1.7.1 Function, SPM Practice (Long Question)

Question 1:
The function f and g is defined by
$\begin{array}{l}f:x\mapsto 2x-3\\ g:x\mapsto \frac{2}{x};x\ne 0\end{array}$
Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f^-1 ,
Calculate the value of x such that ff(x) = gf(x).

Solution:
(a)
$\begin{array}{l}ff\left(x\right)=f\left[f\left(x\right)\right]\\ =f\left(2x-3\right)\\ =2\left(2x-3\right)-3\\ =4x-9\\ ff:x\mapsto 4x-9\end{array}$

(b)
$\begin{array}{l}gf\left(x\right)=g\left[f\left(x\right)\right]\\ =g\left(2x-3\right)\\ =\frac{2}{2x-3}\\ gf:x\mapsto \frac{2}{2x-3}\end{array}$

(c)
$\begin{array}{l}\text{Let }{f}^{-1}\left(x\right)=y,\\ \text{thus }f\left(y\right)=x\\ 2y-3=x\\ y=\frac{x+3}{2}\\ \therefore f^{-1}\left(x\right)=\frac{x+3}{2}\\ f^{-1}:x\mapsto \frac{x+3}{2}\\ \\ \text{When }ff\left(x\right)=gf\left(x\right),\\ 4x-9=\frac{2}{2x-3}\\ \left(4x-9\right)\left(2x-3\right)=2\\ 8x^2-30x+27=2\\ 8x^2-30x+25=0\\ \left(4x-5\right)\left(2x-5\right)=0\\ 4x-5=0\text{ or }2x-5=0\\ x=\frac{5}{4}\text{ or }x=\frac{5}{2}\end{array}$

Question 2:
The function f and g is defined by
$\begin{array}{l}f\left(x\right)=3x-2\\ g\left(x\right)=\frac{3}{x},x\ne 0\\ \text{Find}\\ \text{(a) }{f}^{-1}\left(2\right),\\ \text{(b) g}f\left(-3\right),\\ \text{(c) function }h\text{ if }hf\left(x\right)=3x+2,\\ \text{(d) function }k\text{ if }fk\left(x\right)=4x-7.\end{array}$

Solution:
(a)
$\begin{array}{l}\text{Let }{f}^{-1}\left(2\right)=x,\\ \text{thus }f\left(x\right)=2\\ \text{ 3}x-2=2\\ 3x=4\\ x=\frac{4}{3}\\ f^{-1}\left(2\right)=\frac{4}{3}\end{array}$

(b)
$\begin{array}{l}gf\left(-3\right)=g\left[3\left(-3\right)-2\right]\\ =g\left(-11\right)\\ =-\frac{3}{11}\end{array}$

(c)
$\begin{array}{l}h\left[f\left(x\right)\right]=3x+2\\ h\left(3x-2\right)=3x+2\\ \text{Let }y=3x-2\\ \text{thus }x=\frac{y+2}{3}\\ h\left(y\right)=3\left(\frac{y+2}{3}\right)+2\\ =y+2+2\\ =y+4\\ \therefore h\left(x\right)=x+4\end{array}$

(d)
$\begin{array}{l}f\left[k\left(x\right)\right]=4x-7\\ 3k\left(x\right)-2=4x-7\\ 3k\left(x\right)=4x-5\\ k\left(x\right)=\frac{4x-5}{3}\end{array}$