Pure Mathematics Practice: Functions — Answers

Questions: Functions Drills.

Question 1

(a)

$$\frac{2x+1}{x-3}=\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}.$$

For $x>3$, $x-3>0$, so $7/(x-3)>0$. As $x\to 3^{+}$, $f(x)\to \infty$, while as $x\to \infty$, $f(x)\to 2$ from above. Hence

$$R_f=(2,\infty ).$$

(b) On $x>3$, the term $7/(x-3)$ decreases strictly as $x$ increases. Therefore $f$ is strictly decreasing and hence one-one on its stated domain.

(c) Let $y=(2x+1)/(x-3)$. Then

$$y(x-3)=2x+1,$$

so

$$x(y-2)=3y+1⟹x=\frac{3y+1}{y-2}.$$

Thus

$$f^{-1}(x)=\frac{3x+1}{x-2},D_{f^{-1}}=(2,\infty ).$$

(d)

$$x=f^{-1}(9)=\frac{3(9)+1}{9-2}=4.$$

Question 2

(a) Since $x+2$ runs from $0$ to $9$,

$$R_f=[0,3].$$

(b) Since $R_f=[0,3]=D_g$, $gf$ is defined for every $x\in [-2,7]$. Moreover,

$$gf(x)=4-(\sqrt{x+2}{)}^2=2-x.$$

(c) For $fg(x)$ to exist, $x\in D_g=[0,3]$ and $g(x)\in D_f=[-2,7]$. Thus

$$-2\le 4-x^2\le 7.$$

On $0\le x\le 3$, the upper inequality is automatic, while the lower inequality gives $x^2\le 6$. Hence the largest domain is

$$[0,\sqrt{6}].$$

(d) On this domain,

$$fg(x)=\sqrt{(4-x^2)+2}=\sqrt{6-x^2}.$$

This decreases continuously from $\sqrt{6}$ to $0$, so

$$R_{fg}=[0,\sqrt{6}].$$

Question 3

(a) The least value is $-2$ at $x=1$, and the greatest endpoint value is $7$ at $x=4$. Hence

$$R_h=[-2,7].$$

(b) The graph fails the horizontal-line test. For example,

$$h(-1)=h(3)=2,$$

although $-1\ne 3$. Thus $h$ is not one-one and has no inverse function on the full domain.

(c) For $k$, $1\le x\le 4$. Let

$$y=(x-1{)}^2-2.$$

Since $x\ge 1$,

$$x-1=\sqrt{y+2}.$$

Therefore

$$k^{-1}(x)=1+\sqrt{x+2},D_{k^{-1}}=[-2,7].$$

(d) The range of $k^{-1}$ is the domain of $k$:

$$[1,4].$$

Question 4

(a)

$$p_a(x)=(x-a{)}^2+3-a^2.$$

(b) If $x_1,x_2\ge a$ and $p_a(x_1)=p_a(x_2)$, then

$$(x_1-a{)}^2=(x_2-a{)}^2.$$

Both $x_1-a$ and $x_2-a$ are non-negative, so $x_1-a=x_2-a$ and hence $x_1=x_2$. Therefore $p_a$ is one-one.

(c) The minimum occurs at $x=a$, giving

$$R_{p_a}=[3-a^2,\infty ).$$

(d) From

$$y=(x-a{)}^2+3-a^2,$$

and $x\ge a$,

$$x=a+\sqrt{y-3+a^2}.$$

Thus

$$p_a^{-1}(x)=a+\sqrt{x-3+a^2},D_{p_a^{-1}}=[3-a^2,\infty ).$$

(e) Requiring $[3-a^2,\infty )=[-1,\infty )$ gives

$$3-a^2=-1⟹a^2=4.$$

Hence

$$a=\pm 2.$$