A non-invertible function can have an invertible composite · source reference

Printed question

$f:x\mapsto(x-7)(x+1)$, $x>0$.

(a) Use an algebraic method to find exact solution of $f(x)<(x^2-7x)/(2-x)$. [5]

(b)(i) Determine if $f^{-1}$ exists. [2]

Given $g$ is one-one with domain $(0, \infty)$ and range $(4, \infty)$:

(ii) Find range of $fg$. [2]

(iii) A function $h$ is one-one if for any $\alpha, \beta\in D_h$, $h(\alpha)=h(\beta)$ implies $\alpha=\beta$. Show that $fg$ is one-one. [2]

(iv) If $g(x)=4e^x$ for $x>0$, find exact $(fg)^{-1}(20)$. [4]

[2011/RI/Promo/13]

Printed working

(a)

$$(x-7)(x+1)<\frac{x^2-7x}{2-x}\Rightarrow(x-7)\left(x+1+\frac x{x-2}\right)<0,$$ $$\frac{(x-7)(x^2-2)}{x-2}<0\Rightarrow\frac{(x-7)(x-\sqrt2)(x+\sqrt2)}{x-2}<0.$$

Since $x>0$, $x+\sqrt2>0$. This reduces to $(x-7)(x-\sqrt2)/(x-2)<0$ ⇒ $(x-2)(x-7)(x-\sqrt2)<0$, $x\ne2$ (multiply by positive $(x-2)^2$).

The source sign sketch has open roots $\sqrt2, 2, 7$, negative to the left of $\sqrt2$ and between 2 and 7. Given $x>0$, solution $0<x<\sqrt2$ or $2<x<7$.

(b)(i) Horizontal line $y=-10$ cuts $f$ twice ⇒$f$ not one-one ⇒$f^{-1}$ does not exist. Graph has open $(0, -7)$ and vertex $(3, -16)$.

(ii) $D_g=(0, \infty)$, $R_g=(4, \infty)$; flow $D_g\xrightarrow g R_g\xrightarrow f R_{fg}=(-15, \infty)$. Thus range $(-15, \infty)$.

(iii) $D_{fg}=D_g=(0, \infty)$. Suppose $fg(\alpha)=fg(\beta)$ for $\alpha, \beta>0$: $f[g(\alpha)]=f[g(\beta)]$. Since $R_g=(4, \infty)$, both inner values lie in $(4, \infty)$. From graph $f$ is one-one restricted to $(4, \infty)$ ⇒$g(\alpha)=g(\beta)$. Since $g$ is one-one, $\alpha=\beta$. Hence $fg$ is one-one (shown).

(iv) Let $(fg)^{-1}(20)=x$. Then $f(g(x))=20$ ⇒$[g(x)]^2-6g(x)-7=20$ ⇒$[g(x)]^2-6g(x)-27=0$ ⇒$[g(x)-9][g(x)+3]=0$.

$g(x)=9$, since $g(x)>4$ makes $g(x)+3>0$. Therefore $4e^x=9\Rightarrow x=\ln(9/4)$.

Teacher notes

$g(\alpha), g(\beta)\in D_f$ in the injectivity argument.

Expand $fg(x)=(g(x)-7)(g(x)+1)=[g(x)]^2-6g(x)-7$ before solving inverse-value equation.

Answer key

(a) $0<x<\sqrt2$ or $2<x<7$. (b)(i) No. (ii) $R_{fg}=(-15, \infty)$. The short key labels $\ln(9/4)$ “(iii)” although it answers(iv); printed numbering retained here.

Native diagrams