What is the extreme f(x)=x^4-2x^3-11x^2+12x+36 ?

The extreme f(x)=x^4-2x^3-11x^2+12x+36 is Minimum(-2,0),Maximum(1/2 , 625/16),Minimum(3,0)

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extreme f(x)=x^4-2x^3-11x^2+12x+36

extreme points

f (x) = x^{4} - 2 x^{3} - 11 x^{2} + 12 x + 36

Minimum (- 2, 0), Maximum (\frac{1}{2}, \frac{625}{16}), Minimum (3, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 6 x^{2} - 22 x + 12

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 2

into

x^{4} - 2 x^{3} - 11 x^{2} + 12 x + 36 : 0

Minimum (- 2, 0)

Plug

x = \frac{1}{2}

into

x^{4} - 2 x^{3} - 11 x^{2} + 12 x + 36 : \frac{625}{16}

Maximum (\frac{1}{2}, \frac{625}{16})

Plug

x = 3

into

x^{4} - 2 x^{3} - 11 x^{2} + 12 x + 36 : 0

Minimum (3, 0)

Minimum (- 2, 0), Maximum (\frac{1}{2}, \frac{625}{16}), Minimum (3, 0)

y = 9 x^{3} - 7

x^{3} - 12 x^{2} + 45 x + 8

x^{4} - 4 x^{2} - 2, - 2 \leq x \leq 2

f (x, y) = \frac{y ^{2} - x}{x ^{2} + 1}

f (x) = \frac{8 x}{x ^{2} + 1}

What is the extreme f(x)=x^4-2x^3-11x^2+12x+36 ?
The extreme f(x)=x^4-2x^3-11x^2+12x+36 is Minimum(-2,0),Maximum(1/2 , 625/16),Minimum(3,0)