+6251 Because f(x) is not a 1 to 1 mapping the inverse doesn't exist unless you restrict the domain of x.
if you restrict x to {x:x >= 7} you could find
$$y=x^2+7$$
$$y-7=x^2$$
$$x=\sqrt{y-7}$$
$$f^{-1}(x)=\sqrt{x-7}$$
on the other hand if you restrict x to {x: x <= -7} you get
$$f^{-1}(x)=\sqrt{7-x}$$
+6251 Because f(x) is not a 1 to 1 mapping the inverse doesn't exist unless you restrict the domain of x.
if you restrict x to {x:x >= 7} you could find
$$y=x^2+7$$
$$y-7=x^2$$
$$x=\sqrt{y-7}$$
$$f^{-1}(x)=\sqrt{x-7}$$
on the other hand if you restrict x to {x: x <= -7} you get
$$f^{-1}(x)=\sqrt{7-x}$$
