Read functions and graphs

Let $f(x) = 2x + 1$ and $g(x) = x - 2$. To find $f(g(4))$, start inside: $g(4) = 2$. Feed that result into $f$: $f(2) = 5$.

Reversing the order changes the result. $g(f(4)) = g(9) = 7$. The notation is an instruction about order, not a product.

A transformed function works the same way. For $h(x) = -2f(x - 2) + 3$, evaluate $h(4)$ by finding $f(2) = 5$, then $-2 \times 5 + 3 = -7$. The negative outside multiplier reflects outputs; it does not change the input.

Composition also combines restrictions. Let $r(t)=\sqrt{t-4}$ and $s(x)=(x-1)^2$ for $-3\le x\le5$. A real $r(s(x))$ needs $(x-1)^2\ge4$, so $x\le-1$ or $x\ge3$. Inside the given interval, the allowed integers are $-3,-2,-1,3,4,5$: six inputs. Both boundaries are included because the radicand may equal zero.

Let $f(x) = (x - 1)^{2} + 2$, restricted to $x \le 1$. Finding $f^{-1}(11)$ means finding the allowed input that produces 11.

From $(x - 1)^{2} + 2 = 11$, we get $(x - 1)^{2} = 9$. The unrestricted equation has inputs $-2$ and 4. Only one belongs to $x \le 1$. Select that input below.

Without the restriction, two inputs could produce the same output, so the quadratic would not have an inverse function on all real inputs. Always keep the domain when undoing a function.