What is the critical f(x)= 5/(x^2-49) ?

The critical f(x)= 5/(x^2-49) is x=0

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critical f(x)= 5/(x^2-49)

critical points

f (x) = \frac{5}{x ^{2} - 49}

x = 0

Solution steps

\frac{5}{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{5}{x ^{2} - 49} : x < - 7 or - 7 < x < 7 or x > 7

\frac{5}{x ^{2} - 49}

is not defined at

x = - 7, x = 7

, therefore:

x = 0

f (x) = 2 x^{3} - 3 x^{2} + 7 x - 4

4 x - 3 y = 9

y = 6 x - 2

5 x + 8

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