What is the extreme f(x)= 1/(x+3)-1/((x+3)^2) ?

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

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extreme f(x)= 1/(x+3)-1/((x+3)^2)

extreme points

f (x) = \frac{1}{x + 3} - \frac{1}{( x + 3 ) ^{2}}

Maximum (- 1, \frac{1}{4})

Solution steps

f^{'} (x) = \frac{- x - 1}{( x + 3 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 1

into

\frac{1}{x + 3} - \frac{1}{( x + 3 ) ^{2}} : \frac{1}{4}

Maximum (- 1, \frac{1}{4})

w (x, y) = 6 (x + y)

f (x) = \frac{- cos ( 2 x )}{2} - 2 sin (x)

4 e^{2 x^{2} + 5 x + 16}

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

f (x) = x^{4} + 4 x^{2} - 10, - 5 \leq x \leq 2

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