Exercises 1

This page allows you to practice some exercises on Properties of Functions. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 5.1 - Shifting Graphs - Section 5.2 - New Functions from Old - Section 5.3 - Inverse Functions

Question 1

If $f(x) = 3x - x^3$ and $g(x) = x^{3}$, compute the six expressions $(f + g)(x)$, $(f-g)(x)$, $(fg)(x)$, $(f/g)(x)$, $f(g(1))$ and $g(f(1))$.

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• $(f + g)(x) = f(x) + g(x) = 3x - x^{3} + x^{3} = 3x$.

• $(f - g)(x) = f(x) - g(c) = 3x - x^{3} - x^{3} = 3x - 2x^{3}$.

• $(fg)(x) = f(x) \cdot g(x) = (3x - x^{3})x^{3} = 3x^{4} - x^{6}$.

• $(f/g)(x) = f(x)/g(x) = (3x - x^{3})/x^{3} = 3x^{-2} - 1$.

• $f(g(1)) = f(1^{3}) = f(1) = 3 -1 = 2$.

• $g(f(1)) = g(3 \cdot 1 - 1^{3}) = g(2) = 2^{3} = 8$.

Question 2

Let $f(x) = 3x + 7$. Compute $f(f(x))$, and find the value $x^{\ast}$ at which $f(f(x^{\ast})) = 100$.

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If $f(x) = 3x + 7$, then $f(f(x)) = f(3x + 7) = 3(3x + 7) + 7 = 9x + 28$. The equality $f(f(x^{\ast})) = 100$ requires $9x^{\ast} + 28 = 100$. Thus $x^{\ast} = 8$.

Question 3

Compute $\ln (\ln e)$ and $(\ln e)^{2}$. What do you notice?

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$\ln(\ln e) = \ln 1 = 0$, while $(\ln e)^{2} = 1^{2} = 1$. This result illustrates how, if we define the function $f^2$ by $f^2(x) = (f(x))^{2}$, then in general, $f^{2}(x) \neq f(f(x))$.

Question 4

Demand $D$ as a function of price $P$ is given by $D = \frac{32}{5} - \frac{3}{10}P$. Solve the equation for $P$ and find the inverse function.

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Note that $\frac{3}{10}P = \frac{32}{5} - D$, thus $P = P(D) = \frac{10}{3} (\frac{32}{5} - D) = \frac{1}{3}(64 - 10D)$. This is also the inverse function.

Question 5

Find the domains, ranges, and inverses of the functions given by the following formulas.

(a) $y = -3x \qquad$(b) $y = 1/x \qquad$(c) $y = x^{3} \qquad$(d) $y = \sqrt{ \sqrt{x} - 2}$

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(a) The domain and range are both $\mathbb{R}$. The inverse function is given by $x = -y/3$.

(b) The domain and range are both $\mathbb{R}\setminus \{0\}$. The inverse function is given by $x = 1/y$.

(c) The domain and range are both $\mathbb{R}$. The inverse function is given by $x = y^{1/3}$.

(d) For outer square root function to be defined, $\sqrt{x} - 2 \geq 0$. Thus, the domain is $[4, \infty)$ with corresponding range of $[0, \infty)$. Now, we can find the inverse function in the following way:$y^2 = \sqrt{x} - 2 \implies \sqrt{x} = y^{2} +2 \implies x = (y^{2} + 2)^{2}$.

Question 6

Why does $f(x) = x^2$, for $x$ in $(-\infty, \infty)$, have no inverse function? Show that $f$ restricted to $[0, \infty)$ has an inverse, and find that inverse.

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$f(x)$ is not one-to-one in $(-\infty, \infty)$, as $f(x) = f(-x)$ for $x \in \mathbb{R}$. Therefore, the function has no inverse. If we restrict $f$ to $[0,\infty)$, the function $f$ is strictly increasing, and therefore has the inverse $f^{-1} (x) = \sqrt{x}$. Note how this inverse function is indeed not defined for $x < 0$ (!).

Question 7

Suppose $Q = f(C)$ is the function that tells you how many kilogram of carrots $Q$ you can buy for a specified amount of money $C$. What does the inverse function $f^{-1}$ tell you?

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The inverse function $f^{-1} (Q)$ determines the costs of $Q$ kilograms of carrots.

Question 8

Find inverses of the following functions.

(a) $f(x) = (x^{3} - 1)^{1/3} \qquad$(b) $f(x) = \frac{x+1}{x-2} \qquad$(c) $f(x) = (1-x^{3})^{1/5} + 2$

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(a) Denote $f(x) = y$, then $y^{3} = (x^{3} - 1)$. Now, $x = (y^{3} + 1)^{1/3} = f^{-1}(y)$.

(b) Denote $f(x) = y$, then $y(x-2) = x+1$. Now, $xy - 2y = x+1$ implies $xy - x = 2y +1$. We get $x(y-1) = 2y + 1$ yielding $x = \frac{2y + 1}{y-1} = f^{-1}(y)$.

(c) Denote $f(x) = y$, then $(y - 2)^{5} = (1 - x^{3})$. Thus, $x = (1 - (y-2)^{5})^{1/3} = f^{-1}(y)$.