What is the extreme y=|x+4|+4/(|x|-3) ?

The extreme y=|x+4|+4/(|x|-3) is Minimum(-5,3),Maximum(1,3),Minimum(5,11)

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extreme y=|x+4|+4/(|x|-3)

extreme points

y = ∣ x + 4 ∣ + \frac{4}{∣ x ∣ - 3}

Minimum (- 5, 3), Maximum (1, 3), Minimum (5, 11)

Solution steps

Find the critical points:

x = - 5, x = - 4, x = 0, x = 1, x = 5

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < - 4,

Increasing

: - 4 < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 3,

Decreasing

: 3 < x < 5,

Increasing

: 5 < x < \infty

Plug the extreme point

x = - 5

into

∣ x + 4 ∣ + \frac{4}{∣ x ∣ - 3} \Rightarrow y = 3

Minimum (- 5, 3)

Plug the extreme point

x = 1

into

∣ x + 4 ∣ + \frac{4}{∣ x ∣ - 3} \Rightarrow y = 3

Maximum (1, 3)

Plug the extreme point

x = 5

into

∣ x + 4 ∣ + \frac{4}{∣ x ∣ - 3} \Rightarrow y = 11

Minimum (5, 11)

Minimum (- 5, 3), Maximum (1, 3), Minimum (5, 11)

6 e^{x - 4}

f (x) = \frac{9 x}{x ^{2} + 6}

f (x) = x^{2} - 6 x + 9 - 4 y^{2}

f (x) = x^{\frac{8}{7}} + 8 x^{\frac{1}{7}}

f (x) = \frac{x}{x - 6}

What is the extreme y=|x+4|+4/(|x|-3) ?
The extreme y=|x+4|+4/(|x|-3) is Minimum(-5,3),Maximum(1,3),Minimum(5,11)