# The derivative of a function f is given by f'(x)= (x-3)e^x for x>0 and f(x)=7? a.) The function has a critical point at x=3. At this point, does f have a relative minimum or neither?

Dec 9, 2016

The function $f \left(x\right)$ has a local maximum at $x = 3$. See explanation.

#### Explanation:

To find if a function has a critical point at a place where $f ' \left(x\right) = 0$ you have to check if the derivative changes sign at this point. If the change occurs then #f(x) has:

• Minimum if $f ' \left(x\right)$ changes sign from negative to positive
• Maximum if $f ' \left(x\right)$ changes sign from positive to negative.

To check it you can calculate the second derivative:

$f ' ' \left(x\right) = 1 \cdot {e}^{x} - \left(x - 3\right) {e}^{x} = \left(1 - x + 3\right) {e}^{x} = \left(2 - x\right) {e}^{x}$

$f ' ' \left(3\right) = \left(2 - 3\right) {e}^{3} = - {e}^{3} < 0$

$f ' ' \left(x\right)$ is negative in $x = 3$. This means that $f ' \left(x\right)$ is decreasing at $x = 3$, this finally means that $f \left(x\right)$ has a MAXIMUM at $x = 3$