1.5E: Exercises

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Exercises

In Exercises 1 - 10, use the pair of functions $\ f$ and $\ g$ to find the following values if they exist.

$\ (f + g)(2)$
$\ (f − g)(−1) $
$\ (g − f)(1)$
$\ (f g)\left(\frac{1}{2}\right)$
$\ \left(\frac{f}{g}\right)(0)$
$\ \left(\frac{g}{f}\right)(-2)$

$\ f(x) = 3x + 1$ and $\ g(x) = 4 − x$
$\ f(x)=x^{2}$ and $\ g(x) = −2x + 1$
$\ f(x)=x^{2}-x$ and $\ g(x)=12-x^{2}$
$\ f(x)=2 x^{3}$ and $\ g(x)=-x^{2}-2 x-3$
$\ f(x)=\sqrt{x+3}$ and $\ g(x) = 2x − 1$
$\ f(x)=\sqrt{4-x}$ and $\ g(x)=\sqrt{x+2}$
$\ f(x)=2 x$ and $\ g(x)=\frac{1}{2 x+1}$
$\ f(x)=x^{2}$ and $\ g(x)=\frac{3}{2 x-3}$
$\ f(x)=x^{2}$ and $\ g(x)=\frac{1}{x^{2}}$
$\ f(x)=x^{2}+1$ and $\ g(x)=\frac{1}{x^{2}+1}$

In Exercises 11 - 20, use the pair of functions $\ f$ and $\ g$ to find the domain of the indicated function then find and simplify an expression for it.

$\ (f + g)(x)$
$\ (f − g)(x)$
$\ (fg)(x)$
$\ \left(\frac{f}{g}\right)(x)$

$\ f(x) = 2x + 1$ and $\ g(x) = x − 2$
$\ f(x) = 1 − 4x$ and $\ g(x) = 2x − 1$
$\ f(x)=x^{2}$ and $\ g(x) = 3x − 1$
$\ f(x)=x^{2}-x$ and $\ g(x) = 7x$
$\ f(x)=x^{2}-4$ and $\ g(x) = 3x + 6$
$\ f(x)=-x^{2}+x+6$ and $\ g(x)=x^{2}-9$
$\ f(x)=\frac{x}{2}$ and $\ g(x)=\frac{2}{x}$
$\ f(x) = x − 1$ and $\ g(x)=\frac{1}{x-1}$
$\ f(x) = x$ and $\ g(x)=\sqrt{x+1}$
$\ f(x)=\sqrt{x-5}$ and $\ g(x)=f(x)=\sqrt{x-5}$

In Exercises 21 - 45, find and simplify the difference quotient $\ \frac{f(x+h)-f(x)}{h}$ for the given function.

$\ f(x) = 2x − 5$
[MOM] $\ f(x) = −3x + 5$
[MOM] $\ f(x) = 6$
[MOM] $\ f(x)=3 x^{2}-x$
[MOM] $\ f(x)=-x^{2}+2 x-1$
$\ f(x)=4 x^{2}$
$\ f(x)=x-x^{2}$
$\ f(x)=x^{3}+1$
$\ f(x) = mx + b$ where $\ m \neq 0$
$\ f(x)=a x^{2}+b x+c$ where $\ a \neq 0$
$\ f(x)=\frac{2}{x}$
[MOM] $\ f(x)=\frac{3}{1-x}$
$\ f(x)=\frac{1}{x^{2}}$
[MOM] $\ f(x)=\frac{2}{x+5}$
$\ f(x)=\frac{1}{4 x-3}$
[MOM] $\ f(x)=\frac{3 x}{x+1}$
$\ f(x)=\frac{x}{x-9}$
$\ f(x)=\frac{x^{2}}{2 x+1}$
[MOM] $\ f(x)=\sqrt{x-9}$
$\ f(x)=\sqrt{2 x+1}$
$\ f(x)=\sqrt{-4 x+5}$
$\ f(x)=\sqrt{4-x}$
$\ f(x)=\sqrt{a x+b}$, where $\ a \neq 0$.
$\ f(x)=x \sqrt{x}$
$\ f(x)=\sqrt[3]{x}$. HINT: $\ (a-b)\left(a^{2}+a b+b^{2}\right)=a^{3}-b^{3}$

In Exercises 46 - 50, $\ C(x)$ denotes the cost to produce $\ x$ items and $\ p(x)$ denotes the price-demand function in the given economic scenario. In each Exercise, do the following

Find and interpret $\ C(0)$.
Find and interpret $\ \overline{C}(10)$.
Find and interpret $\ p(5)$.
Find and simplify $\ R(x)$.
Find and simplify $\ P(x)$.
Solve $\ P(x) = 0$ and interpret.

