Question

Find the critical numbers of the function.

36.\(f(x) = {x^{ - 2}}\ln x\).

Short answer

The critical number of the function \(f(x)\) is \(x = {(e)^{\frac{1}{2}}}\).

Step-by-step solution

Given data

The function is \(f(x) = {x^{ - 2}}\ln x\).

Concept of critical number

A critical number of a function\(f\)is a number\(c\), if it satisfies either of the below conditions:

(1)\({f^\prime }(c) = 0\)

(2)\({f^\prime }(c)\), Does not exist.

Obtain the first derivative of the given function

Obtain the first derivative of the given function.

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^{ - 2}}\ln x} \right)\\ &= {x^{ - 2}}\frac{d}{{dx}}(\ln x) + \ln x\frac{d}{{dx}}\left( {{x^{ - 2}}} \right)\\ &= {x^{ - 2}}\left( {\frac{1}{x}} \right) + \ln x\left( { - 2{x^{ - 3}}} \right)\end{aligned}\)

Simplify above equation as shown below.

\(\begin{aligned}{c}{f^\prime }(x) = {x^{ - 3}} + \ln x\left( { - 2{x^{ - 3}}} \right)\\ = \frac{{1 - 2\ln x}}{{{x^3}}}\end{aligned}\)

Obtain the critical numbers

Take \({f^\prime }(x) = 0\)and obtain the critical numbers.

\(\begin{aligned}{c}\frac{{1 - 2\ln x}}{{{x^3}}} &= 0\\1 - 2\ln x &= 0\\2\ln x &= 1\\\ln {x^2} &= 1\end{aligned}\)

Take exponents on both sides.

\(\begin{aligned}{c}{x^2} &= e\\x &= {(e)^{\frac{1}{2}}}\end{aligned}\)

Hence, the solution is \({(e)^{\frac{1}{2}}}\).

The derivative function \({f^\prime }(x)\) is not defined when, \(x = 0\).

Therefore, it fails to satisfy the conditions of the critical numbers.

Thus, the critical number of the function \(f(x)\) is \(x = {(e)^{\frac{1}{2}}}\).