6.1: Operations on functions given by formulas

We can combine functions in many different ways. First, note that we can add, subtract, multiply, and divide functions.

Example $\PageIndex{1}$

Let $f(x)=x^2+5x$ and $g(x)=7x-3$. Find the following functions, and state their domain. $$(f+g)(x), (f-g)(x), (f\cdot g)(x), \text{ and } \left(\dfrac f g\right)(x) \nonumber$$

Solution

The functions are calculated by adding the functions (or subtracting, multiplying, dividing).

$$\begin{aligned} (f+g)(x) &= (x^2+5x) +(7x -3) = x^2+12x -3 \\ (f-g)(x) &= (x^2+5x) -(7x -3) \\ &= x^2+5x-7x +3=x^2-2x+3 \\ (f\cdot g)(x) &= (x^2+5x) \cdot (7x -3)\\ &= 7x^3-3x^2+35x^2 -15x=7x^3+32x^2-15x \\ \left(\dfrac f g\right)(x) &= \dfrac{x^2+5x}{7x-3}\end{aligned}$$

The calculation of these functions was straightforward. To state their domain is also straightforward, except for the domain of the quotient $\dfrac f g$. Note, that $f+g$, $f-g$, and $f\cdot g$ are all polynomials. By the standard convention (See Convention definition) all these functions have the domain $\mathbb{R}$, that is their domain is all real numbers.

Now, for the domain of $\dfrac f g$, we have to be a bit more careful, since the denominator of a fraction cannot be zero. The denominator of $\dfrac f g (x)=\dfrac{x^2+5x}{7x-3}$ is zero, exactly when

$$7x-3=0 \quad\implies \quad 7x=3 \quad\implies\quad x=\dfrac 3 7 \nonumber$$

We have to exclude $\dfrac 3 7$ from the domain. The domain of the quotient $\dfrac f g$ is therefore $\mathbb{R}-\left \{\dfrac 3 7 \right \}$.

We can formally state the observation we made in the previous example.

Observation

Let $f$ be a function with domain $D_f$, and $g$ be a function with domain $D_g$. A value $x$ can be used as an input of $f+g$, $f-g$, and $f\cdot g$ exactly when $x$ is an input of both $f$ and $g$. Therefore, the domains of the combined functions are the intersection of the domains $D_f$ and $D_g$:

$$\begin{aligned} D_{f+g}&= D_f\cap D_g =\{x\quad |\quad x\in D_f\text{ and } x\in D_g\} \\ D_{f-g}&= D_f\cap D_g \\ D_{f\cdot g}&= D_f\cap D_g\end{aligned}$$

For the quotient $\dfrac f g$, we also have to make sure that the denominator $g(x)$ is not zero.

$$D_ {\frac f g}= \{ x \,\, | \,\, x\in D_f, x\in D_g, \text{ and }g(x)\neq 0 \} \nonumber$$

Example $\PageIndex{2}$

Let $f(x)=\sqrt{x+2}$, and let $g(x)=x^2-5x+4$. Find the functions $\dfrac f g$ and $\dfrac g f$ and state their domains.

Solution

First, the domain of $f$ consists of those numbers $x$ for which the square root is defined. In other words, we need $x+2\geq 0$, that is $x\geq -2$, so that the domain of $f$ is $D_f=[-2,\infty)$. On the other hand, the domain of $g$ is all real numbers, $D_g=\mathbb{R}$. Now, we have the quotients

$$\left(\dfrac f g\right) (x)=\dfrac{\sqrt{x+2}}{x^2-5x+4},\quad\quad \text{ and } \quad\quad \left(\dfrac g f\right) (x)=\dfrac{x^2-5x+4}{\sqrt{x+2}} \nonumber$$

For the domain of $\dfrac f g$, we need to exclude those numbers $x$ for which $x^2-5x+4=0$. Thus,

$$\begin{aligned} x^2-5x+4=0 & \implies & (x-1)(x-4)=0 \\ &\implies & x=1, \text{ or }x=4\end{aligned}$$

We obtain the domain for $\dfrac f g$ as the combined domain for $f$ and $g$, and exclude $1$ and $4$. Therefore, $D_{\frac f g}=[-2,\infty)-\{1,4\}$.

