Question 5 of 12

Question 5

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four short function questions; in (d), $f(x)=x^5$ and $(f+g)(x)=x^5+x^3$.

Find. (a) whether $f(n)=\sqrt{n^2-9}$ is a function $\mathbb{Z}\to\mathbb{R}$. (b) the range of $\sqrt{x^2+16}$ on $\mathbb{R}$. (c) whether $5x^5+5$ is one-to-one on $\mathbb{R}$. (d) $f^{-1}(x)$ and $(g\circ f)(x)$.

Approach. (a) a function $\mathbb{Z}\to\mathbb{R}$ must assign a real value to every integer, so test the radicand on all of $\mathbb{Z}$. (b) bound the radicand below. (c) apply the definition $f(x_1)=f(x_2)\Rightarrow x_1=x_2$. (d) subtract to get $g$, invert $f$, then compose.

1. a) Is $f(n)=\sqrt{n^2-9}$ a function from $\mathbb{Z}$ to $\mathbb{R}$? For $n\in\{-2,-1,0,1,2\}$ the radicand $n^2-9$ is negative (e.g. $f(0)=\sqrt{-9}$), so $f(n)$ is not a real number for those integers. A function from $\mathbb{Z}$ to $\mathbb{R}$ must give every element of the domain $\mathbb{Z}$ exactly one value in $\mathbb{R}$. $\boxed{\text{No}}$ — it fails at $n=0,\pm1,\pm2$. (It would be a function from $\{n\in\mathbb{Z}:|n|\ge3\}$ to $\mathbb{R}$.)
2. b) Range of $f(x)=\sqrt{x^2+16}$, $x\in\mathbb{R}$. $x^2\ge0$, so $x^2+16\ge16$ with equality at $x=0$, and $x^2+16\to\infty$ as $|x|\to\infty$; by continuity every value in between is attained. $\boxed{\text{Range}=[4,\infty)}$.
3. c) Is $f(x)=5x^5+5$ one-to-one on $\mathbb{R}$? Suppose $f(x_1)=f(x_2)$. Then $5x_1^5+5=5x_2^5+5\Rightarrow x_1^5=x_2^5$. Since $t\mapsto t^5$ is strictly increasing on $\mathbb{R}$ (odd power; derivative $5t^4\ge0$, zero only at the single point $t=0$), $x_1^5=x_2^5\Rightarrow x_1=x_2$. $\boxed{\text{Yes, one-to-one}}$ (indeed a bijection, with inverse $f^{-1}(y)=\sqrt[5]{(y-5)/5}$).
4. d) $f^{-1}(x)$ and $g\circ f(x)$. $g(x)=(f+g)(x)-f(x)=(x^5+x^3)-x^5=x^3$. $f(x)=x^5$ is a bijection of $\mathbb{R}$ (as in (c)), so its inverse is the real fifth root: $\boxed{f^{-1}(x)=\sqrt[5]{x}=x^{1/5}}$ (defined for all real $x$, including negatives, because 5 is odd). Composition: $(g\circ f)(x)=g(f(x))=g(x^5)=(x^5)^3=\boxed{x^{15}}$.