For this case we must find the inverse of the following function:
![f (x) = - \frac {1} {2} \sqrt {x + 3}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t5v4w2owapyr4viu9eygkhl5yufpa0wxfh.png)
We follow the steps below:
Replace f(x) with y:
![y = -\frac {1} {2} \sqrt {x + 3}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pb3ir1k675hq6u7of88mc2dvj737eamvgw.png)
We exchange the variables:
![x = - \frac {1} {2} \sqrt {y + 3}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/iarb9vckw8zhmt5yu5hdwzv6t60x97mhoa.png)
We solve for "y":
![- \frac {1} {2} \sqrt {y + 3} = x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xtl2tq5ck27t4tmfmzd84mvd5s8xb9uzsr.png)
Multiply by -2 on both sides of the equation:
![\sqrt {y + 3} = - 2x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/j3yfe44jmaxkiem59ttnu2clmc5x891unj.png)
We raise both sides of the equation to the square to eliminate the radical:
![(\sqrt {y + 3}) ^ 2 = (- 2x) ^ 2\\y + 3 = 4x ^ 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7hcfkxtkpgeembn0pdsc3woe3wvdo13axt.png)
We subtract 3 from both sides of the equation:
![y = 4x ^ 2-3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/y6rwto2e401fb8ez7d75ppt4s72knakgm6.png)
We change y by f ^ {- 1} (x):
![f ^ {- 1} (x) = 4x ^ 2-3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/93tf2u09hopimtu8dg9miywvm3usjg57as.png)
Answer:
![f ^ {- 1} (x) = 4x ^ 2-3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/93tf2u09hopimtu8dg9miywvm3usjg57as.png)