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

The critical f(x)=2+3/(1+(x+1)^2) is x=-1

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critical f(x)=2+3/(1+(x+1)^2)

critical points

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

x = - 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - 1

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

4 x - ln (4 x)

(6.1) (7.7)

f (x) = 4 (\frac{1}{5})^{x}

f (x) = x y - 2 y - 3 x = 2

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