What is the extreme f(x)= 1/2 x^4-4x^3+8x^2+46 ?

The extreme f(x)= 1/2 x^4-4x^3+8x^2+46 is Minimum(0,46),Maximum(2,54),Minimum(4,46)

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extreme f(x)= 1/2 x^4-4x^3+8x^2+46

extreme points

f (x) = \frac{1}{2} x^{4} - 4 x^{3} + 8 x^{2} + 46

Minimum (0, 46), Maximum (2, 54), Minimum (4, 46)

Solution steps

f^{'} (x) = 2 x^{3} - 12 x^{2} + 16 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

\frac{1}{2} x^{4} - 4 x^{3} + 8 x^{2} + 46 : 46

Minimum (0, 46)

Plug

x = 2

into

\frac{1}{2} x^{4} - 4 x^{3} + 8 x^{2} + 46 : 54

Maximum (2, 54)

Plug

x = 4

into

\frac{1}{2} x^{4} - 4 x^{3} + 8 x^{2} + 46 : 46

Minimum (4, 46)

Minimum (0, 46), Maximum (2, 54), Minimum (4, 46)

f (x) = - x^{4} + 8 x^{2} + 5

f (x) = 3 x^{2} - 12 x

f (x) = 2 x^{4} + 16 x^{3} + 27

f (x, y) = x^{2} - x y + y^{2} + 3

f (x, y) = 2 x^{3} + y^{3} + 3 x^{2} - 3 y - 12 x - 4

What is the extreme f(x)= 1/2 x^4-4x^3+8x^2+46 ?
The extreme f(x)= 1/2 x^4-4x^3+8x^2+46 is Minimum(0,46),Maximum(2,54),Minimum(4,46)