Hello John,
We are given the function:
$$f(x)=2x^4-4x^2+6$$
And are asked to find the local extrema. In order to do this, we need to equate the first derivative to zero, and finct the critical values:
$$f'(x)=8x^3-8x=8x\left(x^2-1 \right)=8x(x+1)(x-1)=0$$
Hence, our critical values are:
$$x=-1,0,1$$
Observing that the roots here are all off multiplicity 1, and seeing that for $1<x$, we have:
$$f'(x)>0$$
we may then conclude that the sign of the derivative alternates across the 4 intervals made by dividing the real number line at the three critical values, hence we have:
$$f(x)$$ increasing on $$(-1,0)\,\cup\,(1,\infty)$$
$$f(x)$$ decreasing on $$(-\infty,-1)\,\cup\,(0,1)$$
Thus, by the first derivative test, we are led to conclude that we have the following:
Relative minima at $$\left(-1,f(-1) \right)=(-1,4)$$ and $$\left(1,f(1) \right)=(1,4)$$
Relative maximum at $$\left(0,f(0) \right)=(0,6)$$
Here is a plot of the function and its relative extrema:
View attachment 2308