misyjny76

2022-09-20

Let $f(x)=\frac{\mathrm{ln}(x)}{\sqrt{x}}$. Find the critical values of f(x)

How would you find the critical value of this equation? So far... I have gotten that you find the derivative of f(x) which is ${f}^{\prime}(x)=\frac{1}{{x}^{3/2}}-\frac{\mathrm{ln}(x)}{2{x}^{3/2}}.$.

To find the critical values set ${f}^{\prime}(x)=0$? But how would you factor out the derivative to find the critical values?

Also, after that how would you find the intervals of increase and decrease and relative extreme values of f(x).

And lastly the intervals of concavity and inflection points.

How would you find the critical value of this equation? So far... I have gotten that you find the derivative of f(x) which is ${f}^{\prime}(x)=\frac{1}{{x}^{3/2}}-\frac{\mathrm{ln}(x)}{2{x}^{3/2}}.$.

To find the critical values set ${f}^{\prime}(x)=0$? But how would you factor out the derivative to find the critical values?

Also, after that how would you find the intervals of increase and decrease and relative extreme values of f(x).

And lastly the intervals of concavity and inflection points.

wayiswa8i

Beginner2022-09-21Added 6 answers

Explanation:

Since $f}^{\prime}(x)=\frac{2-\mathrm{ln}x}{{x}^{3/2}$, ${f}^{\prime}(x)=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x={e}^{2}$. Furthermore, ${f}^{\prime}(x)<0$ when $x>{e}^{2}$ and ${f}^{\prime}(x)>0$ when $0<x<{e}^{2}$

Since $f}^{\prime}(x)=\frac{2-\mathrm{ln}x}{{x}^{3/2}$, ${f}^{\prime}(x)=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x={e}^{2}$. Furthermore, ${f}^{\prime}(x)<0$ when $x>{e}^{2}$ and ${f}^{\prime}(x)>0$ when $0<x<{e}^{2}$

Julia Chang

Beginner2022-09-22Added 2 answers

Explanation:

Rewrite f′ as ${f}^{\prime}(x)=\frac{1-1/2\mathrm{ln}x}{{x}^{3/2}}$ and to find ${f}^{\prime}=0$ you need to find when $1-1/2\mathrm{ln}x=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}x=2$.

Rewrite f′ as ${f}^{\prime}(x)=\frac{1-1/2\mathrm{ln}x}{{x}^{3/2}}$ and to find ${f}^{\prime}=0$ you need to find when $1-1/2\mathrm{ln}x=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}x=2$.

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