# For the function `f(x) = x^2 - 2` and using the inverse function, represented by `f^(-1)` , find `f(f^-1(-7))` .

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### 1 Answer

`f(x)=x^2-2` , `f(f^(-1)(-7))=?`

First, determine the inverse function of f(x). To do so, replace f(x) with y.

`y=x^2-2`

Then, switch x and y.

`x=y^2-2`

And solve for y. So, add both sided by 2.

`x+2=y^2-2+2`

`x+2=y^2`

Also, take the square root of both sides.

`+-sqrt(x+2)=sqrt(y^2)`

`+-sqrt(x+2)=y`

And, replace y with `f^(-1)(x) ` to indicate that it is the inverse of f(x).

`f^(-1)(x)=+-sqrt(x+2)`

Next, solve for value of `f^(-1)(x)` when x=-7.

`f^(-1)(-7)=+-sqrt(-7+2)=+-sqrt(-5)`

Since the radicand of the square root is negative, to simplify apply `i=sqrt(-1)` .

`f^(-1)(x)=+-isqrt5`

So,when x=-7, the values of the inverse function are:

`f^(-1)(-7)=isqrt5` and `f^(-1)(-7)=-sqrt5`

Next, plug-in the values of the inverse function to f(x) to get `f(f^(-1)(-7))` . This is the same as solving for the values of f(x) when `x=isqrt5` and `x=-isqrt5` .

`f(f^-1)(-7)) = f(isqrt5) ` and `f(f^(-1)(-7)=f(-sqrt5)`

So when `x=isqrt5` , the value of f(x) is:

`f(x)=x^2-2`

`f(isqrt5)=(isqrt5)^2-2=5i^2-2`

Since i^2=-1, then:

`f(isqrt5)=5(-1)-2=-5-2`

`f(isqrt5)=-4`

And when `x=-isqrt5` , the value of f(x) is:

`f(x)=x^2-2`

`f(-isqrt5)=(-isqrt5)-2=5i^2-2=5(-1)-2=-5-2`

`f(-isqrt5)=-7`

**Since the value of f(x) is the same for both values of x, hence `f(f^(-1)(-7))=-7` .**