What is the inverse function of f(x) = -sqrt ( 4 - x^2) for -2<= x < = 0?  

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the function:

f(x) = -sqrt(4-x^2)

We need to find the inverse.

Let us assume that y= - sqrt(4-x^2)

Let us square both sides.

==> y^2 = (4-x^2)

Now we will rewrite:

==> x^2 = 4- y^2

==> x = +- sqrt(4-y^2)

Now we will replace y and x.

==> y= +-sqrt( 4-x^2)

But since -2 =< x =< 0 , then we will not consider sqrt(4-x^2) as an inverse.

Then the inverse function is:

f^-1 ( x) = - sqrt(4-x^2) when  -2 =< x =< 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the inverse function of f(x) = -sqrt (4 – x^2)

Now as -2<= x < = 0, we have -2<= y < = 0, also.

Let y = f(x) = -sqrt (4 - x^2)

=> -y = sqrt (4 – x^2)

=> y^2 = 4 – x^2

=> x^2 = 4 – y^2

=> x = sqrt (4 – y^2) or x = -sqrt (4 – y^2)

We have to select x = -sqrt (4 – y^2), due to the condition imposed earlier.

Interchange y and x

=> y = f^-1(x) = -sqrt (4 – x^2)

Therefore we see that the inverse function f^-1(x) = -sqrt (4 - x^2) for -2<= x < = 0, is the same function.

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