# Is the inverse of a function also a function?If the function is f(x)=square root (4x+8) then is also a function and it is given by f^-1=1/square root(4x+8)?

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You need not to confuse the inverse of the function with the reverse of equation of the function.

You should follow the next steps to evaluate the inverse of the given function, such that:

`y = sqrt(4x + 8) => y^2 = (sqrt(4x + 8))^2`

`y^2 = 4x + 8`

You need to express x in terms of y such that:

`y^2 -8 = 4x => x = (y^2 - 8)/4`

**Hence, evaluating the inverse of the given function `f(x) = sqrt(4x + 8)` yields the function **`f^(-1)(x) = (x^2 - 8)/4.`

The inverse function is also a function but it is not the inverse of the expression of the given function.

First, we'll note y= sqrt(4x+8).

Now we'll change x by y:

x= sqrt(4y+8)

We'll raise to square both sides to eliminate the square root:

x^2 = 4y + 8

We'll isolate y to the left side:

-4y = 8 - x^2

We'll divide by -4

y = x^2/4 - 2

**The inverse function is:**

**f^-1(x) = x^2/4 - 2**

We notice that the expression of the inverse function is also a function but it is not the inversed of the expression of the original function!

Before defining the inverse of a function we need to have the right mental image of function.

At f(2): the regular function f(x) = sqrt(4x+8) will make sqrt(16) = 4.

It helps to think that 2 transforms into 4.

Now that we think of f as "acting on" numbers and transforming them, we can define the inverse of f as the function that "undoes" what f did. In other words, the inverse of f needs to take 4 back to 2.

To find f^-1 or y, we need to:

a) exchange x and y:

x=sqrt(4y+8)

b) solve for y

x^2 = 4y+8

4y = x^2 - 8

y = (x^2 - 8) / 4

**The Inverse of f(x) is f^-1(x) which is (x^2-8)/4.**

**Thus, f^-1 is not equal to 1/square root(4x+8).**

sources used: http://uncw.edu/courses/mat111hb/functions/inverse/inverse.html#sec1