1.2 - Combining functions

Composite function

The composite function $gf$ means $g$ after $f$: first calculate $f(x)$, then use that output as the input to $g$. It is not multiplication of $g(x)$ by $f(x)$.

Order of composition

In general, $gf\ne fg$. Each composition also requires the inner outputs to be permitted inputs to the outer function.

Not in the formula booklet: learn it.

Worked example 1: applying g after f

Question. Functions $f$ and $g$ are defined for all real $x$ by $f(x)=2x-1$ and $g(x)=x^2+3$. Find $gf(x)$ and $gf(2)$.

Plan. Put $2x-1$ into the square in $g$, then simplify.

Answer. $gf(x)=4x^2-4x+4$, and $gf(2)=12$.

Check. Following the original chain, $f(2)=3$ and $g(3)=12$, agreeing with the composite expression.

Worked example 2: applying f after g

Question. For the same functions $f(x)=2x-1$ and $g(x)=x^2+3$, find $fg(x)$ and $fg(2)$.

Plan. Replace the input in $f$ by $x^2+3$.

Answer. $fg(x)=2x^2+5$, and $fg(2)=13$.

Check. $g(2)=7$, then $f(7)=13$. The values $12$ and $13$ confirm that reversing the order can change the result.

Watch out

The wrong line $g(2x-1)=2x^2-1+3$ substitutes into only part of the square. The correct line is $g(2x-1)=(2x-1)^2+3$.

Condition for a composite

The composite $gf$, with domain equal to the stated domain of $f$, exists only when

Here $\subseteq$ means “is contained in”, including the possibility of equality.

Not in the formula booklet: learn it.

Worked example 3: including the boundary

Question. Let $f(x)=x-3$ for $x\ge0$, and $g(x)=\sqrt{x}$ for $x\ge0$. Determine whether $gf$ exists for the stated functions. If it does not, find the largest restriction of the domain of $f$ that permits it, and give the composite rule.

Plan. Compare the whole range of $f$ with the domain of $g$, then require $f(x)\ge0$.

The range of $f$ is $y\ge-3$. It includes negative values, which are outside the domain of $g$; for instance, $f(0)=-3$.

For a restriction that works,

This is already inside the original domain $x\ge0$.

Answer. The original $gf$ does not exist on $x\ge0$. Restricting $f$ to $x\ge3$ permits the rule $gf(x)=\sqrt{x-3}$, with domain $x\ge3$.

Check. At the boundary $x=3$, the chain is $3\mapsto0\mapsto0$, which is defined. Every $x\ge3$ gives a non-negative inner output.

Worked example 4: excluding the boundary

Question. Let $f(x)=x-3$ for $x\ge0$, and $h(x)=1/\sqrt{x}$ for $x>0$. Determine whether $hf$ exists for the stated functions. Find the largest restriction of the domain of $f$ that permits it and give its rule.

Plan. Require the inner output to be strictly positive, because it enters a square root in a denominator.

The range $y\ge-3$ of $f$ is not contained in $x>0$. For the restricted function,

Answer. The original $hf$ does not exist on $x\ge0$. Restricting $f$ to $x>3$ gives $hf(x)=1/\sqrt{x-3}$, with domain $x>3$.

Check. At $x=4$, the chain gives $1$. At $x=3$, it would require $1/0$, confirming why equality must be excluded.

Exam tip

Support an existence decision with the relevant range and domain. A successful numerical example proves only that one chain works; the condition must hold for every input in the inner function's stated domain.

Worked example 5: a non-negative input

Question. Let $f(x)=x^2$ for $x\ge0$, and $g(x)=2x+1$ for $x\in\mathbb R$. Solve $gf(x)=19$.

Plan. Form $g(f(x))$, solve the equation, then apply the domain of $f$.

The range of $f$ is $y\ge0$, contained in the domain of $g$, so the composite exists on $x\ge0$.

The real numbers whose square is $9$ are $3$ and $-3$. Only $3$ is in the domain.

Answer. $x=3$.

Check. $f(3)=9$, then $g(9)=19$. The rejected value $-3$ is not a permitted input to $f$, even though it satisfies the algebraic equation.

Worked example 6: a non-positive input

Question. Let $h(x)=x^2$ for $x\le0$, and $g(x)=2x+1$ for $x\in\mathbb R$. Solve $gh(x)=19$.

Plan. Use the same composite algebra, then keep the candidate allowed by $x\le0$.

Only the negative candidate belongs to the stated domain.

Answer. $x=-3$.

Check. $h(-3)=9$, then $g(9)=19$. The positive candidate would require evaluating $h(3)$, which is undefined.

Watch out

“$x=\pm3$” solves $x^2=9$ over all real numbers. It is not the final answer when the original function restricts $x$ to one side of zero. Test candidates against the original domain before substituting them into the original chain.