Find any critical numbers of the following functions:

Solution (a) Since $f$ is a polynomial, it is differentiable at every real number. So, the critical numbers occur where $f^\prime (x) =0$. $$f^\prime (x) ={3x}^{2}-12x+9=3({x-1})({x-3})$$

$f^\prime (x)=0$ at $x=1$ and $x=3$; the numbers $1$ and $3$ are the critical numbers of $f$.

(b) The domain of $R(x) = \dfrac{1}{x-2}$ is $\{ x\,|\,x\neq 2\}$, and $R^\prime (x)=-\dfrac{\,1}{( x-2) ^{2} }$. $\ R^\prime$ exists for all numbers $x$ in the domain of $R$ (remember, $2$ is not in the domain of $R$). Since $R^\prime$ is never $0$, $R$ has no critical numbers.

(c) The domain of $g(x)=\dfrac{(x\,{-}\,2)^{2/3}}{x}$ is $\{ x\,|\,x\neq 0\} ,$ and the derivative of $g$ is $$\begin{eqnarray*} {{g^\prime }(x)}={\frac{x\cdot \left[ {\dfrac{2}{3}}({x-2})^{-1/3}\right] -1\cdot ({x-2})^{2/3}}{x^{2}}} \underset{\underset{{\color{#0066A7}{\hbox{Multiply by} \dfrac{3(x-2) ^{1/3}}{3( x-2) ^{1/3}}}}}{\color{#0066A7}{\uparrow}}}{=} {\frac{2x-3({x-2})}{3x^{2}({x-2})^{1/3}}=\frac{6-x}{3x^{2}({x-2})^{1/3}}}\\ \end{eqnarray*}$$

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Critical numbers occur where $g^\prime (x)$ $=0$ or where $g^\prime (x)$ does not exist. If $x=6$, then $g^\prime (6) =0$. Next, $g^\prime (x)$ does not exist where $$\begin{eqnarray*} \begin{array}{rl@{\qquad}r@{\qquad}rr} 3x^{2}( x-2) ^{1/3} &= 0 \\ 3x^{2} &= 0 & \hbox{ or } & ( x-2) ^{1/3} &= 0 \\ x &= 0 & \hbox{ or } & x &= 2 \end{array} \end{eqnarray*}$$

We ignore $0$ since it is not in the domain of $g$. The critical numbers of $g$ are $6$ and $2$.

(d) The domain of $G$ is all real numbers. The function $G$ is differentiable on its domain, so the critical numbers occur where $G^\prime (x) =0$. $G^\prime (x)=\cos \,x$ and $\cos x=0$ at $x=\pm \dfrac{\pi }{2}$, ${\pm }\dfrac{3\pi }{2}$, ${\pm }\dfrac{5\pi }{2},\ldots.$ This function has infinitely many critical numbers.