]

\[ \left(\mathrm{d}\right)f(x)=2-\sqrt{4-x^{2}}\qquad-2\leq x\leq0 \]

Ans. (a) $ f^{-1}(x) = $ $ 2 x $ ; (b) $ f^{-1}(x) = - $ $ 2 x $ ;

\[ \left(\mathrm{c}\right)f^{-1}(x)=\sqrt{4x-x^{2}}\qquad0\leq x\leq2;\\\left(\mathrm{d}\right)f^{-1}(x)=-\sqrt{4x-x^{2}}\qquad0\leq x\leq2 \]