What Happens When You Add, Multiply or Divide Two Functions

Numbers can be added and multiplied, and so can functions. Profit is revenue minus cost; average cost is total cost divided by quantity. Each combination is a new function with its own domain, which must be worked out carefully.

This part covers sums and differences, products and quotients, evaluating combined functions, and reading information from graphs.

**For real functions f and g, $(f+g)(x)=f(x)+g(x)$ and $(f-g)(x)=f(x)-g(x)$, and both are defined only where f and g are both defined — on the intersection of their domains, $D_f\cap D_g$.

Rules:**

- $(f+g)(x)=f(x)+g(x)$
- $(f-g)(x)=f(x)-g(x)$
- Domain of each $=D_f\cap D_g$

Worked example. Let $f(x)=\sqrt{x}$ and $g(x)=\sqrt{4-x}$.

- $D_f=[0,\infty )$ and $D_g=(-\infty ,4]$
- $(f+g)(x)=\sqrt{x}+\sqrt{4-x}$, with domain $[0,4]$
- $(f-g)(3)=\sqrt{3}-\sqrt{1}\approx 1.732-1=0.732$

No value outside $[0,4]$ can be used, even though f alone accepts $x=9$.

An everyday example. A tea stall's profit is revenue minus cost, $P(x)=R(x)-C(x)$, where x is the number of cups sold — a difference of two functions.

The substance. The domain is fixed before simplifying — if $f(x)=\sqrt{x}$ and $g(x)=-\sqrt{x}$, then $(f+g)(x)=0$, but only for $x\ge 0$.

**The product $(fg)(x)=f(x)g(x)$ is defined on $D_f\cap D_g$, and the quotient $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$ is defined on $D_f\cap D_g$ with every x for which $g(x)=0$ removed.

Rules:**

- $(fg)(x)=f(x)g(x)$, domain $D_f\cap D_g$
- $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$, domain $\{x\in D_f\cap D_g:g(x)\ne 0\}$
- A scalar multiple $(kf)(x)=kf(x)$ keeps the domain of f

Worked example. Let $f(x)=x^2-1$ and $g(x)=x^2-5x+6$.

- $(fg)(x)=(x^2-1)(x^2-5x+6)$, with domain $\mathbb{R}$
- $g(x)=(x-2)(x-3)$ is zero at $x=2$ and $x=3$
- So $\left(\frac{f}{g}\right)(x)=\frac{x^2-1}{(x-2)(x-3)}$ has domain $\mathbb{R}-\{2,3\}$

An everyday example. A scooter's mileage is distance divided by fuel used — a quotient that has no meaning when no fuel has been used.

The substance. Cancelling a common factor does not restore an excluded point — $\frac{x^2-4}{x-2}$ equals $x+2$ only for $x\ne 2$.

To evaluate a combined function at a point, first check that the point lies in its domain, then compute f and g separately at that point and combine the two numbers.

Worked example. Let $f(x)=2x+3$ and $g(x)=x^2-4$.

At $x=3$: $f(3)=9$ and $g(3)=5$, so

- $(f+g)(3)=9+5=14$
- $(f-g)(3)=9-5=4$
- $(fg)(3)=9\times 5=45$
- $\left(\frac{f}{g}\right)(3)=\frac{9}{5}=1.8$

At $x=2$: $f(2)=7$ and $g(2)=0$, so $(fg)(2)=0$, but $\left(\frac{f}{g}\right)(2)$ is undefined — the domain of $\frac{f}{g}$ is $\mathbb{R}-\{-2,2\}$.

An everyday example. A school canteen's cost per plate — total cost divided by plates served — is worked out from each day's figures, and has no value on a holiday when no plates are served.

The substance. A combined function can be undefined where both parts are defined — f and g both exist at $x=2$, but their quotient does not.

The domain is the spread of the graph along the x-axis and the range is its spread along the y-axis, while rising and falling stretches, breaks, jumps and asymptotes show how the function behaves.

Reading a graph:

- Domain — project the graph onto the x-axis; hollow dots exclude points, solid dots include them
- Range — project the graph onto the y-axis
- Increasing or decreasing — the graph rises or falls from left to right
- Asymptotes — lines the graph approaches but never meets, as in $y=\frac{1}{x}$

Worked example. The graph of $f(x)=|x-2|+1$ is a V with its corner at $(2,1)$.

- Domain $\mathbb{R}$, since the V extends both ways
- Range $[1,\infty )$, since the lowest point has height 1
- Decreasing for $x<2$ and increasing for $x>2$
- $f(0)=3$ and $f(5)=4$

An everyday example. A graph of Delhi's temperature through a day shows the hours covered (domain), the lowest and highest temperatures (range), and when it was warming or cooling.

The substance. Shifting a graph changes its position, not its shape — $|x-2|+1$ is the V of $|x|$ moved 2 units right and 1 unit up.