Function Composition and Inverses: practice solutions, Grade 12

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Practice solutions Grade 12 : Function Composition and Inverses — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

Sommaire

1 Evaluating composites ★★★

$g(3)=9-3=6$, so $(f\circ g)(3)=f(6)=17$.

$f(3)=11$, so $(g\circ f)(3)=g(11)=121-3=118$.

$f(-1)=3$, so $(f\circ f)(-1)=f(3)=11$.

Answers: $17$, $118$ and $11$.

2 Building composite formulas ★★★

$(f\circ g)(x)=f(4x)=4x-7$.

$(g\circ f)(x)=g(x-7)=4(x-7)=4x-28$.

They are not equal (for $x=1$ they give $-3$ and $-24$). Composition is not commutative.

3 A simple domain ★★★

$(f\circ g)(x)=\sqrt{x+6}$. We need $x+6\ge0$, so the domain is $[-6,\infty)$.

$(g\circ f)(x)=\sqrt{x}+6$. The inner function $\sqrt{x}$ requires $x\ge0$, so the domain is $[0,\infty)$.

4 Horizontal line test ★★★

$f$: a non-constant line, every horizontal line meets it once: one-to-one.

$g$: the line $y=9$ meets the graph at $x=\pm3$: not one-to-one.

$h$: increasing everywhere: one-to-one.

$k$: the line $y=2$ meets it at $x=\pm2$: not one-to-one.

$m$: each horizontal line $y=c$ with $c\neq0$ meets it once at $x=1/c$: one-to-one.

5 A linear inverse ★★★

Set $y=5x-8$. Then $5x=y+8$ and $x=\dfrac{y+8}{5}$. So $f^{-1}(x)=\dfrac{x+8}{5}$.

Check: $f^{-1}(7)=\dfrac{15}{5}=3$ and $f(3)=15-8=7$.

6 Even, odd or neither ★★★

$f(-x)=x^6-2x^2=f(x)$: even.

$g(-x)=-4x^3-x=-g(x)$: odd.

$h(1)=4$ and $h(-1)=1-3=-2$. Since $-2$ is neither $4$ nor $-4$, $h$ is neither.

7 Absolute values and pieces ★★★

(a) $x+4=9$ gives $x=5$; $x+4=-9$ gives $x=-13$. Solutions: $5$ and $-13$.

(b) $q(-2)=3(-2)-1=-7$. $q(2)=3(2)-1=5$ (the first rule, since $2\le2$). $q(4)=16-1=15$.

8 Payroll at a repair shop ★★★

$P(h(d))=22(6d)+15=132d+15$.

For $d=5$: $132(5)+15=675$. Check: $h(5)=30$ hours and $P(30)=660+15=675$.

The technician earns $675.

9 A hidden restriction ★★★

$(f\circ g)(x)=f(x+5)=\dfrac{1}{(x+5)-2}=\dfrac{1}{x+3}$.

The inner function $g$ accepts every real number. The outer function needs $g(x)\neq2$, that is $x+5\neq2$, so $x\neq-3$.

Domain: all real numbers except $-3$.

10 Inverse of a rational function ★★★

Set $y=\dfrac{4x-1}{x+3}$. Then $y(x+3)=4x-1$, so $xy+3y=4x-1$, $xy-4x=-1-3y$ and $x(y-4)=-(3y+1)$. Hence $x=\dfrac{3y+1}{4-y}$.

$f^{-1}(x)=\dfrac{3x+1}{4-x}$, with $x\neq4$ (the value $4$ is not in the range of $f$).

Check: $f(0)=-\dfrac13$ and $f^{-1}\!\left(-\dfrac13\right)=\dfrac{0}{4+1/3}=0$.

11 A cubic inverse ★★★

The cube function is increasing, so $f$ is increasing and therefore one-to-one.

Set $y=2x^3-16$. Then $x^3=\dfrac{y+16}{2}$ and $x=\sqrt[3]{\dfrac{y+16}{2}}$.

$f^{-1}(x)=\sqrt[3]{\dfrac{x+16}{2}}$, defined for all real numbers. Check: $f(2)=0$ and $f^{-1}(0)=\sqrt[3]{8}=2$.

