0
$\begingroup$

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 $$f'(x) = \frac{1}{x^{3/2}} - \frac{\ln(x)}{2x^{3/2}}.$$ To find the critical values set $f'(x) = 0$? 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.

Thank you for your help.

$\endgroup$
  • $\begingroup$ Your questions sound like "study the function". $\endgroup$ – farruhota Aug 6 '17 at 17:16
2
$\begingroup$

Since $\displaystyle f'(x)=\frac{2-\ln x}{x^{3/2}}$, $f'(x)=0\iff x=e^2$. Furthermore, $f'(x)<0$ when $x>e^2$ and $f'(x)>0$ when $0<x<e^2$.

$\endgroup$
0
$\begingroup$

Rewrite $f'$ as

$$ f'(x) = \frac{1 - 1/2 \ln x}{x^{3/2}} $$

and to find $f'=0$ you need to find when $1 - 1/2 \ln x = 0 \implies \ln x = 2$.

$\endgroup$
0
$\begingroup$

just hint

$$f (x)=\frac {\ln (x)}{\sqrt {x}} $$

$$\lim_{0^+}f (x)=-\lim_{+\infty}\ln (x)\sqrt {x}=-\infty $$

$$f (1)=0$$ $$\lim_{+\infty}f (x)=0$$

thus there is a maximum at $x=a $ on $(1,+\infty) $ and an inflection point at $x=b>a $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.