# The special case of x⁴

## Key Questions

• The stationary point is $\left(0 , 0\right)$ and it is a minimum point.

The first task in finding stationary points is to find the critical points, that is, where $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ DNE:

$f ' \left(x\right) = 4 {x}^{3}$ using the power rule
$4 {x}^{3} = 0$
$x = 0$ is the only solution

There are 2 ways to test for a stationary point, the First Derivative Test and the Second Derivative Test.

The First Derivative Test checks for a sign change in the first derivative: on the left the derivative is negative and on the right the derivative is positive, so this critical point is a minimum.

The Second Derivative Test checks for the sign of the second derivative:

$f ' ' \left(x\right) = 12 {x}^{2}$
$f ' ' \left(0\right) = 0$

There is no sign, so the second derivative doesn't tell us anything in this case.

Since there are no other minimums and ${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to \infty} f \left(x\right) = \infty$. The stationary point is an absolute minimum.