# On what interval is f concave upward? f(x)=xln(x)

## $f \left(x\right) = x \ln \left(x\right)$ is concave up when $f ' ' \left(x\right) > 0$

Oct 25, 2016

$f \left(x\right) = x \ln \left(x\right)$ is concave up on the interval (0,∞)

#### Explanation:

To start off, we must realize that a function $f \left(x\right)$ is concave upward when $f ' ' \left(x\right)$ is positive.

To find $f ' \left(x\right)$, the Product Rule must be used and the derivative of the natural logarithmic function must be known:

Product Rule (Simplified format of the rule): (Derivative of first)(Second function) + (First function)(Derivative of the second function)
Derivative of the natural logarithmic function: $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

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

Now we need to take the derivative of this in order to find $f ' ' \left(x\right)$. Remember that the derivative of a constant is equal to zero.

$f ' ' \left(x\right) = \frac{1}{x}$

Remember that this question is asking us to look for when the function is concave up, meaning where $f ' ' \left(x\right) > 0$.

$f ' ' \left(x\right) = \frac{1}{x} > 0$
Since $1$ is a positive value, $\frac{1}{x}$ is positive whenever $x > 0$ (a fraction is positive if the numerator and denominator have the same sign).

The final answer is that the function $f \left(x\right) = x \ln x$ is concave up on the interval (0,∞), which is when $x > 0$.