SOLUTION: For the function below, find the value(s) of x in which f'(x)=0. f(x)=(x^2-2)(x^2-√2) (round to three decimal places as needed.)

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Question 1195274: For the function below, find the value(s) of x in which f'(x)=0.
f(x)=(x^2-2)(x^2-√2)
(round to three decimal places as needed.)
Found 2 solutions by math_tutor2020, MathLover1:
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Let
g(x) = x^2 - 2
h(x) = x^2 - sqrt(2)

The derivative of each is
g ' (x) = 2x
h ' (x) = 2x
Both arrive at the same derivative because the constants go to 0 (because they don't have any rate of change to them).

Apply the product rule:
f(x) = (x^2 - 2) * (x^2 - sqrt(2))
f(x) = g(x)*h(x)
f ' (x) = g ' (x)*h(x) + g(x)*h ' (x)
f ' (x) = 2x*(x^2 - sqrt(2)) + (x^2-2)*2x
f ' (x) = 2x*(x^2 - sqrt(2) + x^2-2)
f ' (x) = 2x*(2x^2 - sqrt(2)-2)

Set this equal to zero and solve for x.
f ' (x) = 0
2x*(2x^2 - sqrt(2)-2) = 0
2x = 0 or 2x^2 - sqrt(2)-2 = 0
x = 0 or 2x^2 = sqrt(2)+2
x = 0 or x^2 = 0.5sqrt(2)+1
x = 0 or x = sqrt( 0.5sqrt(2)+1 ) or x = -sqrt( 0.5sqrt(2)+1 )
x = 0 or x = 1.30656 or x = -1.30656
x = 0 or x = 1.307 or x = -1.307

Answers:
x = -1.307, x = 0, and x = 1.307

Answer by MathLover1(20849) (Show Source):
You can put this solution on YOUR website!

.......expand

' ........factor
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now find '
.......will be true if

or

if =>
if => => => or
so, ' for , ,

SOLUTION: For the function​ below, find the​ value(s) of x in which f'(x)=0. f(x)=(x^2-2)(x^2-√2) (round to three decimal places as needed.)

SOLUTION: For the function below, find the value(s) of x in which f'(x)=0. f(x)=(x^2-2)(x^2-√2) (round to three decimal places as needed.)