What is the critical ((x+4)^2)/((x-3)^2) ?

The critical ((x+4)^2)/((x-3)^2) is x=-4

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critical ((x+4)^2)/((x-3)^2)

critical points

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

x = - 4

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{( x + 4 ) ^{2}}{( x - 3 ) ^{2}} : x < 3 or x > 3

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

is not defined at

x = 3

, therefore:

x = - 4

f (x) = 2 cos (4 x)

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

ψ (x) = x^{2} + 3 x, - 2 \leq x \leq 1

f (x) = 3 x (x - 4)

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

What is the critical ((x+4)^2)/((x-3)^2) ?
The critical ((x+4)^2)/((x-3)^2) is x=-4