# How do you find the second derivative test to find extrema for f(x) = 2x^2lnx-5x^2?

Sep 27, 2015

See the explanation.

#### Explanation:

${D}_{f} = {R}^{+}$

$f ' = 4 x \ln x + 2 {x}^{2} \cdot \frac{1}{x} - 10 x = 4 x \ln x - 8 x = 4 x \left(\ln x - 2\right)$

$f ' ' = 4 \ln x + 4 x \cdot \frac{1}{x} - 8 = 4 \ln x - 4 = 4 \left(\ln x - 1\right)$

$f ' = 0 \iff 4 x = 0 \vee \ln x - 2 = 0$

$x = 0 \notin {D}_{f}$
$\ln x = 2 \implies x = {e}^{2}$

$f ' ' \left({e}^{2}\right) = 4 \left(\ln {e}^{2} - 1\right) = 4 \left(2 - 1\right) = 4 > 0$ and hence function has a minimum at $x = {e}^{2}$.

${f}_{\min} = f \left({e}^{2}\right) = 2 {\left({e}^{2}\right)}^{2} \ln {e}^{2} - 5 {\left({e}^{2}\right)}^{2} = 4 {e}^{4} - 5 {e}^{4} = - {e}^{4}$