Question

What are the values and types of the critical points, if any, of `f(x)=7x^4-6x^2+1`?

Answer 1

`x=0` or `x=+-sqrt(3/7)`

Explanation:

Finding the critical values requires that

• `f'(c)=0`, or
• `f'(c)` is undefined, and
• `c in` the domain of `f(x)`

Find the derivative of `f(x)`.

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

Set `f'(x)=0` and solve for `x`.

`28x^3-12x=0`

Factor the left hand side.

`4x(7x^2-3)=0`

Divide both sides by 4 to simplify.

`x(7x^2-3)=0`

Solve each term in the product separately.

`x=0" "` or `" "7x^2-3=0`
`" "7x^2=3`
`" "x^2=3/7`
`" "x=+-sqrt(3/7)`

Determine the types of critical points by looking at the graph of `f(x)`

graph{7x^4-6x^2+1 [-1.3, 1.3, -0.5, 1.5]}

When `x=0`, the critical point is a local maximum
When `x=+-sqrt(3"/"7)`, the critical points are local minimums

All three critical points stationary points because they have horizontal tangents at those critical points.