What is the extreme f(x)=0.2(x^4+4x^3-16x-16)+5 ?

The extreme f(x)=0.2(x^4+4x^3-16x-16)+5 is Saddle(-2,5),Minimum(1,-0.4)

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extreme f(x)=0.2(x^4+4x^3-16x-16)+5

extreme points

f (x) = 0.2 (x^{4} + 4 x^{3} - 16 x - 16) + 5

Saddle (- 2, 5), Minimum (1, - 0.4)

Solution steps

Find the critical points:

x = - 2, x = 1

Domain of

0.2 (x^{4} + 4 x^{3} - 16 x - 16) + 5 : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = - 2

into

0.2 (x^{4} + 4 x^{3} - 16 x - 16) + 5 \Rightarrow y = 5

Saddle (- 2, 5)

Plug the extreme point

x = 1

into

0.2 (x^{4} + 4 x^{3} - 16 x - 16) + 5 \Rightarrow y = - 0.4

Minimum (1, - 0.4)

Saddle (- 2, 5), Minimum (1, - 0.4)

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

f (x) = - x^{7} ln (3 x), (0, 4)

\frac{x ^{2}}{- x ^{2} + 4}

y = x^{x} I n (x^{5} + x)

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

What is the extreme f(x)=0.2(x^4+4x^3-16x-16)+5 ?
The extreme f(x)=0.2(x^4+4x^3-16x-16)+5 is Saddle(-2,5),Minimum(1,-0.4)