# If f(x)=square root (6x+12) then f^-1=1/square root(6x+12)?

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### 2 Answers

f(x) = (6x+12)^(1/2).

To find the inverse function of f(x).

Let f^(-1) (x) = y be the inverse function of f(x).

The x = f(y).

By definition, f(y) = put y in place of x iin (6x+12)^1/2).

=> f(y) = (6y+12)^(1/2).

=> x = (6y+12)^(1/2).

We square both sides:

x^2 = 6y+12.

x^2-12 = 6y.

Therefore y = 1/6(x^2-12).

Therefore f^(-1) (x) = y = (1/6)x^2 - 2 is the inverse of f(x). f^(1) (x) = y i= not 1/sqrt(6x+12).

Absolutely not!

f^-1(x) is the inverse function of f(x).

First, we'll note y= sqrt(6x+12).

Now we'll change x by y:

x= sqrt(6y+12)

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

x^2 = 6y + 12

We'll isolate y to the left side:

-6y = 12 - x^2

We'll divide by -6:

y = x^2/6 - 2

**The inverse function is:**

**f^-1(x) = x^2/6 - 2**

We notice that the expression of the inverse function is not the inversed of the expression of the original function!