The cost, in dollars, to produce $\ x$ “I’d rather be a Sasquatch” T-Shirts is $\ C(x) = 2x + 26$, $\ x ≥ 0$ and the price-demand function, in dollars per shirt, is $\ p(x) = 30 − 2x$, $\ 0 ≤ x ≤ 15$.
The cost, in dollars, to produce $\ x$ bottles of 100% All-Natural Certified Free-Trade Organic Sasquatch Tonic is $\ C(x) = 10x + 100$, $\ x ≥ 0$ and the price-demand function, in dollars per bottle, is $\ p(x) = 35 − x, 0 ≤ x ≤ 35$.
The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is $\ C(x) = 18x + 240$, $\ x ≥ 0$ and the price-demand function, in cents per cup, is $\ p(x) = 90 − 3x, 0 ≤ x ≤ 30$.
The daily cost, in dollars, to produce $\ x$ Sasquatch Berry Pies $\ C(x) = 3x + 36$, $\ x ≥ 0$ and the price-demand function, in dollars per pie, is $\ p(x) = 12 − 0.5x$, $\ 0 ≤ x ≤ 24$.
The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is $\ C(x) = 20x + 1000$, $\ x ≥ 0$ and the price-demand function, in hundreds of dollars per scooter, is $\ p(x) = 140 − 2x$, $\ 0 ≤ x ≤ 70$.

In Exercises 51 - 62, let f be the function defined by

$\ f=\{(-3,4),(-2,2),(-1,0),(0,1),(1,3),(2,4),(3,-1)\}$

and let $\ g$ be the function defined

$\ g=\{(-3,-2),(-2,0),(-1,-4),(0,0),(1,-3),(2,1),(3,2)\}$

Compute the indicated value if it exists.

$\ (f + g)(−3)$
$\ (f − g)(2)$
$\ (fg)(−1)$
$\ (g + f)(1)$
$\ (g − f)(3)$
$\ (gf)(−3)$
$\ \left(\frac{f}{g}\right)(-2)$
$\ \left(\frac{f}{g}\right)(-1)$
$\ \left(\frac{f}{g}\right)(2)$
$\ \left(\frac{g}{f}\right)(-1)$
$\ \left(\frac{g}{f}\right)(3)$
$\ \left(\frac{g}{f}\right)(-3)$

