# If f(x) = sqrt(3x-5) what is f^-1

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The function given is f(x) = sqrt(3x-5). We have to find the inverse function.

Let y = f(x) = sqrt(3x-5)

=> y^2 = 3x - 5

=> 3x = y^2 + 5

=> x = y^2 / 3 + 5/3

Interchange x and y

=> y = x^2/3 + 5/3

**Therefore the inverse of f(x) = sqrt(3x-5) is f^-1(x) = x^2/3 + 5/3.**

Given the function f(x) = sqrt(3x-5).

We need to find f^-1 (x) which is the inverse of the function f(x).

First, we will assume that y= sqrt(3x-5).

Now we will try to isolate x on one sides.

==> y= sqrt( 3x -5)

We will square both sides:

==> y^2 = 3x - 5

Now we will add 5 to both sides.

==> y^2 + 5 = 3x

Now we will divide by 3.

==> x = ( y^2 + 5) /3

Now we will rewrite x and y.

==> y= (x^2 + 5) /3

Then the inverse if the function f(x) is:

**f^-1 (x) = (x^2 + 5) /3 **

f(x) = sqrt(3x-5)

To find the inverse of the function 3x-5.

We know 3x-5 > 5,

So for x> 5/3, we assume f^-1(x) = y.

So f(y) = x.

=> sqrt(3y-5) = x.

=> 3y-5 = x^2.

=> 3y = x^2+5.

=> y = (x^2+5)/3.

Therefore y = (x^2+5)/3 is the inverse of y = sqrt(3x-5).