What is the extreme f(x)=-4x^3+6x^2+1 ?

The extreme f(x)=-4x^3+6x^2+1 is Minimum(0,1),Maximum(1,3)

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extreme f(x)=-4x^3+6x^2+1

extreme points

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

Minimum (0, 1), Maximum (1, 3)

Solution steps

f^{'} (x) = - 12 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

- 4 x^{3} + 6 x^{2} + 1 : 1

Minimum (0, 1)

Plug

x = 1

into

- 4 x^{3} + 6 x^{2} + 1 : 3

Maximum (1, 3)

Minimum (0, 1), Maximum (1, 3)

- 2 cos (5 x)

f (x) = - x^{3} + 3 x - 6

f (x) = 5 sin (\frac{1}{4} π x - π) - 3

3 cos (π x - 2) + 5

f (x) = x^{2} + 7 x + 1

What is the extreme f(x)=-4x^3+6x^2+1 ?
The extreme f(x)=-4x^3+6x^2+1 is Minimum(0,1),Maximum(1,3)