Answers

For $\ f(x) = 3x + 1$ and $\ g(x) = 4 − x$
- $\ (f + g)(2) = 9$
- $\ (f − g)(−1) = −7$
- $\ (g − f)(1) = −1$
- $\ (f g)\left(\frac{1}{2}\right)=\frac{35}{4}$
- $\ \left(\frac{f}{g}\right)(0)=\frac{1}{4}$
- $\ \left(\frac{g}{f}\right)(-2)=-\frac{6}{5}$
For $\ f(x)=x^{2}$ and $\ g(x)=-2 x+1$
- $\ (f + g)(2) = 1$
- $\ (f − g)(−1) = − 2$
- $\ (g − f)(1) = − 2$
- $\ (f g)\left(\frac{1}{2}\right)=0$
- $\ \left(\frac{f}{g}\right)(0)=0$
- $\ \left(\frac{g}{f}\right)(-2)=\frac{5}{4}$
For $\ f(x)=x^{2}-x$ and $\ g(x)=12-x^{2}$
- $\ (f + g)(2) = 10$
- $\ (f − g)(−1) = − 9$
- $\ (g − f)(1) = 11$
- $\ (f g)\left(\frac{1}{2}\right)=-\frac{47}{16}$
- $\ \left(\frac{f}{g}\right)(0)=0$
- $\ \left(\frac{g}{f}\right)(-2)=\frac{4}{3}$
For $\ f(x)=2 x^{3}$ and $\ g(x)=-x^{2}-2 x-3$
- $\ (f + g)(2) = 5$
- $\ (f − g)(−1) = 0$
- $\ (g − f)(1) = − 8$
- $\ (f g)\left(\frac{1}{2}\right)=-\frac{17}{16}$
- $\ \left(\frac{f}{g}\right)(0)=0$
- $\ \left(\frac{g}{f}\right)(-2)=\frac{3}{16}$
For $\ f(x)=\sqrt{x+3}$ and $\ g(x) = 2x - 1$
- $\ (f+g)(2)=3+\sqrt{5}$
- $\ (f-g)(-1)=3+\sqrt{2}$
- $\ (g − f)(1) = -1$
- $\ (f g)\left(\frac{1}{2}\right)=0$
- $\ \left(\frac{f}{g}\right)(0)=-\sqrt{3}$
- $\ \left(\frac{g}{f}\right)(-2)=-5$
For $\ f(x)=\sqrt{4-x}$ and $\ g(x)=\sqrt{x+2}$
- $\ (f+g)(2)=2+\sqrt{2}$
- $\ (f-g)(-1)=-1+\sqrt{5}$
- $\ (g − f)(1) = 0$
- $\ (f g)\left(\frac{1}{2}\right)=\frac{\sqrt{35}}{2}$
- $\ \left(\frac{f}{g}\right)(0)=\sqrt{2}$
- $\ \left(\frac{g}{f}\right)(-2)=0$
For $\ f(x) = 2x$ and $\ g(x)=\frac{1}{2 x+1}$
- $\ (f+g)(2)=\frac{21}{5}$
- $\ (f − g)(−1) = −1$
- $\ (g-f)(1)=-\frac{5}{3}$
- $\ (f g)\left(\frac{1}{2}\right)=\frac{1}{2}$
- $\ \left(\frac{f}{g}\right)(0)=0$
- $\ \left(\frac{g}{f}\right)(-2)=\frac{1}{12}$
For $\ f(x)=x^{2}$ and $\ g(x)=\frac{3}{2 x-3}$
- $\ (f + g)(2) = 7$
- $\ (f-g)(-1)=\frac{8}{5}$
- $\ (g − f)(1) = −4$
- $\ (f g)\left(\frac{1}{2}\right)=-\frac{3}{8}$
- $\ \left(\frac{f}{g}\right)(0)=0$
- $\ \left(\frac{g}{f}\right)(-2)=-\frac{3}{28}$
For $\ f(x)=x^{2}$ and $\ g(x)=\frac{1}{x^{2}}$
- $\ (f+g)(2)=\frac{17}{4}$
- $\ (f − g)(−1) = 0$
- $\ (g − f)(1) = 0$
- $\ (f g)\left(\frac{1}{2}\right)=1$
- $\ \left(\frac{f}{g}\right)(0)$ is undefined.
- $\ \left(\frac{g}{f}\right)(-2)=\frac{1}{16}$
For $\ f(x)=x^{2}+1$ and $\ g(x)=\frac{1}{x^{2}+1}$
- $\ (f+g)(2)=\frac{26}{5}$
- $\ (f-g)(-1)=\frac{3}{2}$
- $\ (g-f)(1)=-\frac{3}{2}$
- $\ (f g)\left(\frac{1}{2}\right)=1$
- $\ \left(\frac{f}{g}\right)(0)=1$
- $\ \left(\frac{g}{f}\right)(-2)=\frac{1}{25}$
For For $\ f(x) = 2x + 1$ and $\ g(x) = x - 2$
- $\ (f + g)(x) = 3x - 1$
  Domain: $\ (-\infty, \infty)$
- ˆ$\ (f − g)(x) = x + 3$
  Domain: $\ (-\infty, \infty)$
- $\ (f g)(x)=2 x^{2}-3 x-2$
  Domain: $\ (-\infty, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{2 x+1}{x-2}$
  Domain: $\ (-\infty, 2) \cup(2, \infty)$
For $\ f(x) = 1 − 4x$ and $\ g(x) = 2x - 1$
- ˆ (f + g)(x) = -2x
  Domain: $\ (-\infty, \infty)$
- ˆ $\ (f − g)(x) = 2 - 6x$
  Domain: $\ (-\infty, \infty)$
- $\ (f g)(x)=-8 x^{2}+6 x-1$
  Domain: $\ (-\infty, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{1-4 x}{2 x-1}$
