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

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Answer 1 · 2015-05-05T21:25:21

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 $x=0, x= -sqrt(3/7), " and " x=sqrt(3/7)$

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