Is the function concave up or down if #f(x)= (lnx)^2#?

To determine if the function $f(x) = (\ln x)^2$ is concave up or down, we need to analyze its second derivative.

The first derivative of $f(x)$ can be found as follows:

$f'(x) = 2\ln(x) \cdot \frac{1}{x} = \frac{2\ln(x)}{x}$

Now, let's find the second derivative:

$f''(x) = \frac{d}{dx} \left(\frac{2\ln(x)}{x}\right)$

Using the quotient rule, we get:

$f''(x) = \frac{2 \cdot \frac{1}{x} - 2\ln(x) \cdot \frac{1}{x^2}}{x^2}$

Simplify this expression:

$f''(x) = \frac{2 - 2\ln(x)}{x^3}$

To determine concavity, we need to examine the sign of $f''(x)$. Since $x$ is always positive in the domain of $f(x)$, we only need to consider the sign of $2 - 2\ln(x)$.

$2 - 2\ln(x)$ is negative for $x > e$, where $e$ is the Euler's number. This means $f''(x)$ is negative for $x > e$, indicating that the function $f(x)$ is concave down in this interval.

Therefore, the function $f(x) = (\ln x)^2$ is concave down for $x > e$.