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# How do you find the intervals of concavity and increasing or decreasing for f(x)=x^(5)ln(x)?

Oct 5, 2017

Decreasing: $\left(0 , {e}^{- \frac{1}{5}}\right)$. Increasing: $\left({e}^{- \frac{1}{5}} , \infty\right)$
Concave down: $\left(0 , {e}^{- \frac{9}{20}}\right)$. Convex/Concave Up: $\left({e}^{- \frac{9}{20}} , \infty\right)$

#### Explanation:

The simplest means of doing this is to graph the function.

graph{x^5lnx [-10, 10, -5, 5]}

The function exists on the interval from $\left(0 , \infty\right)$. On this interval ${x}^{5}$ is always positive, and $\ln \left(x\right)$ is negative until $x = 1$. Looking at the graph, we know that the function will be concave upwards and increasing after $x = 1$, but via taking the derivatives we can find when exactly the change from decreasing to increasing occurs.

We must use the product rule for this, which states that for $f \left(x\right) = g \left(x\right) h \left(x\right) , f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$. Thus here...

$f \left(x\right) = {x}^{5} \ln \left(x\right) \to f ' \left(x\right) = 5 {x}^{4} \ln x + {x}^{5} / x = 5 {x}^{4} \ln x + {x}^{4} = {x}^{4} \left(5 \ln x + 1\right)$.

We can find where the change from $f ' \left(x\right) < 0$ to $f ' \left(x\right) > 0$ occurs by finding where f'(x)=0. From our equation, this cannot happen at $x = 0$ because the natural log is undefined there. Therefore, setting the equation equal to zero...

$f ' \left(x\right) = {x}^{4} \left(5 \ln x + 1\right) = 0 \to 5 \ln x + 1 = 0 \to \ln x = - \frac{1}{5} \to x = {e}^{- \frac{1}{5}}$ is our critical point. Plugging in values slightly larger and slightly smaller than this for x reveals that for $x < {e}^{- \frac{1}{5}} , f ' \left(x\right) < 0$, and for $x > {e}^{- \frac{1}{5}} , f ' \left(x\right) > 0$. Therefore the function is decreasing on $\left(0 , {e}^{- \frac{1}{5}}\right)$, and increasing on $\left({e}^{- \frac{1}{5}} , \infty\right)$

To find the concavity, we must instead find $f ' ' \left(x\right)$ in the same way as we found the first derivative:

f''(x) = d/dx(5x^4lnx + x^4) = (20x^3lnx + 5x^3 + 4x^3 = 20x^3lnx + 9x^3 = x^3(20lnx + 9).

The concavity can only switch at a point where this second derivative is equal to zero. As above, we know that x=0 is not a zero of this second derivative because the natural log is undefined there...

$f ' ' \left(x\right) = 0 = {x}^{3} \left(20 \ln x + 9\right) = 20 \ln x + 9 \to 20 \ln x = - 9 \to \ln x = - \frac{9}{20} \to x = {e}^{- \frac{9}{20}}$

This is our zero for the second derivative. As above, plugging in a slightly smaller x (i.e. ${e}^{- \frac{1}{2}}$) yields a negative value ($\ln x = - \frac{1}{2} < - \frac{9}{20} \to 20 \ln x + 9 < 0 \to {x}^{3} \left(20 \ln x + 9\right) < 0$), and using a larger x yields a positive value, i.e. $x = e \to \ln e = 1 \succ \frac{9}{20} \to {e}^{3} \left(20 \ln e + 9\right) = {e}^{3} \left(20 + 9\right) = {e}^{3} \left(29\right) > 0$. Thus, the function is concave down for $\left(0 , {e}^{- \frac{9}{20}}\right)$, and concave up for $\left({e}^{- \frac{9}{20}} , \infty\right)$

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