What is the extreme f(x)=-4x^4+120x^2+120 ?

The extreme f(x)=-4x^4+120x^2+120 is Maximum(-sqrt(15),1020),Minimum(0,120),Maximum(sqrt(15),1020)

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extreme f(x)=-4x^4+120x^2+120

extreme points

f (x) = - 4 x^{4} + 120 x^{2} + 120

Maximum (- 15, 1020), Minimum (0, 120), Maximum (15, 1020)

Solution steps

f^{'} (x) = - 16 x^{3} + 240 x

Find intervals:

Increasing

: - \infty < x < - 15,

Decreasing

: - 15 < x < 0,

Increasing

: 0 < x < 15,

Decreasing

: 15 < x < \infty

Plug

x = - 15

into

- 4 x^{4} + 120 x^{2} + 120 : 1020

Maximum (- 15, 1020)

Plug

x = 0

into

- 4 x^{4} + 120 x^{2} + 120 : 120

Minimum (0, 120)

Plug

x = 15

into

- 4 x^{4} + 120 x^{2} + 120 : 1020

Maximum (15, 1020)

Maximum (- 15, 1020), Minimum (0, 120), Maximum (15, 1020)

\frac{x ^{2} - 63}{x - 8}

f (x) = 9 x^{3} + 2 1^{2} + 8 x - 3

- 2 x^{2} + 8 x - 6

f (x) = 9 + 2 x^{2} - x^{4}

f (x) = 3 x^{2} - 6 x + 4, - 1 \leq x \leq 3

What is the extreme f(x)=-4x^4+120x^2+120 ?
The extreme f(x)=-4x^4+120x^2+120 is Maximum(-sqrt(15),1020),Minimum(0,120),Maximum(sqrt(15),1020)