Question

(i) Find the critical numbers of $x^{\frac{3}{5}}(4-x)$
(ii) Determine the domain of convexity of $y=e^x$

Answer 1

$\left(1\right)$ To determine the critical point, first find the derivative of the given function $f\left(x\right)$,

$f^{\prime }\left(x\right)=\frac{d}{dx}{x}^{3/5}(4-x)=\frac{4\times 3}{5x^{2/5}}-\frac{8}{5}{x}^{3/5}=\frac{12-8x^{}}{5x^{2/5}}$

Now, put $f^{\prime }\left(x\right)=0\Rightarrow x=\frac{3}{2}$

$\therefore$ Critical number for $x$ is $3$.

$\left(2\right)$ A function $f\left(x\right)$ is convex, if $f^{\prime \prime }\left(x\right)>0$

$\Rightarrow y=e^x\Rightarrow y^{\prime \prime }=e^x$, which is always greater than zero

$\therefore$ domain of convexity $\in R$