On what intervals the following equation is concave up, concave down and where it's inflection point is (x,y) f(x)=x^8(ln(x))?

• if $0 < x < {e}^{- \frac{15}{56}}$ then $f$ is concave down;
• if $x > {e}^{- \frac{15}{56}}$ then $f$ is concave up;
• $x = {e}^{- \frac{15}{56}}$ is a (falling) inflection point

Explanation:

To analyze concavity and inflection points of a twice differentiable function $f$, we can study the positivity of the second derivative. In fact, if ${x}_{0}$ is a point in the domain of $f$, then:

• if $f ' ' \left({x}_{0}\right) > 0$, then $f$ is concave up in a neighborhood of ${x}_{0}$;
• if $f ' ' \left({x}_{0}\right) < 0$, then $f$ is concave down in a neighborhood of ${x}_{0}$;
• if $f ' ' \left({x}_{0}\right) = 0$ and the sign of $f ' '$ on a sufficiently small right-neighborhood of ${x}_{0}$ is opposite to the sign of $f ' '$ on a sufficiently small left-neighborhood of ${x}_{0}$, then $x = {x}_{0}$ is called an inflection point of $f$.

In the specific case of $f \left(x\right) = {x}^{8} \ln \left(x\right)$, we have a function whose domain has to be restricted to the positive reals ${\mathbb{R}}^{+}$.
The first derivative is
$f ' \left(x\right) = 8 {x}^{7} \ln \left(x\right) + {x}^{8} \frac{1}{x} = {x}^{7} \left[8 \ln \left(x\right) + 1\right]$
The second derivative is
$f ' ' \left(x\right) = 7 {x}^{6} \left[8 \ln \left(x\right) + 1\right] + {x}^{7} \frac{8}{x} = {x}^{6} \left[56 \ln \left(x\right) + 15\right]$

Let's study the positivity of $f ' ' \left(x\right)$:

• ${x}^{6} > 0 \iff x \ne 0$
• $56 \ln \left(x\right) + 15 > 0 \iff \ln \left(x\right) > - \frac{15}{56} \iff x > {e}^{- \frac{15}{56}}$

So, considering that the domain is ${\mathbb{R}}^{+}$, we get that

• if $0 < x < {e}^{- \frac{15}{56}}$ then $f ' ' \left(x\right) < 0$ and $f$ is concave down;
• if $x > {e}^{- \frac{15}{56}}$ then $f ' ' \left(x\right) > 0$ and $f$ is concave up;
• if $x = {e}^{- \frac{15}{56}}$ then $f ' ' \left(x\right) = 0$. Considering that on the left of this point $f ' '$ is negative and on the right it is positive, we conclude that $x = {e}^{- \frac{15}{56}}$ is a (falling) inflection point