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

Feb 3, 2017

$x = 0$ or $x = \pm \sqrt{\frac{3}{7}}$

#### Explanation:

Finding the critical values requires that

• $f ' \left(c\right) = 0$, or
• $f ' \left(c\right)$ is undefined, and
• $c \in$ the domain of $f \left(x\right)$

Find the derivative of $f \left(x\right)$.

$f ' \left(x\right) = 28 {x}^{3} - 12 x$

Set $f ' \left(x\right) = 0$ and solve for $x$.

$28 {x}^{3} - 12 x = 0$

Factor the left hand side.

$4 x \left(7 {x}^{2} - 3\right) = 0$

Divide both sides by 4 to simplify.

$x \left(7 {x}^{2} - 3\right) = 0$

Solve each term in the product separately.

$x = 0 \text{ }$ or $\text{ } 7 {x}^{2} - 3 = 0$
$\text{ } 7 {x}^{2} = 3$
$\text{ } {x}^{2} = \frac{3}{7}$
$\text{ } x = \pm \sqrt{\frac{3}{7}}$

Determine the types of critical points by looking at the graph of $f \left(x\right)$

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 $x = \pm \sqrt{3 \text{/} 7}$, the critical points are local minimums

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