What is the critical f(x)=((2x-1))/(x^2-4x-5) ?

The critical f(x)=((2x-1))/(x^2-4x-5) is None

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critical f(x)=((2x-1))/(x^2-4x-5)

critical points

f (x) = \frac{( 2 x - 1 )}{x ^{2} - 4 x - 5}

None

Solution steps

\frac{2 x - 1}{x ^{2} - 4 x - 5}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 5

Identify critical points not in

f (x)

domain

Domain of

\frac{2 x - 1}{x ^{2} - 4 x - 5} : x < - 1 or - 1 < x < 5 or x > 5

\frac{2 x - 1}{x ^{2} - 4 x - 5}

is not defined at

x = - 1, x = 5

, therefore:

None

F (t) = 2 t \frac{2}{3} + t \frac{5}{3}

f (x) = 7 (x - 2)^{\frac{2}{3}}

\frac{d}{d x} (\frac{2 x y}{x ^{2} + y ^{2}})

x^{3} - 9 x^{2} + 15 x - 5

f (x, y) = 2 x^{2} + y^{2}

What is the critical f(x)=((2x-1))/(x^2-4x-5) ?
The critical f(x)=((2x-1))/(x^2-4x-5) is None