Q.Let $f: \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}$ and $g: \{3, 4, 5, 9\} \to \{7, 11, 15\}$ be functions defined as $f(2) = 3$, $f(3) = 4$, $f(4) = f(5) = 5$ and $g(3) = g(4) = 7$ and $g(5) = g(9) = 11$. Find $gof$.

Concept: Domain Of Composite Function — $gof$ is defined only for those $x$ in the domain of $f$ for which $f(x)$ lies in the domain of $g$.

Step 1: Domain of $f$ is $\{2,3,4,5\}$. Compute $f(x)$ for each:

• $f(2)=3$, $f(3)=4$, $f(4)=5$, $f(5)=5$.

Step 2: Domain of $g$ is $\{3,4,5,9\}$. All outputs $3,4,5$ are in this set, so $gof$ is defined on the whole domain of $f$.

Step 3: Apply $g$ to each $f(x)$:

• $g(f(2)) = g(3) = 7$
• $g(f(3)) = g(4) = 7$
• $g(f(4)) = g(5) = 11$
• $g(f(5)) = g(5) = 11$

Why this approach works

When you compose two functions, you're essentially applying one after the other: first $f$, then $g$. But there's a catch — the output of $f$ must be a valid input for $g$. That means the range of $f$ (the set of all values $f$ actually produces) must be a subset of the domain of $g$ (the set of values $g$ can accept). If any $f(x)$ lands outside $g$'s domain, then $g(f(x))$ is simply not defined for that $x$.

Here, both functions are given explicitly as finite sets of ordered pairs, so we can compute $g(f(x))$ for each $x$ in the domain of $f$ by direct substitution — but we must check each time that $f(x)$ actually belongs to the domain of $g$.

Step-by-step computation

1. Identify the domains and ranges

• Domain of $f$: $\{2, 3, 4, 5\}$
• Codomain of $f$: $\{3, 4, 5, 9\}$ (but the actual range is $\{3, 4, 5\}$ since $f(4)=f(5)=5$)
• Domain of $g$: $\{3, 4, 5, 9\}$
• Codomain of $g$: $\{7, 11, 15\}$

Notice that every value in the range of $f$ — namely $3, 4, 5$ — is indeed in the domain of $g$. So $g \circ f$ will be defined for all $x \in \{2,3,4,5\}$.

2. Compute $g(f(2))$

$f(2) = 3$. Now $g(3) = 7$. So $g(f(2)) = 7$.

3. Compute $g(f(3))$

$f(3) = 4$. Then $g(4) = 7$. So $g(f(3)) = 7$.

4. Compute $g(f(4))$

$f(4) = 5$. Then $g(5) = 11$. So $g(f(4)) = 11$.

5. Compute $g(f(5))$

$f(5) = 5$. Then $g(5) = 11$. So $g(f(5)) = 11$.

6. Write the composite as a set of ordered pairs

The composite $g \circ f$ is the function from $\{2,3,4,5\}$ to $\{7,11,15\}$ given by:

Method: Computing a composite $g\circ f$ of finite functions

Use this when $f$ and $g$ are given as explicit input-output lists and you must find $g\circ f$.

Steps

Step 1: Check the composition is defined

$g\circ f$ needs every output of $f$ to be a valid input of $g$, i.e. $\text{range}(f)\subseteq\text{domain}(g)$. Verify this before computing.

Step 2: Apply the functions in order, right-to-left

$(g\circ f)(x)=g(f(x))$: first read off $f(x)$, then feed that value into $g$. Do this for each $x$ in the domain of $f$.

Step 3: Write the composite as ordered pairs

Collect the results into $\{(x,\ g(f(x)))\}$. Note that a composite can be many-to-one even if you weren't expecting it — just record what the computation gives, and don't assume $g\circ f$ is injective.

Common Mistakes

Mistake 1: Applying the functions in the wrong order

Why it's wrong: $g\circ f$ means $g(f(x))$ — $f$ acts first, then $g$; students sometimes compute $f(g(x))$. Correct approach: read $g\circ f$ right-to-left, e.g. $(g\circ f)(2)=g(f(2))=g(3)=7$.

Mistake 2: Not checking that $f$'s outputs lie in $g$'s domain

Why it's wrong: if some $f(x)$ were outside $\text{domain}(g)$, the composite would be undefined there. Correct approach: confirm $\text{range}(f)=\{3,4,5\}\subseteq\text{domain}(g)=\{3,4,5,9\}$ before composing.