# How do you determine whether the function f(x)= (lnx)^2 is concave up or concave down and its intervals?

Jan 22, 2016

Concave up on $\left(0 , e\right)$; concave down on $\left(e , + \infty\right)$

#### Explanation:

The concavity of a function is determined by the sign of the second derivative of the function:

• If $f ' ' \left(a\right) < 0$, then $f \left(x\right)$ is concave down at $x = a$.
• If $f ' ' \left(a\right) > 0$, then $f \left(x\right)$ is concave up at $x = a$.

Find the second derivative of the function. But first, we must find the first derivative, which will require the chain rule: $\frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = u ' \cdot 2 u$, and here $u = \ln x$. Thus,

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\ln x\right] \cdot 2 \ln x = \frac{1}{x} \left(2 \ln x\right) = \frac{2 \ln x}{x}$

To differentiate this, use the quotient rule.

$f ' ' \left(x\right) = \frac{x \frac{d}{\mathrm{dx}} \left[2 \ln x\right] - 2 \ln x \frac{d}{\mathrm{dx}} \left[x\right]}{x} ^ 2 = \frac{x \left(\frac{2}{x}\right) - 2 \left(1\right) \ln x}{x} ^ 2 = \frac{2 - 2 \ln x}{x} ^ 2$

Now, to determine the intervals of concavity, we have to find when the second derivative is positive and when it's negative.

The sign of the second derivative could change, that is, go from positive or negative or vice versa, when it's equal to $0$. Set $f ' ' \left(x\right) = 0$ to find a possible point where the concavity could shift (called a possible point of inflection).

$\frac{2 - 2 \ln x}{x} ^ 2 = 0$

$2 - 2 \ln x = 0$

$\ln x = 1$

$x = e$

Thus the concavity could shift when $x = e$. We can test the sign of the second derivative around this point.

When $m a t h b f \left(0 < x < e\right)$:

$f ' ' \left(1\right) = \frac{2 - 2 \ln \left(1\right)}{1} ^ 2 = \frac{2 - 0}{1} = 2$

Since this is positive, we know that $f \left(x\right)$ is concave up on $\left(0 , e\right)$. Note that the interval is not $\left(- \infty , e\right)$ since $\ln x$ is only defined for $x > 0$.

When $m a t h b f \left(x > e\right)$:

$f ' ' \left({e}^{2}\right) = \frac{2 - 2 \ln \left({e}^{2}\right)}{{e}^{2}} ^ 2 = \frac{2 - 2 \left(2\right)}{e} ^ 4 = - \frac{2}{e} ^ 4$

This is obviously negative. Since it is negative, we know this whole interval will be negative. Thus, $f \left(x\right)$ is concave down on $\left(e , + \infty\right)$.

We can check a graph of $f \left(x\right)$: the concave up portion should resemble the $\cup$ shape, and the concave down portion should resemble the $\cap$ shape. The small circle is where the concavity shifts, which is the point of inflection: $\left(e , 1\right)$.

graph{((lnx)^2-y)((x-e)^2+(y-1)^2-.01)=0 [-4.52, 15, -1.314, 6.586]}