What is the extreme f(x)=6x^4-8x^3-24x^2+1 ?

The extreme f(x)=6x^4-8x^3-24x^2+1 is Minimum(-1,-9),Maximum(0,1),Minimum(2,-63)

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extreme f(x)=6x^4-8x^3-24x^2+1

extreme points

f (x) = 6 x^{4} - 8 x^{3} - 24 x^{2} + 1

Minimum (- 1, - 9), Maximum (0, 1), Minimum (2, - 63)

Solution steps

f^{'} (x) = 24 x^{3} - 24 x^{2} - 48 x

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

6 x^{4} - 8 x^{3} - 24 x^{2} + 1 : - 9

Minimum (- 1, - 9)

Plug

x = 0

into

6 x^{4} - 8 x^{3} - 24 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 2

into

6 x^{4} - 8 x^{3} - 24 x^{2} + 1 : - 63

Minimum (2, - 63)

Minimum (- 1, - 9), Maximum (0, 1), Minimum (2, - 63)

f (x, y) = y + x e^{y}

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

2 x^{3} - 3 x y + 3 y^{3}

f (x) = 9 x - x^{3}

f (x) = x^{2} + 2 + \frac{243}{x}

What is the extreme f(x)=6x^4-8x^3-24x^2+1 ?
The extreme f(x)=6x^4-8x^3-24x^2+1 is Minimum(-1,-9),Maximum(0,1),Minimum(2,-63)