$f(x) = 2(x-1.5)^2 + 0.5$ for $x \le 3/2$ Find $f^{-1}(x)$ and state it's domain.
So I'm getting: $f^{-1}(x) = [(x-0.5)/2]^{\frac{1}{2}} + 1.5$ for $x ≥ 1/2$ But according to the marking scheme, I'm supposed to take the negative square root, not the positive. Need an explanation. Graphs would be a great help.
The function $f$ is defined for $x\leq \frac32$. You need to make sure that that matches the range of $f^{-1}$. In other words, you want $\frac32$ to be the largest possible value you can get from applying $f^{-1}$. That is exactly what happens if you choose $$ f^{-1}(y) = -\sqrt{\frac{y-0.5}{2}} +\frac32 $$ whereas if you choose the positive square root, then you wouldn't get the inverse of $f$. You would get the inverse of $g(x)$ defined as $$ g(x) = 2\left(x-\frac32\right)^2 + \frac12, \quad x\geq \frac32 $$ Here is a simple graph:
The red part is $f$, but you have found the inverse of the blue part, which is what I called $g$ above.