# Comparing Graphs of Functions

Alignments to Content Standards: F-LE.A

1. How do the values of the three functions $f(x) = 2x$, $g(x) = x^2$ and $h(x) = 2^x$ compare for large positive and negative values of $x$?
2. Explain your findings from part (a) for $f$ and $g$ by studying the quotients $\frac{g(x)}{f(x)}$.
3. Explain your findings from part (a) for $g$ and $h$ by studying $\frac{g(x+1)}{g(x)}$ and $\frac{h(x+1)}{h(x)}$ for different values of $x$.

## IM Commentary

The goal of this task is to use appropriate tools to compare graphs of several functions. In addition, students are asked to study the structure of the different expressions to explain why these functions grow as they do. Two different approaches for part (a) are presented: one producing a table of values and one using graphing technology. It is important, for the latter method, to choose an appropriate domain and range. This is difficult when we have an exponential function and a linear function because the exponential grows so much faster that it either disappears from the screen or the linear function is hard to distinguish from the $x$-axis.

## Solutions

Solution: 1 Making a table

1. Below is a table showing some values of $x$ and the corresponding values of the the functions $f(x) = 2x$, $g(x)=x^2$, and $h(x)=2^x$:

$x$ $f(x)$ $g(x)$ $h(x)$
$-100$ $-200$ $10,000$ $7.89 \times 10^{-31}$ (approximate)
$-10$ $-20$ $100$ $\frac{1}{1024}$
$-5$ $-10$ $25$ $\frac{1}{32}$
$-2$ $-4$ $4$ $\frac{1}{4}$
$0$ $0$ $0$ $1$
$2$ $4$ $4$ $4$
$5$ $10$ $25$ $32$
$10$ $20$ $100$ $1024$
$100$ $200$ $10,000$ $1.27 \times 10^{30}$ (approximate)

We can observe several different trends from the table. First, when we substitute negative values of $x$, the functions $g$ and $h$ take positive values while $f$ takes negative values. Secondly, the values of $g$ get larger and larger as $x$ decreases while the values of $h$ appear to get close to $0$. For positive values of $x$ all three functions take positive values and they all grow as $x$ grows. The function $f$ appears to grow the most slowly, followed by $g$, and then $h$ seems to grow the most rapidly. It is also interesting to note that all three functions have the same value when $x = 2$.

2. The function $f(x) = 2x$ defines a line with slope $2$ and $y$ intercept $0$. It takes large positive values when $x$ is large and positive, and large negative values when $x$ is large and negative. The function $x^2$ never takes negative values. We also have $(-x)^2 = x^2$ so it is symmetric about the $y$-axis. When $x$ is large, $x^2$ is also large: to compare $x^2$ with $2x$ we can look at the quotient

\begin{align} \frac{g(x)}{f(x)} &= \frac{x^2}{2x}\\ &= \frac{x}{2}. \end{align}

This expression grows larger as $x$ grows. This means that $g$ takes larger values than $f$ when $x$ is large and positive.

3. Comparing $g$ and $h$ for positive values note that:

\begin{align} \frac{h(x+1)}{h(x)} &= \frac{2^{x+1}}{2^x}\\ &= 2. \end{align}

So the function $h$ doubles each time the input $x$ is replaced by $x+1$. For $g$ we have

\begin{align} \frac{g(x+1)}{g(x)} &= \frac{(x+1)^2}{x^2}\\ &= \frac{x^2 + 2x +1}{x^2}. \end{align}

For large values of $x$ this is close to but larger than 1. So the value of $g$ is multiplied by a factor closer and closer to 1 when the input $x$ is replaced by $x+1$. This means that the values of $h$ eventually overtake the values of $g$: in fact, $h(x) \gt g(x)$ as soon as $x \gt 2$.

The only feature of these graphs that remains to be explained is the behavior of $h$ for large negative values of $x$. We have $$h(-x) = 2^{-x} = \frac{1}{2^x}.$$ Since $2^x$ is very large and positive when $x$ is large and positive, this means that $2^{-x}$ is very small and positive when $x$ is large and negative.

Solution: 2 Graphing technology for part (a)

Below is a sketch using Desmos which shows graphs of $f$, $g$, and $h$:

The domain of the graph can be changed after clicking on the graph. The shapes of the graphs are difficult to evaluate, however, with a larger domain as the graphs of $g$ and $h$ ''appear'' almost vertical: the scales on the axes would need to be changed in order remedy this, but then the graph of $f$ would look essentially horizontal.