HA7 - Composite and inverse functions

Functions and allowed inputs

A function assigns exactly one output to each allowed input. In $f(x)=3x-2$, $f(x)$ means the output of function $f$ at input $x$; it does not mean $f\times x$. The domain is the set of allowed inputs. The range is the set of outputs actually produced from that domain.

Worked example 1: numerical and algebraic inputs

Question. Let $f(x)=3x-2$, with all real inputs allowed. Find $f(-4)$, $f(2/3)$ and $f(a+1)$.

Plan. Substitute the whole input for $x$, then apply the rule's operations.

The brackets in the final line ensure that both terms in $a+1$ are multiplied by three.

Answer. $-14$, $0$ and $3a+1$, respectively.

Check. Reversing the numerical operations gives $(-14+2)/3=-4$ and $(0+2)/3=2/3$. At $a=0$, the original input is one and $f(1)=1$, matching $3a+1=1$.

Worked example 2: a restricted domain

Question. Let $g(x)=12/(x-3)$. State its real domain and find $g(-1)$, $g(5)$ and, if possible, $g(3)$.

Plan. Identify a zero denominator before evaluating the permitted inputs.

The denominator is zero when $x-3=0$, so three is excluded. Then

At three the denominator would be zero; division by zero has no defined real value.

Answer. Domain: all real $x$ with $x\ne3$. The outputs are $g(-1)=-3$, $g(5)=6$; $g(3)$ is undefined.

Check. Output times denominator recovers twelve: $(-3)(-4)=12$ and $6(2)=12$. No output multiplied by zero could recover twelve, confirming why the excluded input cannot work.

Watch out

$f(2)$ means apply the rule to two. Writing $2f(x)$ instead means double the output at $x$, which is a different instruction.

Composition

The composite function $fg(x)$ means $f(g(x))$: apply $g$ first, then apply $f$ to its output. Read the function nearest the input first. Both stages must be defined: $x$ must be in the domain of $g$, and $g(x)$ must be an allowed input of $f$.

Worked example 3: comparing the two orders

Question. Let $f(x)=2x-3$ and $g(x)=x^2+1$, both with all real inputs. Find $fg(x)$, $gf(x)$, $fg(2)$ and $gf(2)$.

Plan. Write the nested brackets first, replacing the outer rule's input by the complete inner output.

Both composite domains are all real numbers. At two, these give $2(2)^2-1=7$ and $4(2)^2-12(2)+10=2$.

Answer. $fg(x)=2x^2-1$, $gf(x)=4x^2-12x+10$; $fg(2)=7$, $gf(2)=2$.

Check. Direct machine routes give $2\to5\to7$ for $g$ then $f$, and $2\to1\to2$ for $f$ then $g$. This checks the order independently of the expanded expressions.