12 Are they inverses? ★★★

$f(g(x))=2\cdot\dfrac{x-3}{2}+3=(x-3)+3=x$.

$g(f(x))=\dfrac{(2x+3)-3}{2}=\dfrac{2x}{2}=x$.

Both compositions give $x$, so $f$ and $g$ are inverses.

13 Reading a graph and its inverse ★★★

(a) Since $f(2)=4$, $f^{-1}(4)=2$. Since $f(-2)=-4$, $f^{-1}(-4)=-2$.

(b) Swap the coordinates: $(4,2)$ and $(-4,-2)$.

14 Describing a transformation ★★★

Shift right 3, reflect in the x-axis, shift up 5.

Vertex: $(3,5)$, and the graph opens downward.

y-intercept: $g(0)=-|-3|+5=2$.

x-intercepts: $|x-3|=5$, so $x=8$ or $x=-2$.

15 Symmetry proofs ★★★

$f(-x)=\dfrac{-x}{(-x)^2+1}=-\dfrac{x}{x^2+1}=-f(x)$: odd.

$g(-x)=|-x|+(-x)^2=|x|+x^2=g(x)$: even.

$h(1)=2$ and $h(-1)=-1+1=0$. Since $0\neq2$ and $0\neq-2$, $h$ is neither even nor odd.

16 Absolute value as pieces ★★★

(a) $3-2x\ge0$ when $x\le\dfrac32$. So $|3-2x|=3-2x$ if $x\le\dfrac32$, and $|3-2x|=2x-3$ if $x\gt\dfrac32$.

(b) $3-2x=7$ gives $x=-2$; $3-2x=-7$ gives $x=5$. Check: $|3+4|=7$ and $|3-10|=7$.

17 Converting temperatures ★★★

(a) Set $F=\dfrac95C+32$. Then $C=\dfrac59(F-32)$, so $F^{-1}(x)=\dfrac59(x-32)$.

(b) $F^{-1}(98.6)=\dfrac59(66.6)=37$, so $98.6\,^{\circ}\text{F}=37\,^{\circ}\text{C}$.

18 Domain traps with a root ★★★

(a) Take $g(x)=x^2-9$ (inner) and $f(x)=\sqrt{x}$ (outer).

(b) We need $x^2-9\ge0$, so $|x|\ge3$: domain $(-\infty,-3]\cup[3,\infty)$.

(c) $(g\circ f)(x)=(\sqrt{x})^2-9=x-9$, but $x$ must be in the domain of $f$, so the domain is $[0,\infty)$. The formula $x-9$ alone would suggest all real numbers: the inner function restricts the domain.

19 Both orders, both domains ★★★

$(f\circ g)(x)=\sqrt{4-x^2}$. We need $4-x^2\ge0$, so $-2\le x\le2$: domain $[-2,2]$.

$(g\circ f)(x)=(\sqrt{4-x})^2=4-x$. The inner function needs $4-x\ge0$, so $x\le4$: domain $(-\infty,4]$.

20 Inverse after restriction ★★★

(a) $f(x)=(x+3)^2-8$. For $x\ge-3$ the range is $[-8,\infty)$.

(b) $y=(x+3)^2-8$ with $x+3\ge0$ gives $x=-3+\sqrt{y+8}$. So $f^{-1}(x)=-3+\sqrt{x+8}$, with domain $[-8,\infty)$ and range $[-3,\infty)$.

(d) Then $x+3\le0$, so $x=-3-\sqrt{y+8}$ and $f^{-1}(x)=-3-\sqrt{x+8}$.

21 Inverse of a piecewise function ★★★

(a) For $x\le0$, $f(x)=x+4\le4$, range $(-\infty,4]$. For $x\gt0$, $f(x)=2x+6\gt6$, range $(6,\infty)$. Each piece is increasing and the ranges do not overlap, so $f$ is one-to-one.

(b) From $y=x+4$: $x=y-4$. From $y=2x+6$: $x=\dfrac{y-6}{2}$. So $f^{-1}(x)=x-4$ if $x\le4$, and $f^{-1}(x)=\dfrac{x-6}{2}$ if $x\gt6$. The domain is $(-\infty,4]\cup(6,\infty)$.