Question 5 of 12

Question 5 (10 marks)

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 function definitions as stated, with stated domains/codomains.

Find. (a) Whether $f$ is well-defined as a function $\mathbb{Z}\to\mathbb{R}$. (b) Whether $f$ is injective. (c) Whether $f$ is surjective. (d) The composite $(g^{-1}\circ f)(x)$.

Approach. (a) check the square root is always defined and real-valued on the stated domain; (b) test whether $f(x_1)=f(x_2)\Rightarrow x_1=x_2$; (c) test whether every integer is hit; (d) recover $g$ from $(f+g)(x)=f(x)+g(x)$, invert it, then substitute $f(x)$.

1. (a) Is $f(n)=\sqrt{n^3+n^2}$ a function $\mathbb{Z}\to\mathbb{R}$? Factor $n^3+n^2=n^2(n+1)$. For $n=-1$: $n^2(n+1)=1\cdot 0=0\ge 0$, fine. But for e.g. $n=-2$: $n^2(n+1)=4\cdot(-1)=-4\lt 0$, so $\sqrt{-4}$ is not a real number — $f(-2)$ is undefined in $\mathbb{R}$. A function must assign exactly one output IN THE CODOMAIN to every input in the domain; since some integers give a negative radicand, $f$ fails to be defined for all of $\mathbb{Z}$. $$\boxed{\text{NOT a function }\mathbb{Z}\to\mathbb{R}\text{ (undefined at, e.g., } n=-2\text{)}}$$
2. (b) Is $f(x)=3x^3+1$ one-to-one? Suppose $f(x_1)=f(x_2)$: $3x_1^3+1=3x_2^3+1 \Rightarrow x_1^3=x_2^3$. Since $t\mapsto t^3$ is strictly increasing (hence injective) over all of $\mathbb{R}$, $x_1^3=x_2^3\Rightarrow x_1=x_2$. $$\boxed{f\text{ is one-to-one}}$$
3. (c) Is $f(x)=\lceil 1.5x\rceil$ onto $\mathbb{Z}$? Given any target integer $m$, choose $x=m/1.5=2m/3$; then $1.5x=m$ exactly, so $\lceil 1.5x\rceil=m$. Since a real preimage exists for every integer $m$: $$\boxed{f\text{ is onto}}$$
4. (d) Compute $(g^{-1}\circ f)(x)$. Since $(f+g)(x)=f(x)+g(x)$, recover $g$ by subtraction: $$g(x)=(f+g)(x)-f(x)=(x^3+x^2)-(x^2+1)=x^3-1$$ $g(x)=x^3-1$ is a bijection $\mathbb{R}\to\mathbb{R}$ (the cube is strictly increasing and takes every real value), so $g^{-1}$ exists. Solving $y=x^3-1$ for $x$ gives $g^{-1}(y)=\sqrt[3]{y+1}$. Then $$(g^{-1}\circ f)(x)=g^{-1}(x^2+1)=\sqrt[3]{(x^2+1)+1}$$ $$\boxed{(g^{-1}\circ f)(x)=\sqrt[3]{x^2+2}}$$ Check: $g\big(\sqrt[3]{x^2+2}\big)=(x^2+2)-1=x^2+1=f(x)$, as required.