Question

Find the inverse of the given function.

f(x) = -1/2√x + 3, x ≥ -3

Answer 1

Find the inverse of the given function.

f(x) = -1/2√x + 3, x ≥ -3

I will have to assume that you meant f(x) = -(1/2)sqrt(x) + 3.  If you actually meant f(x) = -(1/2)sqrt(x+3), then obviously the correct result would be different.

1.  Replace "f(x)" by "y:"  y = -(1/2)sqrt(x) + 3
 2.  Interchange x and y:  x = -(1/2)sqrt(y) + 3
3.  Solve for y:  x-3=-(1/2)sqrt(y), so that 2(3-x)= sqrt(y) and y=+sqrt(2[3-x])
4.  Replace "y" with 

           -1
        f      (x) = sqrt(2[3-x])

Here, there are restrictions on x, since the domain of the sqrt function does not include - numbers.  The domain here is (-infinity,3]

Answer 2

Answer:

The inverse of the function is $f^{-1}(x)=4x^2-3$, for $x\leq 0$

Step-by-step explanation:

Given : Function $f(x)=-\frac{1}{2}\sqrt{x+3}$

To find : The inverse of the given function?

Solution :  

To find the inverse of the function we replace the value of x and y and then find y in terms of x which is the inverse of the function.

Let f(x)=y

$y=-\frac{1}{2}\sqrt{x+3}$                

Replace the value of x and y.

$x=-\frac{1}{2}\sqrt{y+3}$                  

Now, we solve in terms of x the value of y

$-2x=\sqrt{y+3}$  

(x must be negative)

Squaring both side,

$(-2x)^2=(\sqrt{y+3})^2$  

$4x^2=y+3$  

$4x^2-3=y$  

So, The inverse of the function is $f^{-1}(x)=4x^2-3$, for $x\leq 0$