# Intro to combining functions

CCSS Math: HSF.BF.A.1b
Become familiar with the idea that we can add, subtract, multiply, or divide two functions together to make a new function.
Just like we can add, subtract, multiply, and divide numbers, we can also add, subtract, multiply, and divide functions.

# The sum of two functions

## Part 1: Creating a new function by adding two functions

Let's add ${f(x)=x+1}$ and ${g(x)=2x}$ together to make a new function.
Let's call this new function $h$. So we have:
${h(x)}={f(x)}+{g(x)}{=3x+1}$

## Part 2: Evaluating a combined function

We can also evaluate combined functions for particular inputs. Let's evaluate function $h$ above for $x=2$. Below are two ways of doing this.
Method 1: Substitute $x=2$ into the combined function $h$.
\begin{aligned}h(x)&=3x+1\\\\ h(2)&=3(2)+1\\\\ &=\greenD{7} \end{aligned}
Method 2: Find $f(2)$ and $g(2)$ and add the results.
Since $h(x)=f(x)+g(x)$, we can also find $h(2)$ by finding $f(2) +g(2)$.
First, let's find $f(2)$:
\begin{aligned}f(x)&= {x + 1}\\\\ f(2)&=2+1 \\\\ &=3\end{aligned}
Now, let's find $g(2)$:
\begin{aligned}g(x)&={2x}\\\\ g(2)&=2\cdot 2 \\\\ &=4\end{aligned}
So $f(2)+g(2)=3+4=\greenD7$.
Notice that substituting $x =2$ directly into function $h$ and finding $f(2) + g(2)$ gave us the same answer!

# Now let's try some practice problems.

In problems 1 and 2, let $f(x)=3x+2$ and $g(x)=x-3$.

#### Problem 1

Find $f(x)+g(x)$.

#### Problem 2

Evaluate $f(-1)+g(-1)$.

Here are a couple of ways to find the answer.
Method 1: Find $f(-1)$ and $g(-1)$ separately. Then add the two.
First, let's find $f(-1)$:
\begin{aligned} f(x) &=3x+2\\\\ f(-1)&= 3(-1)+2\\\\ &= -1\end{aligned}
Next, let's find $g(-1)$:
\begin{aligned} g(x) &=x-3\\\\ g(-1)&= -1-3\\\\ &= -4\end{aligned}
So $f(-1)+g(-1)=-1+(-4)=\greenD{-5}$
Method 2: Find $f(x)+g(x)$ first, and then evaluate this expression when $x=-1$.
When $x=-1$, this expression is equal to $4(-1)-1$ or $\greenD{-5}$.

# A graphical connection

We can also understand what it means to add two functions by looking at graphs of the functions.
The graphs of $y=m(x)$ and $y=n(x)$ are shown below. In the first graph, notice that $m(4)=2$. In the second graph, notice that $n(4)=5$.
Let $p(x)=m(x)+n(x)$. Now look at the graph of $y=p(x)$. Notice that $p(4)=\blueD 2+\maroonD 5=\purpleD7$.
Challenge yourself to see that $p(x) = m(x) + n(x)$ for every value of $x$ by looking at the three graphs.

## Let's practice.

#### Problem 3

The graphs of $y=f(x)$ and $y=g(x)$ are shown below.
Which is the best approximation of $f(3)+g(3)$?
Choose 1 answer:
Choose 1 answer:

We can use the graphs of $y=f(x)$ and $y=g(x)$ to find $f(3)$ and $g(3)$.
The graph below shows that $f(3)=6$.
The graph below shows that $g(3)=3$.
So $f(3)+g(3)=6+3=9$.

# Other ways to combine functions

All of the examples we've looked at so far create a new function by adding two functions, but you can also subtract, multiply, and divide two functions to make new functions!
For example, if $f(x)=x+3$ and $g(x)=x-2$, then we can not only find the sum, but also ...
... the difference.
\begin{aligned}f(x)-g(x)&=(x+3)-(x-2)~~~~~~~\small{\gray{\text{Substitute.}}}\\\\ &=x+3-x+2~~~~~~~~~~~~~\small{\gray{\text{Distribute negative sign.}}}\\\\ &=5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Combine like terms.}}}\end{aligned}
... the product.
\begin{aligned}f(x)\cdot g(x)&=(x+3)(x-2)~~~~~~~~~~~~\small{\gray{\text{Substitute.}}}\\\\ &=x^2-2x+3x-6~~~~~~~~\small{\gray{\text{Distribute.}}}\\\\ &=x^2+x-6~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Combine like terms.}}}\end{aligned}
... the quotient.
\begin{aligned}f(x)\div g(x)&=\dfrac{f(x)}{g(x)} \\\\ &=\dfrac{(x+3)}{(x-2)}~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Substitute.}}} \end{aligned}
In doing so, we have just created three new functions!

# Challenge problem

$p(t) = t + 2$
$q(t) = t - 1$
$r(t) = t$
Evaluate $p(3) \cdot q(3) \cdot r(3) - p(3)$.

Let's first find $p(3)$, $q(3)$, and $r(3)$ by substituting $t=3$ into each of the formulas.
First, let's find $p(3)$:
\begin{aligned}p(t) &=t+2\\\\ p(3)&= 3+2\\\\ &=\blueD5\end{aligned}
Next, let's find $q(3)$:
\begin{aligned} q(t) &=t-1\\\\ q(3)&= 3-1\\\\ &=\purpleC2\end{aligned}
Finally, let's find $r(3)$:
\begin{aligned} r(t) &=t\\\\ r(3)&= 3\\\\ &=\goldD3\end{aligned}
We can now find $p(3) \cdot q(3) \cdot r(3) - p(3)$ as follows:
\begin{aligned} p(3) \cdot q(3) \cdot r(3) - p(3)&= \blueD5\cdot \purpleC2\cdot \goldD3-\blueD5 \\\\ &= 30-5 \\\\ &= 25 \end{aligned}
We could also find $p(t) \cdot q(t) \cdot r(t) - p(t)$ first, and then find the value of this expression when $t=3$. This would be more complicated, though, since the result would be a third degree polynomial!