Answer :
The critical numbers of the function is [tex]1.18$.[/tex]
As per the question we have to determine the critical numbers of the function.
The number “c” has to be in the domain of the original function (the one you took the derivative of). If the number isn’t in the domain (for example, if there is a removable discontinuity at x = 0), then that number isn’t a critical number.
We have given,
[tex]$f(x)=x^{-6} \ln x$[/tex]
For critical numbers [tex]\mathrm{f}^{\prime}(\mathrm{x})=0$ of $\mathrm{f}^{\prime}(\mathrm{x})$[/tex] is undefined.
Now,
[tex]$\begin{aligned}\mathrm{f}^{\prime}(\mathrm{x}) & =\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{-6} \ln x\right) \\& =\mathrm{x}^{-6} \times \frac{\mathrm{d}}{\mathrm{dx}}(\ln \mathrm{x})+\ln \mathrm{x} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{-6}\right) \\& =\mathrm{x}^{-6} \times \frac{1}{\mathrm{x}}+\ln \mathrm{x} \times\left(-6 \mathrm{x}^{-7}\right) \\& =\mathrm{x}^{-7}(1-6 \ln x)\end{aligned}$[/tex]
We will use product rule,
[tex]\frac{d}{d x}(f . g)=f \cdot g^{\prime}+g \cdot f^{\prime}$[/tex]
Now,
[tex]$\begin{aligned}\mathrm{f}^{\prime}(\mathrm{x}) & =0 \\\mathrm{x}^{-7}(1-6 \ln \mathrm{x}) & =0 \\1-6 \ln \mathrm{x} & =0 \\\mathrm{x} & =\mathrm{e}^{\frac{1}{6}}=1.18\end{aligned}$[/tex]
Therefore, critical number is [tex]\mathrm{x}=\mathrm{e}^{\frac{1}{6}}=1.18$.[/tex]
For more questions on critical numbers
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