Question

41.Find the critical numbers of the function.

41. \(F\left( x \right) = {x^{\frac{{\bf{4}}}{{\bf{5}}}}}{\left( {x - {\bf{4}}} \right)^{\bf{2}}}\)

Short answer

The critical numbers are\(x = 0,\frac{8}{7}{\rm{ and 4}}\).

Step-by-step solution

Critical Numbers of a function

TheCritical numbersfor any function \(F\left( x \right)\) are obtained by putting \(F'\left( x \right) = 0\).

Differentiating the function for critical numbers:

The given function is:

\(F\left( x \right) = {x^{\frac{4}{5}}}{\left( {x - 4} \right)^2}\)

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

\(\begin{aligned}{c}F'\left( x \right) &= \frac{d}{{dx}}\left( {{x^{\frac{4}{5}}}{{\left( {x - 4} \right)}^2}} \right)\\ &= {x^{\frac{4}{5}}}\frac{d}{{dx}}\left( {{{\left( {x - 4} \right)}^2}} \right) + {\left( {x - 4} \right)^2}\frac{d}{{dx}}\left( {{x^{\frac{4}{5}}}} \right)\\ &= {x^{\frac{4}{5}}}\left( {2\left( {x - 4} \right)} \right) + \frac{4}{5}{\left( {x - 4} \right)^2}{x^{ - \frac{1}{5}}}\\ &= \frac{{\left( {x - 4} \right)\left( {14x - 16} \right)}}{{5{x^{\frac{1}{5}}}}}\\ &= \frac{2}{5}\left\{ {\frac{{\left( {x - 4} \right)\left( {7x - 8} \right)}}{{5{x^{\frac{1}{5}}}}}} \right\}\end{aligned}\)

Solving for critical numbers:

\(\begin{aligned}{c}F'\left( x \right) &= 0\\\frac{2}{5}\left\{ {\frac{{\left( {x - 4} \right)\left( {7x - 8} \right)}}{{5{x^{\frac{1}{5}}}}}} \right\} &= 0\\x &= 0,\frac{8}{7}{\rm{ and }}4\end{aligned}\)

Hence, the critical numbers are \(x = 0,\frac{8}{7}{\rm{ and 4}}\).