What is the extreme f(x)=|x|-3|x+1| ?

The extreme f(x)=|x|-3|x+1| is Maximum(-1,1)

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extreme f(x)=|x|-3|x+1|

extreme points

f (x) = ∣ x ∣ - 3 ∣ x + 1 ∣

Maximum (- 1, 1)

Solution steps

Find the critical points:

x = - 1, x = 0

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < \infty

Plug the extreme point

x = - 1

into

∣ x ∣ - 3 ∣ x + 1 ∣ \Rightarrow y = 1

Maximum (- 1, 1)

f (x) = 2 cos (x) + cos^{2} (x)

f (x) = - x^{3} - 4.5 x^{2} - 6 x + 2

f (x, y) = x^{4} + y^{4} + 4 x y

f (x) = x e^{1 - \frac{x}{2}}

5 x + 1 + \frac{2}{x - 2}

What is the extreme f(x)=|x|-3|x+1| ?
The extreme f(x)=|x|-3|x+1| is Maximum(-1,1)