How do you use the first derivative test to determine whether the critical point is a local maximum, local minimum, or neither for y=x^4 - 8x^2?

Mar 28, 2015

First evaluate the first derivative:
$y ' = 4 {x}^{3} - 16 x$
Second, set it equal to zero;
$4 {x}^{3} - 16 x = 0$
you get:
$4 x \left({x}^{2} - 4\right) = 0$
and so:
${x}_{1} = 0$
${x}_{2} = + 2$
${x}_{3} = - 2$
This gives you the critical points (where the inclination of your curve changes).
Third set your derivative >0, and study the combined signs of the derivatives: