What is the critical y=(x^2)/(x^2-16) ?

The critical y=(x^2)/(x^2-16) is x=0

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critical y=(x^2)/(x^2-16)

critical points

y = \frac{x ^{2}}{x ^{2} - 16}

x = 0

Solution steps

\frac{x ^{2}}{x ^{2} - 16}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x ^{2} - 16} : x < - 4 or - 4 < x < 4 or x > 4

\frac{x ^{2}}{x ^{2} - 16}

is not defined at

x = - 4, x = 4

, therefore:

x = 0

f (x) = x^{2} + 4 y^{2} + 6

f (x, y) = 7 x^{2} + 2 x y^{2} - 6 x + 1

f (x) = e^{5 x} (75 x^{2} + 3)

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

x^{4} + 4 x^{3}