  Domain: $\ \left(-\infty, \frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)$
For $\ f(x)=x^{2}$ and $\ g(x) = 3x − 1$
- $\ (f+g)(x)=x^{2}+3 x-1$
  Domain: $\ (-\infty, \infty)$
- $\ (f-g)(x)=x^{2}-3 x+1$
  Domain: $\ (-\infty, \infty)$
- $\ (f g)(x)=3 x^{3}-x^{2}$
  Domain: $\ (-\infty, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{x^{2}}{3 x-1}$
  Domain: $\ \left(-\infty, \frac{1}{3}\right) \cup\left(\frac{1}{3}, \infty\right)$
For $\ f(x)=x^{2}-x$ and $\ g(x) = 7x$
- $\ (f+g)(x)=x^{2}+6 x$
  Domain: $\ (-\infty, \infty)$
- $\ (f-g)(x)=x^{2}-8 x$
  Domain: $\ (-\infty, \infty)$
- $\ (f g)(x)=7 x^{3}-7 x^{2}$
  Domain: $\ (-\infty, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{x-1}{7}$
  Domain: $\ (-\infty, 0) \cup(0, \infty)$
For $\ f(x)=x^{2}-4$ and $\ g(x) = 3x + 6$
- $\ (f+g)(x)=x^{2}+3 x+2$
  Domain: $\ (-\infty, \infty)$
- $\ (f-g)(x)=x^{2}-3 x-10$
  Domain: $\ (-\infty, \infty)$
- $\ (f g)(x)=3 x^{3}+6 x^{2}-12 x-24$
  Domain: $\ (-\infty, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{x-2}{3}$
  Domain: $\ (-\infty,-2) \cup(-2, \infty)$
For $\ f(x)=-x^{2}+x+6$ and $\ g(x)=x^{2}-9$
- ˆ$\ (f + g)(x) = x − 3$
  Domain: $\ (-\infty, \infty)$
- $\ (f-g)(x)=-2 x^{2}+x+15$
  Domain: $\ (-\infty, \infty)$
- $\ (f g)(x)=-x^{4}+x^{3}+15 x^{2}-9 x-54$
  Domain: $\ (-\infty, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=-\frac{x+2}{x+3}$
  Domain: $\ (-\infty,-3) \cup(-3,3) \cup(3, \infty)$
For $\ f(x)=\frac{x}{2}$ and $\ g(x)=\frac{2}{x}$
- $\ (f+g)(x)=\frac{x^{2}+4}{2 x}$
  Domain: $\ (-\infty, 0) \cup(0, \infty)$
- $\ (f-g)(x)=\frac{x^{2}-4}{2 x}$
  Domain: $\ (-\infty, 0) \cup(0, \infty)$
- ˆ$\ (fg)(x) = 1$
  Domain: $\ (-\infty, 0) \cup(0, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{x^{2}}{4}$
  Domain: $\ (-\infty, 0) \cup(0, \infty)$
For $\ f(x) = x − 1$ and $\ g(x)=\frac{1}{x-1}$
- $\ (f+g)(x)=\frac{x^{2}-2 x+2}{x-1}$
  Domain: $\ (-\infty, 1) \cup(1, \infty)$
- $\ (f-g)(x)=\frac{x^{2}-2 x}{x-1}$
  Domain: $\ (-\infty, 1) \cup(1, \infty)$
- $\ (fg)(x) = 1$
  Domain: $\ (-\infty, 1) \cup(1, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=x^{2}-2 x+1$
  Domain: $\ (-\infty, 1) \cup(1, \infty)$
For $\ f(x) = x$ and $\ g(x)=\sqrt{x+1}$
- $\ (f+g)(x)=x+\sqrt{x+1}$
  Domain: $\ [-1, \infty)$
- $\ (f-g)(x)=x-\sqrt{x+1}$
  Domain: $\ [-1, \infty)$
- $\ (f g)(x)=x \sqrt{x+1}$
  Domain: $\ [-1, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=\frac{x}{\sqrt{x+1}}$
  Domain: $\ (-1, \infty)$
For $\ f(x)=\sqrt{x-5}$ and $\ g(x)=f(x)=\sqrt{x-5}$
- $\ (f+g)(x)=2 \sqrt{x-5}$
  Domain: $\ [5, \infty)$
- (f − g)(x) = 0
  Domain: $\ [5, \infty)$
- $\ (fg)(x) = x − 5$
  Domain: $\ [5, \infty)$
- $\ \left(\frac{f}{g}\right)(x)=1$
  Domain: $\ (5, \infty)$
2
[MOM]
[MOM]
[MOM]
[MOM]
$\ 8x + 4h$
$\ −2x − h + 1$
$\ 3 x^{2}+3 x h+h^{2}$
$\ m$
$\ 2ax + ah + b$
$\ \frac{-2}{x(x+h)}$
[MOM]
$\ \frac{-(2 x+h)}{x^{2}(x+h)^{2}}$
[MOM]
$\ \frac{-4}{(4 x-3)(4 x+4 h-3)}$
[MOM]
$\ \frac{-9}{(x-9)(x+h-9)}$
$\ \frac{2 x^{2}+2 x h+2 x+h}{(2 x+1)(2 x+2 h+1)}$
[MOM]
$\ \frac{2}{\sqrt{2 x+2 h+1}+\sqrt{2 x+1}}$
$\ \frac{-4}{\sqrt{-4 x-4 h+5}+\sqrt{-4 x+5}}$
$\ \frac{-1}{\sqrt{4-x-h}+\sqrt{4-x}}$
