What is the critical y=(sqrt(1-x^2))/x ?

The critical y=(sqrt(1-x^2))/x is x=-1,x=1

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critical y=(sqrt(1-x^2))/x

critical points

y = \frac{1 - x ^{2}}{x}

x = - 1, x = 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{1 - x ^{2}}{x} : - 1 \leq x < 0 or 0 < x \leq 1

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

is not defined at

x = 0

, therefore:

x = - 1, x = 1

[\frac{x}{3}] - [\frac{y}{5}] = 1

y = - 2 x^{2} - 8 x - 6

y = - \frac{1}{3} x - 2

f (x) = 4 + 5^{x}

2 x + 6

What is the critical y=(sqrt(1-x^2))/x ?
The critical y=(sqrt(1-x^2))/x is x=-1,x=1