Now, for $\dfrac g f(x)=\dfrac{x^2-5x+4}{\sqrt{x+2}}$, the denominator becomes zero exactly when

$$x+2=0, \quad \implies \quad x=-2 \nonumber$$

Therefore, we need to exclude $-2$ from the domain, that is

$$D_{\frac g f}=[-2,\infty)-\{-2\}=(-2,\infty) \nonumber$$

Example $\PageIndex{3}$

To form the quotient $\dfrac{f}{g}(x)$ where $f(x)=x^2-1$ and $g(x)=x+1$ we write $\dfrac{f}{g}(x)=\dfrac{x^2-1}{x+1}=\dfrac{(x+1)(x-1)}{x+1}=x-1$. One might be tempted to say that the domain is all real numbers. But it is not! The domain is all real numbers except $-1$, and the last step of the simplification performed above is only valid for $x\not=-1$.

Another operation we can perform is the composition of two functions.

Definition: Composition of a Function

Let $f$ and $g$ be functions, and assume that $g(x)$ is in the domain of $f$. Then define the composition of $f$ and $g$ at $x$ to be

$$(f\circ g)(x):= f(g(x)) \nonumber$$

We can take any $x$ as an input of $f\circ g$ which is an input of $g$ and for which $g(x)$ is an input of $f$. Therefore, if $D_f$ is the domain of $f$ and $D_g$ is the domain of $g$, the domain of $D_{f\circ g}$ is

$$D_{f\circ g}=\{\,\, x \,\, | \,\, x\in D_g, \,\, g(x)\in D_f\,\, \} \nonumber$$

Example $\PageIndex{4}$

Let $f(x)=2x^2+5x$ and $g(x)=2-x$. Find the following compositions:

1. $f(g(3))$
2. $g(f(3))$
3. $f(f(1))$
4. $f(2\cdot g(5))$
5. $g(g(4)+5)$

Solution

We evaluate the expressions, one at a time, as follows:

1. $f(g(3)) = f(2-3) = f(-1) = 2\cdot (-1)^2+5\cdot(-1) = 2-5 = -3$
2. $g(f(3)) = g(2\cdot 3^2+5\cdot 3) = g(18+15) = g(33) = 2-33 = -31$
3. $f(f(1)) = f(2\cdot 1^2+5\cdot 1) = f(2+5)=f(7) = 2\cdot 7^2+5\cdot 7 = 98+35 = 133$
4. $f(2 \cdot g(5)) =f(2 \cdot(2-5))=f(2 \cdot(-3))=f(-6) =2 \cdot(-6)^{2}+5 \cdot(-6)=72-30=42$
5. $g(g(4)+5) = g((2-4)+5) = g((-2)+5) = g(3) = 2-3 = -1$

We can also calculate composite functions for arbitrary $x$ in the domain.

Example $\PageIndex{5}$

Let $f(x)=x^2+1$ and $g(x)=x+3$. Find the following compositions:

1. $(f\circ g)(x)$
2. $(g\circ f)(x)$
3. $(f\circ f)(x)$
4. $(g\circ g)(x)$

Solution

1. There are essentially two ways to evaluate $(f\circ g)(x)=f(g(x))$. We can either first use the explicit formula for $f(x)$ and then the one for $g(x)$, or vice versa. We will evaluate $f(g(x))$ by substituting $g(x)$ into the formula for $f(x)$: $$\begin{aligned} (f\circ g)(x)&= f(g(x))=(g(x))^2+1 =(x+3)^2+1 \\ &= x^2+6x+9+1 = x^2+6x+10 \end{aligned} \nonumber$$

Similarly, we evaluate the other expressions (b)-(d):

2. $(g\circ f)(x) = g(f(x)) = f(x)+3 = x^2+1+3 = x^2+4$
3. $(f\circ f)(x) = f(f(x)) = (f(x))^2+1 = (x^2+1)^2+1 = x^4+2x^2+1+1=x^4+2x^2+2$
4. $(g\circ g)(x) = (g(x)) = g(x)+3 = x+3+3=x+6$

Example $\PageIndex{6}$

Find $(f\circ g)(x)$ and $(g\circ f)(x)$ for the following functions, and state their domains.