$\ \frac{a}{\sqrt{a x+a h+b}+\sqrt{a x+b}}$
$\ \frac{3 x^{2}+3 x h+h^{2}}{(x+h)^{3 / 2}+x^{3 / 2}}$
$\ \frac{1}{(x+h)^{2 / 3}+(x+h)^{1 / 3} x^{1 / 3}+x^{2 / 3}}$
- $\ C(0)=26$, so the fixed costs are $26.
- $\ \overline{C}(10)=4.6$, so when 10 shirts are produced, the cost per shirt is $4.60.
- $\ p(5) = 20$, so to sell 5 shirts, set the price at $20 per shirt.
- $\ R(x)=-2 x^{2}+30 x$, $\ 0 ≤ x ≤ 15$
- $\ P(x)=-2 x^{2}+28 x-26$, $\ 0 ≤ x ≤ 15$
- $\ P(x) = 0$ when $\ x = 1$ and $\ x = 13$. These are the ‘break even’ points, so selling 1 shirt or 13 shirts will guarantee the revenue earned exactly recoups the cost of production.
- $\ C(0)=100$, so the fixed costs are $100.
- $\ \overline{C}(10)=20$, so when 10 bottles of tonic are produced, the cost per bottle is $20.
- $\ p(5) = 30$, so to sell 5 bottles of tonic, set the price at $30 per bottle.
- $\ R(x)=-x^{2}+35 x, 0 \leq x \leq 35$
- $\ P(x)=-x^{2}+25 x-100,0 \leq x \leq 35$
- $\ P(x)=0$, when $\ x = 5$ and $\ x = 20$. These are the ‘break even’ points, so selling 5 bottles of tonic or 20 bottles of tonic will guarantee the revenue earned exactly recoups the cost of production.
- $\ C(0)=240$, so the fixed costs are 240¢ or $2.40.
- $\ \overline{C}(10)=42$, so when 10 cups of lemonade are made, the cost per cup is 42¢.
- $\ p(5) = 75$, so to sell 5 cups of lemonade, set the price at 75¢ per cup.
- $\ R(x)=-3 x^{2}+90 x, 0 \leq x \leq 30$
- $\ P(x)=-3 x^{2}+72 x-240,0 \leq x \leq 30$
- $\ P(x) = 0$ when $\ x = 4$ and $\ x = 20$. These are the ‘break even’ points, so selling 4 cups of lemonade or 20 cups of lemonade will guarantee the revenue earned exactly recoups the cost of production.
- $\ C(0) = 36$, so the daily fixed costs are $36.
- $\ \overline{C}(10)=6.6$, so when 10 pies are made, the cost per pie is $6.60.
- $\ p(5) = 9.5$, so to sell 5 pies a day, set the price at $9.50 per pie.
- $\ R(x)=-0.5 x^{2}+12 x, 0 \leq x \leq 24$
- $\ P(x)=-0.5 x^{2}+9 x-36,0 \leq x \leq 24$
- $\ P(x) = 0$ when $\ x = 6$ and $\ x = 12$. These are the ‘break even’ points, so selling 6 pies or 12 pies a day will guarantee the revenue earned exactly recoups the cost of production.
- $\ C(0) = 1000$, so the monthly fixed costs are 1000 hundred dollars, or $100,000.
- $\ \overline{C}(10)=120$, so when 10 scooters are made, the cost per scooter is 120 hundred dollars, or $12,000.
- $\ p(5) = 130$, so to sell 5 scooters a month, set the price at 130 hundred dollars, or $13,000 per scooter.
- $\ R(x)=-2 x^{2}+140 x, 0 \leq x \leq 70$
- $\ P(x)=-2 x^{2}+120 x-1000,0 \leq x \leq 70$
- P(x) = 0 when x = 10 and x = 50. These are the ‘break even’ points, so selling 10 scooters or 50 scooters a month will guarantee the revenue earned exactly recoups the cost of production.
$\ (f + g)(−3) = 2$
$\ (f − g)(2) = 3$
$\ (fg)(−1) = 0$
$\ (g + f)(1) = 0$
$\ (g − f)(3) = 3$
$\ (gf)(−3) = −8$
$\ \left(\frac{f}{g}\right)(-2)$ does not exist
$\ \left(\frac{f}{g}\right)(-1)=0$
$\ \left(\frac{f}{g}\right)(2)=4$
$\ \left(\frac{g}{f}\right)(-1)$ does not exist
$\ \left(\frac{g}{f}\right)(3)=-2$
$\ \left(\frac{g}{f}\right)(-3)=-\frac{1}{2}$

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