Find any critical numbers of the following functions:
\(f^\prime (x)=0\) 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.