What is the extreme f(x)=(x+1)^5-5x-2 ?

The extreme f(x)=(x+1)^5-5x-2 is Maximum(-2,7),Minimum(0,-1)

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extreme f(x)=(x+1)^5-5x-2

extreme points

f (x) = (x + 1)^{5} - 5 x - 2

Maximum (- 2, 7), Minimum (0, - 1)

Solution steps

f^{'} (x) = 5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

(x + 1)^{5} - 5 x - 2 : 7

Maximum (- 2, 7)

Plug

x = 0

into

(x + 1)^{5} - 5 x - 2 : - 1

Minimum (0, - 1)

Maximum (- 2, 7), Minimum (0, - 1)

f (x) = 5 x^{5} - 3 x^{3}

f (x) = \frac{- 5 x ^{2} + 16 x}{2 4 - x}

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

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

f (x, y) = x^{2} + x y + \frac{1}{2} y^{2} - 5 x + y

What is the extreme f(x)=(x+1)^5-5x-2 ?
The extreme f(x)=(x+1)^5-5x-2 is Maximum(-2,7),Minimum(0,-1)