What is the extreme f(x)=6x^4-8x^3-24x^2+35 ?

The extreme f(x)=6x^4-8x^3-24x^2+35 is Minimum(-1,25),Maximum(0,35),Minimum(2,-29)

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extreme f(x)=6x^4-8x^3-24x^2+35

extreme points

f (x) = 6 x^{4} - 8 x^{3} - 24 x^{2} + 35

Minimum (- 1, 25), Maximum (0, 35), Minimum (2, - 29)

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} + 35 : 25

Minimum (- 1, 25)

Plug

x = 0

into

6 x^{4} - 8 x^{3} - 24 x^{2} + 35 : 35

Maximum (0, 35)

Plug

x = 2

into

6 x^{4} - 8 x^{3} - 24 x^{2} + 35 : - 29

Minimum (2, - 29)

Minimum (- 1, 25), Maximum (0, 35), Minimum (2, - 29)

f (x) = x + \frac{13}{x}

f (x) = \frac{2 x}{x ^{2} + 9}

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

y (x, t) = x (3 t)

f (x) = 3 x^{\frac{2}{3}} - 2 x [- 1.1]

What is the extreme f(x)=6x^4-8x^3-24x^2+35 ?
The extreme f(x)=6x^4-8x^3-24x^2+35 is Minimum(-1,25),Maximum(0,35),Minimum(2,-29)