Functions — FAQ

The order of operations is different, which usually leads to a different final function. For $f(g(x))$, you apply the function $g$ first, then apply $f$ to that result. For $g(f(x))$, you apply $f$ first, then apply $g$. Think of it as putting on socks then shoes ($f(g(x))$) versus shoes then socks ($g(f(x))$) – the order matters!

The function $f(x)=x^2$ is not one-to-one because different inputs can give the same output (e.g., $f(2)=4$ and $f(-2)=4$). An inverse must map an output back to a single unique input. By restricting the domain to, for example, $x \ge 0$, we ensure that every output comes from only one input, which allows an inverse function, $f^{-1}(x) = \sqrt{x}$, to exist.

The best method is to complete the square to write the quadratic $ax^2+bx+c$ in the form $a(x-h)^2+k$. The vertex is at $(h, k)$. If the parabola opens upwards (i.e., $a>0$), the range is $y \ge k$. If it opens downwards (i.e., $a<0$), the range is $y \le k$. Always remember to check if the domain is restricted, as this might alter the range.

No, this is a very important distinction. $f^{-1}(x)$ denotes the inverse function, which reverses the mapping. $\frac{1}{f(x)}$ is the reciprocal of the function's value, sometimes written as $[f(x)]^{-1}$. They are completely different concepts.