Question

Find the critical numbers of the function.

\(h\left( x \right) = {x^{ - \frac{{\bf{1}}}{{\bf{3}}}}}\left( {x - {\bf{2}}} \right)\)

Short answer

The critical number is \(x = - 1\).

Step-by-step solution

Critical Numbers of a function

TheCritical numbers for any function \(h\left( x \right)\) are obtained by putting\(h'\left( x \right) = 0\).

Differentiating the function for critical numbers:

The given function is:

\(h\left( x \right) = {x^{ - \frac{1}{3}}}\left( {x - 2} \right)\)

On differentiating with respect to\(x\)as:

\(\begin{aligned}{c}h'\left( x \right) &= \frac{d}{{dx}}\left( {{x^{ - \frac{1}{3}}}\left( {x - 2} \right)} \right)\\ &= {x^{ - \frac{1}{3}}}\frac{d}{{dx}}\left( {x - 2} \right) + \left( {x - 2} \right)\frac{d}{{dx}}\left( {{x^{ - \frac{1}{3}}}} \right)\\ &= {x^{ - \frac{1}{3}}}\left( 1 \right) - \frac{1}{3}\left( {x - 2} \right){x^{ - \frac{4}{3}}}\\ &= \frac{{2\left( {x + 1} \right)}}{{3{x^{\frac{4}{3}}}}}\end{aligned}\)

Solving for critical numbers:

\(\begin{aligned}{c}h'\left( x \right) &= 0\\\frac{{2\left( {x + 1} \right)}}{{3{x^{\frac{4}{3}}}}} &= 0\\x + 1 &= 0\\x &= - 1\end{aligned}\)

Here, \(x \ne 0\) as 0 is not in the domain of \(h\). Hence, the critical number is \(x = - 1\).