Exercises 1

This page allows you to practice some exercises on Functions of One Variable. 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 4.1 - Introduction - Section 4.2 - Definitions - Section 4.3 - Graphs of Functions

Question 1

Let $f(x) = x^2 + 1$. For what values of $x$ is it true that (a) $f(x) = f(-x)$

(b) $f(x + 1) = f(x) + f(1)$

(c) $f(2x) = 2f(x)$

Show answer

(a) For any $x \in \mathbb{R}$ we have that $-x^2=x^2$. Hence, $f(x) = x^2 + 1 = (-x)^2 + 1 = f(-x)$ for $x \in \mathbb{R}$.

(b) Notice that $f(1) = 1^2 + 1 = 2$. Hence, $f(x) + f(1) = x^2 + 3$. Also notice that $f(x+1) = (x+1)^2 + 1 = x^2 + 2x + 2$. Thus, $x^2 + 2x + 2 = x^2 + 3 \Leftrightarrow 2x+2 = 3 \Leftrightarrow 2x = 1 \Leftrightarrow x = \frac{1}{2}$

(c) Notice that $f(2x) = (2x)^2 + 1 = 4x^2 + 1$ and $2f(x) = 2(x^2 + 1) = 2x^2 + 2$. Thus, $f(2x)=2f(x) \Leftrightarrow 4x^2 + 1 = 2x^2 + 2 \Leftrightarrow 2x^2 = 1 \Leftrightarrow x^2 = \frac{1}{2} \Leftrightarrow x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

Remark relevant for (c): for fractions of the form $\frac{a}{b \sqrt{c}}$, for integers $a$ and $b$ and natural number $c$ a widely used convention is to rewrite such a fraction as $\frac{a \sqrt{c}}{b c}$. Please double-check if you understand why these two forms are equivalent.

Question 2

Let $f(x) = x/(1+x^2)$.

(a) Show that $f(-x) = -f(x)$ for all $x$.

(b) Show that $f(1/x) = f(x)$ for $x \neq 0$.

Show answer

(a) First, notice, by substitution (i.e., replacing $x$ by transformed input $-x$) that $f(-x) = -x / (1 + (-x)^2)$. Also, from Question 1.a, recall that $(-x)^2 = x^2$. Hence, $f(-x) = - x / (1+ x^2) = -f(x)$.

(b) Again using substitution (now replacing $x$ by transformed input $1/x$), we have that $f(1/x) = (1/x)/(1+(1/x)^2)$. Evaluating $f(1/x)$ for $x=0$ is fairly tricky (do you see why?), but as indicated in the question, we may ignore that case. For $x \neq 0$, $f(1/x)$ is easy to evaluate. By multiplying numerator and denominator in our expression for $f(1/x)$ by $x^2$, we find that $f(1/x) = (x^2/x )/(x^2 + x^2 \cdot (1/x)^2) = x / (x^2 + 1) = x/(1+x^2) =f(x)$.

Question 3

Show the following results. (a) If $f(x) = 100x^2$, then $f(tx) = t^2f(x)$ for all $t$.

(b) If $P(x) = x^{1/2}$, then $P(tx) = t^{1/2}P(x)$ for all $t \geq 0$.

Show answer

(a) $f(tx) = 100 (tx)^2 = 100 t^2 x^2 = t^2 100 x^2 = t^2 f(x)$.

(b) $P(tx) = (tx)^{1/2} = t^{1/2} x^{1/2} = t^{1/2} P(x)$.

Question 4

Show the following results.

(a) If $f(x) = Ax$, show that $f(a + b) = f(a) + f(b)$ for all numbers $a$ and $b$.

(b) If $f(x) = 10^{x}$, show that $f(a+b) = f(a) \cdot f(b)$ for all natural numbers $a$ and $b$.

Show answer

(a) Notice that $f(a)=Aa$ (warning! $A$ and $a$ do not have to be the same numbers; uppercase and lowercase letters matter in math) and $f(b)=Ab$. Thus, $f(a) + f(b) = Aa + Ab = A(a+b)$. In addition, again using substitution, notice that $f(a+b) = A (a + b)$. Hence, we conclude $f(a+b) = f(a) + f(b)$

(b) By substitution, notice that $f(a+b) = 10^{a+b} = 10^a \cdot 10^b$. Moreover, since $f(a)=10^a$ and $f(b)=10^b$, it follows that $f(a) \cdot f(b) = 10^a \cdot 10^b = f( a + b)$. Since this holds true for all $a,b \in \mathbb{R}$, surely this also holds true for all $a,b \in \mathbb{N}$, since $\mathbb{N} \subset \mathbb{R}$.

Question 5

Find the domains of the following functions.

(a) $y = \sqrt{5-x}$

(b) $y = (2x - 1)/(x^2 - x)$

(c) $y = \sqrt{\frac{x-1}{(x-2)(x+3)}}$

Show answer

(a) In real analysis (i.e., only considering numbers in $\mathbb{R}$), the square root is only defined for nonnegative numbers. Thus, for $y$ to be defined, $5-x$ has to be nonnegative. To write down this requirement mathematically: $5-x \geq 0$. Adding $x$ to both sides of this inequality yields $x \leq 5$. In other words, the domain $D$ of this function is given by the set of all real numbers less than or equal to five. We can write this down in set-builder notation as well as interval notation as follows:

1. $D = \{ x \in \mathbb{R} : x \leq 5\}$ and

2. $D = ( - \infty, 5]$.

(b) Both the numerator and denominator yield real-valued numbers of any $x \in \mathbb{R}$ (verify yourself!). Thus, the only cases where $y$ may not be a real-valued number is when the denominator equals zero. More specifically, any value for $x$ s.t. that numerator is nonzero while the denominator is zero cannot be part of the domain for $y$. Cases where both numerator and denominator are zero require more attention (using the so-called L'Hôpital's rule, which you will learn about later).

For now, we look for $x$ s.t. $x^2 - x=0$. Factoring out $x$, we have that $x^2 - x = x(x-1)$. Hence, $x = 0 \lor x = 1$. Interestingly, for those two values, the numerator is nonzero. Hence, we can immediately conclude that $y$ is not defined for those values of $x$. Thus, using interval notation, we conclude that the domain is given by $D=(-\infty, 0) \cup (0,1) \cup (1, \infty)$.

Question 6

Find the domain and the range of $g(x) = 1 - \sqrt{x+2}$.

Show answer

Using logic that is analogous to that used in Question 5a, we have that the domain is given by $D = [-2, \infty)$. Also notice that $x+2$ increases continuously with $x$. Similarly, $\sqrt{x+2}$ increases continuously with $x$. Hence, $g(x) = 1 - \sqrt{x+2}$ decreases continuously with $x$. Therefore, $g(x)$ attains its highest value for $x=-2$. At that point, we have $g(-2)=1$. Moreover, as $x \rightarrow \infty$, we have that $g(x) \rightarrow - \infty$ (check yourself!). Consequently, the range is given by $R = [1,\infty)$.