# Decide if g(x)=square root(x-3) has inverse for x>=0

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Let us replace g(x) with y.

=> y = sqrt (x - 3)

square the two sides

=> y^2 = x - 3

add 3 to both the sides

=> x = y^2 + 3

exchange x and y

=> y = x^2 + 3

We see that y is defined for all values of x > = 0

**Therefore the function has an inverse g^-1(x) = x^2 + 3.**

g(x) = sqrt(x-3)..

For x<3, x-3 is negative.

Therefore g(x) is not defined for x<3.

Therefore there is no inverse of g(x) for x< 3.

For x> 3, g(x) has the inverse..

Let g^-(x) = y be the inverse of g(x).

Then x = g(y).

As g(x) sqrt(x-3), g(y) = sqrt(y-3).

Therefore g(y) = x => sqrt(y-3) = x.

=> y-3 = x^2.

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

The first step is to put:

g(x) = y

y = sqrt(x - 3)

We'll change x by y:

x = sqrt(y - 3)

We'll square both sides:

x^2 = y - 3

We'll add 3 both sides, to isolate y:

x^2 + 3 = y

**So, the inverse function is:**

**g^-1(x) = x^2 + 3, for x>=0**