# How do you use the second derivative test how do you find the local maxima and minima of f(x) = 12 + 2x^2 - 4x^4?

The function $f \left(x\right) = 12 + 2 {x}^{2} - 4 {x}^{4}$ has derivative $f ' \left(x\right) = 4 x - 16 {x}^{3}$ and second derivative $f ' ' \left(x\right) = 4 - 48 {x}^{2}$.
The critical points occur where $f ' \left(x\right) = 4 x \left(1 - 4 {x}^{2}\right) = 0$, which are $x = 0$ and $x = \setminus \pm \frac{1}{2}$.
Since $f ' ' \left(\setminus \pm \frac{1}{2}\right) = 4 - 48 \setminus \cdot \frac{1}{4} = 4 - 12 = - 8 < 0$, the second derivative test says the critical points at $x = \setminus \pm \frac{1}{2}$ are local maxima (the graph of $f$ is concave down near $x = \setminus \pm \frac{1}{2}$).
Since $f ' ' \left(0\right) = 4 > 0$, the second derivative test says the critical point at $x = 0$ is a local minimum (the graph of $f$ is concave up near $x = 0$).