What is the extreme f(x)=|-4x-5|,-5<x<1 ?

The extreme f(x)=|-4x-5|,-5<x<1 is Minimum(-5/4 ,0)

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extreme f(x)=|-4x-5|,-5<x<1

extreme points

f (x) = ∣ - 4 x - 5 ∣, - 5 < x < 1

Minimum (- \frac{5}{4}, 0)

Solution steps

Find the critical points:

x = - \frac{5}{4}

Find intervals:

Decreasing

: - 5 < x < - \frac{5}{4},

Increasing

: - \frac{5}{4} < x < 1

Plug the extreme point

x = - \frac{5}{4}

into

∣ - 4 x - 5 ∣ \Rightarrow y = 0

Minimum (- \frac{5}{4}, 0)

f (x, y) = 3 x^{6} y - 5 x y^{2}

f (x) = x^{3} e^{- 4 x}, 0 \leq x \leq 2

f (x) = 3 x^{2} + 12 x y + 9 x^{2} + y^{3}

V (u, w) = \frac{1}{2} (3 u + w)

f (x) = (4 - x^{2})^{2}

What is the extreme f(x)=|-4x-5|,-5<x<1 ?
The extreme f(x)=|-4x-5|,-5<x<1 is Minimum(-5/4 ,0)