What is the extreme f(x)= 1/(1+|x|)+1/(1+|x-2|) ?

The extreme f(x)= 1/(1+|x|)+1/(1+|x-2|) is Maximum(0, 4/3),Minimum(1,1),Maximum(2, 4/3)

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extreme f(x)= 1/(1+|x|)+1/(1+|x-2|)

extreme points

f (x) = \frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 2 ∣}

Maximum (0, \frac{4}{3}), Minimum (1, 1), Maximum (2, \frac{4}{3})

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug the extreme point

x = 0

into

\frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 2 ∣} \Rightarrow y = \frac{4}{3}

Maximum (0, \frac{4}{3})

Plug the extreme point

x = 1

into

\frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 2 ∣} \Rightarrow y = 1

Minimum (1, 1)

Plug the extreme point

x = 2

into

\frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 2 ∣} \Rightarrow y = \frac{4}{3}

Maximum (2, \frac{4}{3})

Maximum (0, \frac{4}{3}), Minimum (1, 1), Maximum (2, \frac{4}{3})

f (x) = \frac{3 x}{x ^{2} + 9}

f (x, y) = 10 x y - x^{3} - 5 y^{2}

f (x) = ln (7 - 2 x^{2})

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

\frac{x}{x ^{2} + 13 x + 36}

What is the extreme f(x)= 1/(1+|x|)+1/(1+|x-2|) ?
The extreme f(x)= 1/(1+|x|)+1/(1+|x-2|) is Maximum(0, 4/3),Minimum(1,1),Maximum(2, 4/3)