Composite functions: domain, range and when fg exists

In $fg(x)$, the function next to $x$ acts first. So $fg(x)=f(g(x))$: put $x$ into $g$, then put the answer into $f$.

Building fg and gf

Take $f(x)=2x+1$ and $g(x)=x^2$.

These are different functions. Their ranges differ too: $fg(x)\ge 1$, while $gf(x)\ge 0$.

When does fg exist?

$fg$ exists only when the range of $g$ lies inside the domain of $f$. Every output of $g$ must be something $f$ can accept.

Take $f(x)=\sqrt{x+3}$ for $x\ge -3$, and $g(x)=x^2+4x$.

• If $g$ is defined for all real $x$, its smallest value is $g(-2)=-4$. But $f$ cannot take $-4$, because it needs inputs of at least $-3$. So $fg$ does not exist.
• If $g$ is restricted to $x\ge 0$, then $g$ only takes values $g(x)\ge 0$, which $f$ accepts. Now $fg$ exists.

Domain and range of fg

With $g$ restricted to $x\ge 0$:

• Domain of $fg$: the domain of $g$, so $x\ge 0$.
• Range of $fg$: feed the range of $g$ into $f$. $fg(x)=\sqrt{x^2+4x+3}$, smallest at $x=0$, where it is $\sqrt{3}$. So $fg(x)\ge \sqrt{3}$.

Try it: does the range of g fit the domain of f?

Predict each round first, then watch the range of $g$ drop into the domain of $f$.

Mistakes that cost marks

1. Applying $f$ first in $fg$. Read it as $f(g(x))$.
2. Giving the domain of $f$ as the domain of $fg$. It is the domain of $g$ (restricted further if needed).
3. Saying $fg$ exists without checking that the range of $g$ fits inside the domain of $f$.

Practise with exam questions: 9709 functions practice.