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

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Answer 1 · 2017-02-03T19:56:05

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

Finding the critical values requires that

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.

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