What is the critical f(x)=(x^2)/(x^2-49) ?

The critical f(x)=(x^2)/(x^2-49) is x=0

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critical f(x)=(x^2)/(x^2-49)

critical points

f (x) = \frac{x ^{2}}{x ^{2} - 49}

x = 0

Solution steps

\frac{x ^{2}}{x ^{2} - 49}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 7, x = 7

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x ^{2} - 49} : x < - 7 or - 7 < x < 7 or x > 7

\frac{x ^{2}}{x ^{2} - 49}

is not defined at

x = - 7, x = 7

, therefore:

x = 0

y = x ln (5 x)

x^{2} - x

- x^{3} + 6 x^{2} - 12 x + 30

f (x) = e^{- x} - e^{- 2 x}

f (x) = \frac{x ^{3}}{3} + x^{2} - 8 x - 3

What is the critical f(x)=(x^2)/(x^2-49) ?
The critical f(x)=(x^2)/(x^2-49) is x=0