1. $f(x)=\dfrac{3}{x+2}$, and $g(x)=x^2-3x$
2. $f(x)=|3x-2|-6x+4$, and $g(x)=5x+1$
3. $f(x)=\sqrt{\dfrac1 2 \cdot (x-4)}$, and $g(x)=2 x^2+4$

Solution

1. Composing $f\circ g$, we obtain

   $$(f\circ g)(x)=f(g(x))=\dfrac 3 {g(x)+2}=\dfrac 3 {x^2-3x+2} \nonumber$$

   The domain is the set of numbers $x$ for which the denominator is non-zero.

   $$\begin{aligned} x^2-3x+2=0 & \implies & (x-2)(x-1)=0 \\ & \implies & x=2 \text{ or } x=1\\ & \implies & D_{f\circ g}=\mathbb{R}-\{1,2\}.\end{aligned}$$

   Similarly,

   $$\begin{aligned} (g\circ f)(x) &= g(f(x))=f(x)^2-3f(x) =\left( \dfrac 3 {x+2}\right)^2-3\dfrac 3 {x+2}\\ &= \dfrac 9 {(x+2)^2}-\dfrac 9 {x+2} = \dfrac{9-9(x+2)}{(x+2)^2}\\ &= \dfrac{9-9x-18}{(x+2)^2}=\dfrac{-9x-9}{(x+2)^2}=\dfrac{-9\cdot (x+1)}{(x+2)^2}\end{aligned}$$

   The domain is all real numbers except $x=-2$, that is $D_{g\circ f}=\mathbb{R}-\{-2\}$.

2. We calculate the compositions as follows.

   $$\begin{aligned} (f\circ g)(x) &= f(g(x))= |3g(x)-2|-6g(x)+4 \\ &= |3(5x+1)-2|-6(5x+1)+4 = |15x+1|-30x-2,\\ (g\circ f)(x) &= g(f(x))= 5f(x)+1 =5\cdot (|3x-2|-6x+4)+1 \\ &= 5\cdot |3x-2|-30x+20+1= 5\cdot |3x-2|-30x+21 \end{aligned} \nonumber$$

   Since the domains of $f$ and $g$ are all real numbers, so are also the domains for both $f\circ g$ and $g\circ f$.

3. Again we calculate the compositions.

   $$\begin{aligned} (f\circ g)(x)&= f(g(x))=\sqrt{\dfrac 1 2 \cdot (g(x)-4)}=\sqrt{\dfrac 1 2 \cdot (2x^2+4-4)} \\ &= \sqrt{\dfrac 1 2 \cdot 2x^2}=\sqrt{x^2}=|x|\end{aligned} \nonumber$$

   The domain of $g$ is all real numbers, and the outputs $g(x)=2x^2+4$ are all $\geq 4$, (since $2x^2\geq 0$). Therefore, $g(x)$ is in the domain of $f$, and we have a combined domain of $f\circ g$ of $D_{f\circ g}=\mathbb{R}$. On the other hand,

   $$\begin{aligned} (g\circ f)(x)&= g(f(x))=2(f(x))^2+4=2\cdot \bigg(\sqrt{\dfrac 1 2 \cdot (x-4)}\,\,\bigg)^2+4\\ &= 2\cdot \bigg(\dfrac 1 2\cdot (x-4)\bigg)+4 =(x-4)+4=x\end{aligned} \nonumber$$

   The domain of $g\circ f$ consists of all numbers $x$ which are in the domain of $f$ and for which $f(x)$ is in the domain of $g$. Now, the domain of $f$ consists of all real numbers $x$ that give a non-negative argument in the square-root, that is: $\dfrac 1 2 (x-4)\geq 0$. Therefore we must have $x-4\geq 0$, so that $x\geq 4$, and we obtain the domain $D_f=[4,\infty)$. Since the domain $D_g=\mathbb{R}$, the composition $g\circ f$ has the same domain as $f$:

   $$D_{g\circ f}=D_f=[4,\infty) \nonumber$$

We remark that at a first glance, we might have expected that $(g\circ f)=x$ has a domain of all real numbers. However, the composition $g(f(x))$ can only have those inputs that are also allowed inputs of $f$. We see that the domain of a composition is sometimes smaller than the domain that we use via the standard convention (See Convention definition).