Question

How do I find the critical points for the function `f(x)=7x^4-6x^2+1`?

Answer 1

The critical points for a continuous function occur at those points where the derivative is zero.

Given `f(x) =7x^4-6x^2 + 1`

`f'(x) = 28x^3 -12x`

If `f'(x) = 0`
then
`28x^3-12x graph{7x^4-6x^2+1 [-2.982, 3.178, -0.404, 2.675]} = x(28x^2-12) = 0`
and
either `x=0`
or `(28x^2-12)=0`

If `28x^2-12 = 0`
then `x=+-sqrt(12/28) = +-sqrt(3/7)`

The critical points occur at `x=0, x= -sqrt(3/7), " and " x=sqrt(3/7)`