What is the extreme f(x)=2x^5+2x^4-x^3+6x^2+8x+4 ?

The extreme f(x)=2x^5+2x^4-x^3+6x^2+8x+4 is Maximum(-1.35388…,4.27058…),Minimum(-0.55481…,1.66354…)

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extreme f(x)=2x^5+2x^4-x^3+6x^2+8x+4

extreme points

f (x) = 2 x^{5} + 2 x^{4} - x^{3} + 6 x^{2} + 8 x + 4

Maximum (- 1.35388 \dots, 4.27058 \dots), Minimum (- 0.55481 \dots, 1.66354 \dots)

Solution steps

Find the critical points:

x \approx - 1.35388 \dots, x \approx - 0.55481 \dots

Domain of

2 x^{5} + 2 x^{4} - x^{3} + 6 x^{2} + 8 x + 4 : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 1.35388 \dots,

Decreasing

: - 1.35388 \dots < x < - 0.55481 \dots,

Increasing

: - 0.55481 \dots < x < \infty

Plug the extreme point

x = - 1.35388 \dots

into

2 x^{5} + 2 x^{4} - x^{3} + 6 x^{2} + 8 x + 4 \Rightarrow y = 4.27058 \dots

Maximum (- 1.35388 \dots, 4.27058 \dots)

Plug the extreme point

x = - 0.55481 \dots

into

2 x^{5} + 2 x^{4} - x^{3} + 6 x^{2} + 8 x + 4 \Rightarrow y = 1.66354 \dots

Minimum (- 0.55481 \dots, 1.66354 \dots)

Maximum (- 1.35388 \dots, 4.27058 \dots), Minimum (- 0.55481 \dots, 1.66354 \dots)

f (x) = \frac{- 1}{x - 4}

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

x^{3} \cdot e^{- x}

f (x) = 2 x^{2} - 4 x + y^{2} - 4 y + 6

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

What is the extreme f(x)=2x^5+2x^4-x^3+6x^2+8x+4 ?
The extreme f(x)=2x^5+2x^4-x^3+6x^2+8x+4 is Maximum(-1.35388…,4.27058…),Minimum(-0.55481…,1.66354…)