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

The extreme f(x)=1+6x^2-x^3 is Minimum(0,1),Maximum(4,33)

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

extreme points

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

Minimum (0, 1), Maximum (4, 33)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

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

Minimum (0, 1)

Plug

x = 4

into

1 + 6 x^{2} - x^{3} : 33

Maximum (4, 33)

Minimum (0, 1), Maximum (4, 33)

2 x + 10

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

f (x, y) = 72 y^{2} + x^{2} - x^{2} y

f (x, y) = x^{2} - 6 x y + y^{3} + 3 x + 6 y - 2

f (x) = 3 x^{\frac{2}{3}}

What is the extreme f(x)=1+6x^2-x^3 ?
The extreme f(x)=1+6x^2-x^3 is Minimum(0,1),Maximum(4,33)