Functions: DHS 2024 Promo Q4

(a) Since $f(1)=f(4)=5$, f is not one-one. Hence, f does not have an inverse.

(b) Let $y=f(x)=\dfrac{x^2+4}{x}$, $0<x\le2$ $$\begin{aligned} xy&=x^2+4\\ x^2-yx+4&=0\\ \left(x-\frac y2\right)^2-\frac{y^2}{4}+4&=0\\ \left(x-\frac y2\right)^2-\frac{y^2-16}{4}&=0\\ \left(x-\frac y2\right)^2&=\frac{y^2-16}{4}\\ x-\frac y2&=\pm\sqrt{\frac{y^2-16}{4}}\\ x&=\frac y2\pm\sqrt{\frac{y^2-16}{4}}\\ x&=\frac{y\pm\sqrt{y^2-16}}2 \end{aligned}$$ Since $0<x\le2$, $x=\dfrac{y-\sqrt{y^2-16}}2$

Hence, $f^{-1}(x)=\dfrac{x-\sqrt{x^2-16}}2$, $x\ge4$.

$D_{f^{-1}}=[4,\infty)$

Alternative (to check plus or minus):

When $x=1$, $f(x)=\dfrac{1^2+4}{1}=5$

When $y=5$, $$x=\frac{5+\sqrt{5^2-16}}2=4\ \text{(rej +ve)},\qquad x=\frac{5-\sqrt{5^2-16}}2=1$$ Hence $f^{-1}(x)=\dfrac{x-\sqrt{x^2-16}}2$, $x\ge4$.