# How do you determine whether the function f(x)=x^8(ln(x)) is concave up or concave down and its intervals?

Oct 15, 2015

See the explanation.

#### Explanation:

The intervals of concavity are determined by 2nd derivative. If 2nd derivative changes sign in the points where it is equal to zero, those points are inflection points.

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

$f ' \left(x\right) = 8 {x}^{7} \ln x + {x}^{8} \frac{1}{x} = 8 {x}^{7} \ln x + {x}^{7} = {x}^{7} \left(8 \ln x + 1\right)$

$f ' ' \left(x\right) = 7 {x}^{6} \left(8 \ln x + 1\right) + {x}^{7} \frac{8}{x} = 7 {x}^{6} \left(8 \ln x + 1\right) + 8 {x}^{6}$

$f ' ' \left(x\right) = {x}^{6} \left(56 \ln x + 15\right)$

$f ' ' \left(x\right) = 0 \iff {x}^{6} \left(56 \ln x + 15\right) = 0 \iff$

$\iff \left({x}^{6} = 0 \vee 56 \ln x + 15 = 0\right)$

${x}^{6} = 0 \iff x = 0 \notin {D}_{f}$

$56 \ln x + 15 = 0 \iff 56 \ln x = - 15 \iff \ln x = - \frac{15}{56} \iff$

$\iff x = {e}^{- \frac{15}{56}} \iff x = \frac{1}{\sqrt[56]{{e}^{15}}} \approx 0.765 > 0 \in {D}_{f}$

$\forall x \in \left(0 , \frac{1}{\sqrt[56]{{e}^{15}}}\right) : f ' ' \left(x\right) < 0$ function is concave down

$\forall x \in \left(\frac{1}{\sqrt[56]{{e}^{15}}} , \infty\right) : f ' ' \left(x\right) > 0$ function is concave up

Note: the sign of $f ' ' \left(x\right)$ depends only on $56 \ln x + 15$ since ${x}^{6}$ is always greater then zero for $x > 0$.