What is the extreme f(x)=5+6x^2-4x^3 ?

The extreme f(x)=5+6x^2-4x^3 is Minimum(0,5),Maximum(1,7)

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extreme f(x)=5+6x^2-4x^3

extreme points

f (x) = 5 + 6 x^{2} - 4 x^{3}

Minimum (0, 5), Maximum (1, 7)

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

5 + 6 x^{2} - 4 x^{3} : 5

Minimum (0, 5)

Plug

x = 1

into

5 + 6 x^{2} - 4 x^{3} : 7

Maximum (1, 7)

Minimum (0, 5), Maximum (1, 7)

f (x) = - 5 x^{2} + 8 x - 7

14 x (x^{2} - 4)^{6}

x^{4} - \frac{4}{3} x^{3} - 12 x^{2}

x^{2} - 4 x + 12

f (x, y) = 7 + 4 x^{2} + 2 y^{2}

What is the extreme f(x)=5+6x^2-4x^3 ?
The extreme f(x)=5+6x^2-4x^3 is Minimum(0,5),Maximum(1,7)