SPM Practice (Long Question)

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