Calculus: Early Transcendentals 8th Edition

$c=\frac{4}{9}$ and $c=0$ are critical numbers of $h$.
How to find the critical numbers of a function $f$ according to definition 1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist. 2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not. - If $c_0$ lies in $D_f$, $c_0$ is a critical number of $f$. - If $c_0$ does not lie in $D_f$, $c_0$ is not a critical number of $f$. $$h(t)=t^{3/4}-2t^{1/4}$$ $$h(t)=\sqrt[4]{t^3}-2\sqrt[4]t.$$ For $h(t)$ to be defined, $$t^3\ge0\hspace{0.5cm}and\hspace{0.5cm}t\ge0.$$ That means, $$t\ge0.$$ Therefore, $$D_h=[0,+\infty).$$ 1) Now, we find $h'(t)$ $$h'(t)=\frac{3}{4}t^{-1/4}-2\frac{1}{4}t^{-3/4}$$ $$h'(t)=\frac{3}{4}t^{-1/4}-\frac{1}{2}t^{-3/4}$$ $$h'(t)=\frac{3}{4}\sqrt[4]{\frac{1}{t}}-\frac{1}{2}\sqrt[4]{\frac{1}{t^3}}.$$ We find $h'(c)=0$, $$h'(c)=0$$ $$\frac{3}{4}c^{-1/4}-\frac{1}{2}c^{-3/4}=0$$ $$\frac{3}{4c^{1/4}}-\frac{1}{2c^{3/4}}=0$$ $$\frac{3c^{1/2}-2}{4c^{3/4}}=0$$ $$3c^{1/2}-2=0$$ $$3\sqrt c-2=0$$ $$\sqrt c=\frac{2}{3}$$ $$c=\frac{4}{9}$$ Looking at $$h'(t)=\frac{3}{4}\sqrt[4]{\frac{1}{t}}-\frac{1}{2}\sqrt[4]{\frac{1}{t^3}}$$, we see that $h'(t)$ does not exist as $t\le0$. So for $c\le0$, $h'(c)$ does not exist. 2) Examine whether $c$ lies in $D_h$ or not. We see that $c=\frac{4}{9}\in[0,+\infty)$, so it lies in $D_h$. For $c=(-\infty,0]$, only $c=0\in[0,+\infty)$, so only $c=0$ lies in $D_h$ We conclude that $c=\frac{4}{9}$ and $c=0$ are critical numbers of $h$.