What is the extreme f(x)=7x^4-42x^2 ?

The extreme f(x)=7x^4-42x^2 is Minimum(-sqrt(3),-63),Maximum(0,0),Minimum(sqrt(3),-63)

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extreme f(x)=7x^4-42x^2

extreme points

f (x) = 7 x^{4} - 42 x^{2}

Minimum (- 3, - 63), Maximum (0, 0), Minimum (3, - 63)

Solution steps

f^{'} (x) = 28 x^{3} - 84 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

7 x^{4} - 42 x^{2} : - 63

Minimum (- 3, - 63)

Plug

x = 0

into

7 x^{4} - 42 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

7 x^{4} - 42 x^{2} : - 63

Minimum (3, - 63)

Minimum (- 3, - 63), Maximum (0, 0), Minimum (3, - 63)

f (x) = x^{2} - 49 - 49

\frac{( x ^{3} - x )}{1 + x ^{2}}

f (x) = \frac{6 x - 12 x ^{9}}{3 x ^{3} + 7}

f (x) = \frac{( 2 x - 1 )}{x + 3}

h (x) = x - 10

What is the extreme f(x)=7x^4-42x^2 ?
The extreme f(x)=7x^4-42x^2 is Minimum(-sqrt(3),-63),Maximum(0,0),Minimum(sqrt(3),-63)