Functions: Question 7

Worked solution

Part (a): Finding $fg(x)$

$fg(x)$ means $f(g(x))$, apply $g$ first, then $f$.

$fg(x) = f(g(x)) = f(2x+1)$

Substitute $2x+1$ into $f(\cdot) = \dfrac{1}{(\cdot)-3}$:

$fg(x) = \frac{1}{(2x+1)-3} = \frac{1}{2x-2}$

Part (b): Domain of $fg$

$fg(x) = \dfrac{1}{2x-2}$ is undefined only when the denominator is zero:

$2x - 2 = 0 \implies x = 1$

This matches the underlying rule: $fg$ is undefined exactly when $g(x)=3$ (the value excluded from $f$‘s domain), and $g(x)=3$ gives $2x+1=3$, i.e. $x=1$.

So the domain of $fg$ is: $x \in \mathbb{R}, \quad x \ne 1$

Part (c): Finding $gf(x)$

$gf(x)$ means $g(f(x))$, apply $f$ first, then $g$: the opposite order to part (a).

$gf(x) = g(f(x)) = g\left(\frac{1}{x-3}\right)$

Substitute $\dfrac{1}{x-3}$ into $g(\cdot) = 2(\cdot)+1$:

$gf(x) = 2\left(\frac{1}{x-3}\right)+1 = \frac{2}{x-3}+1$

Part (d): Domain of $gf$

Since $g$ itself has domain $x\in\mathbb{R}$ (no restriction), the only restriction on $gf$ comes from needing $f(x)$ to be defined in the first place, i.e. $x \ne 3$.

So the domain of $gf$ is: $x \in \mathbb{R}, \quad x \ne 3$

Numerical check at $x=2$: $g(2)=5$, $f(5)=\dfrac{1}{5-3}=\dfrac{1}{2}$, so $fg(2)=\dfrac{1}{2}$. Using the formula: $\dfrac{1}{2(2)-2}=\dfrac{1}{2}$ (matches. Also $f(2)=\dfrac{1}{2-3}=-1$, $g(-1)=2(-1)+1=-1$, so $gf(2)=-1$. Using the formula: $\dfrac{2}{2-3}+1 = -2+1=-1$) matches.

Final answers

• (a) $fg(x) = \dfrac{1}{2x-2}$
• (b) Domain of $fg$: $x\in\mathbb{R}, x\ne 1$
• (c) $gf(x) = \dfrac{2}{x-3}+1$
• (d) Domain of $gf$: $x\in\mathbb{R}, x\ne 3$