$f(2)$ means "substitute 2 in" — a calculation. $f(x)=2$ means "find the $x$ that gives output 2" — an equation to solve. These are commonly confused but are opposite directions of the same mapping.

Work from the inside out: the function closest to the bracket happens first. For $f(x)=2x+1$, $g(x)=\dfrac1x$: find $f(2)=5$ first, then $g(5)=\dfrac15$. So $gf(2)=\dfrac15$.

Substitute the entire expression for the inner function into every $x$ of the outer function.

$f(x)=x+2,\ g(x)=x^2+1$:

Note $fg(x)\neq gf(x)$ in general.

Worked example: $f(x)=4x-1$, $g(x)=kx^2$. Given $fg(2)=12$, find $k$.

$g(2)=4k$, so $f(4k)=4(4k)-1=16k-1=12 \Rightarrow k=\dfrac{13}{16}$.

"Swap, rearrange, rename": write $y=f(x)$, swap every $x$ and $y$, solve for $y$, then rename it $f^{-1}(x)$.

$f(x)=3x-5$: $y=3x-5 \to x=3y-5 \to y=\dfrac{x+5}{3} \to f^{-1}(x)=\dfrac{x+5}{3}$.

Clear the denominator, expand, collect every $y$-term on one side, factorise $y$ out, then isolate it.

Worked example: $f(x)=\dfrac{3x+1}{x-2}$.

1. $y=\dfrac{3x+1}{x-2} \to$ swap: $x=\dfrac{3y+1}{y-2}$.
2. Clear fraction: $x(y-2)=3y+1 \to xy-2x=3y+1$.
3. Collect: $xy-3y=2x+1$.
4. Factorise: $y(x-3)=2x+1$.
5. Isolate: $f^{-1}(x)=\dfrac{2x+1}{x-3}$.

A quadratic is many-to-one, so it only has an inverse over a restricted domain. Complete the square first, then use that restricted domain to choose the correct $\pm$ sign.

Worked exam-style question: $f(x)=5+6x-x^2$ for $x\le3$. Find $f^{-1}(x)$, then the values of $x$ for which $f^{-1}(x)>0$.

1. Complete the square: $f(x)=14-(x-3)^2$.
2. Swap: $x=14-(y-3)^2 \Rightarrow (y-3)^2=14-x \Rightarrow y=3\pm\sqrt{14-x}$.
3. Choose the sign using the domain/range swap rule: domain of $f$ ($x\le3$) becomes the range of $f^{-1}$, so $f^{-1}(x)\le3$ — this forces the minus sign: $f^{-1}(x)=3-\sqrt{14-x}$.
4. Solve $3-\sqrt{14-x}>0 \Rightarrow 3>\sqrt{14-x}\Rightarrow 9>14-x\Rightarrow x>5$.
5. Combine with the domain of $f^{-1}$ (range of $f$ is $f(x)\le14$, so domain of $f^{-1}$ is $x\le14$): final answer $5<x\le14$.

Since $y=f(x)$ and $y=f^{-1}(x)$ are reflections in the line $y=x$, any intersection point must lie on $y=x$ — so just solve $f(x)=x$ instead of the much messier $f(x)=f^{-1}(x)$.

• Reading $fg(x)$ left-to-right instead of applying the inner function first.
• Treating $f^{-1}(x)$ as a reciprocal.
• Trying to rearrange a quadratic for $x$ without completing the square first.
• Forgetting to use the original domain to pick the correct sign after taking a square root.