Functions: NYJC 2024 Promo Q7

(a)

$$\begin{aligned} R_{\mathrm f}&=[0,\infty)\\D_{\mathrm g}&=\mathbb R^+\\R_{\mathrm f}&\not\subset D_{\mathrm g} \end{aligned}$$ Thus $\mathrm{gf}$ does not exist.

(b)

Method 1:

Consider the different scenarios:

If $\alpha>0$, $\mathrm g$ will have a turning point at $x=\sqrt{\alpha}$, making $\mathrm g$ not an one-one function and $\mathrm g^{-1}$ will not exist.

Hence, $\alpha\leq 0$.

Method 2:

We need $\mathrm g'(x)=1-\dfrac{\alpha}{x^2}>0$ for $\mathrm g^{-1}$ to exist.

Since $x^2>0$, $\alpha<0$

Also, when $\alpha=0$, $\mathrm g(x)=x$ is also an one-one function.

Hence, $\alpha\leq 0$

(c)

$$\begin{aligned} \mathrm g(x)&=x+\frac4x\\\mathrm g'(x)&=1-\frac4{x^2}=0\\x&=\pm2 \end{aligned}$$

From graph, least $\beta=2$ since we need $\mathrm g$ to be one-one function.

(or since $x>0$, $\beta=2$) $$\begin{aligned} y&=x+\frac4x\\xy&=x^2+4\\x&=\frac{y\pm\sqrt{y^2-4(1)(4)}}2\\ \text{since }x\geq2,\quad x&=\frac{y+\sqrt{y^2-16}}2\\ \mathrm g^{-1}(x)&=\frac12\left(x+\sqrt{x^2-16}\right)\\ D_{\mathrm g^{-1}}&=R_{\mathrm g}=[4,\infty) \end{aligned}$$