From the day you're born until you're 20 years old, your body continues to grow. As you get older, your bones grow and change, increasing your height. You also begin to put on weight in order to support your growing body. During puberty, the rate at which you grow taller and gain weight peaks, meaning you grow faster than you ever have before.
We can measure the rate at which you grow taller and gain weight as a function of time! But how does the rate at which your height changes compare to the rate at which you gain weight? Calculus and L'Hôpital's Rule provide us with a way to compare the growth rate of functions. Comparing growth rates of functions is useful in a variety of fields, including child growth development, assessing and predicting a company's performance, and the study of population growth.
Meaning of the Growth Rate of Functions
Calculus is all about change. One of the main objects studied in calculus are functions, which are mathematical objects that relate two or more quantities. It is of particular interest knowing how functions change in terms of the variables used to define them. This is where the growth rate of a function is defined.
A function's growth rate is a measure of how much a function changes as its independent variable increases.
Within this context, the growth rate of a function alone is equivalent to its rate of change. However, the growth rate of functions is usually used within the context of comparing the growth of two functions. Keep reading this article for more information about this topic!
Keeping on the subject of the growth rate (or rate of change) of a function, take for instance the function
\[ f(x) = 2x.\]
If you plug in \( 3\) into the above function, you will get
\[ \begin{align} f(3) &= 2(3) \\ &= 6, \end{align}\]
then, if you plug in \( 4 \) you will get
\[ \begin{align} f(4) &= 2(4) \\ &= 8. \end{align}\]
By increasing the value of \(x \) from \( 3 \) to \( 4\), the function \( f(x) \) increased from \( 6 \) to \(8\).
Now consider
\[ g(x) = x+8.\]
By plugging in \( 3 \) you get
\[ \begin{align} g(3) &= (3)+8 \\ &= 11, \end{align}\]
and if you use \( 4 \) you will get
\[ \begin{align} g(4) &= (4)+8 \\ &= 12. \end{align}\]
This time by increasing the value of \(x \) the function \( g(x) \) increased from \( 11 \) to \(12\). Despite your obtained higher values using \( g(x) \), the growth rate of \( f(x) \) is greater because the values of the function had a greater change.
In order to find the growth rate \( G\) of a function you can use the formula
\[ G = \frac{f(x_2)-f(x_1)}{x_2-x_1}.\]
Consider the function
\[ f(x) = 3x-5.\]
Find its growth rate in the interval from \(x=3\) to \(x=10\).
Solution:
You can find the growth rate \( G\) of the given function in the requested interval by using the formula
\[ G = \frac{f(x_2)-f(x_1)}{x_2-x_1}.\]
Here, \(x_2=10\) and \(x_1=3\), so begin by finding \( f(x_2)\) and \( f(x_1)\), that is
\[ \begin{align} f(x_2) &= f(10) \\ &= 3(10)-5 \\ &= 25 \end{align}\]
and
\[ \begin{align} f(x_1) &= f(3) \\ &= 3(3)-5 \\ &= 4. \end{align}\]
Now, you can use the formula, so
\[ \begin{align} G &= \frac{ f(x_2)-f(x_1)}{x_2-x_1} \\ &= \frac{25-4}{10-3} \\ &= \frac{21}{7} \\ &= 3. \end{align}\]
So, the growth rate of the function over the given interval is equal to \(3\).
In the above example, you were looking at linear functions, which are characterized by the fact that they have a constant rate of change. This will not always be the case.
Consider the function
\[ f(x)= x^2.\]
If you were to find how much the function changes from \(x=1\) to \(x=2\) you would need to find
\[ \begin{align} f(1) &= (1)^2 \\ &= 1 \end{align} \]
and
\[ \begin{align} f(2) &= (2)^2 \\&=4.\end{align}\]
This means that the function increased by \(3\) units in the interval from \(1\) to \(2\). However, if you were to calculate how much the function changes from \( x=2 \) to \(x=3\), you would find that
\[ \begin{align} f(3) &= (3)^2 \\ &=9, \end{align}\]
which now means that the function increased by \( 5 \) units in the interval from \(2\) to \(3\).
Within the context of calculus, rather than finding the growth rate of a function over an interval, you will need to find the instant growth rate of the function at a certain point. This can be done by finding the derivative of the function.
Let \( f(x) \) be a differentiable function. Its derivative \( f'(x) \) is a function that describes the growth rate of \( f(x) \).
Usually, the word instant is omitted, so it is assumed that a growth rate refers to an instant growth rate or instant rate of change.
Consider the function
\[ g(x)=x^3+2x^2-1.\]
Find a function that describes its growth rate.
Solution:
Since the derivative of a function is a function that describes its growth rate, you need to find the derivative of \( g(x)\). You can achieve this by using the Power Rule, so
\[ \begin{align} g'(x) &= 3x^2+2(2)x \\ &= 3x^2+4x. \end{align}\]
This means that the function
\[ g'(x) = 3x^2+4x \]
describes the growth rate of the function
\[ g(x) = x^3 +2x^2.\]
Simple, right?
It is also possible to have a negative growth rate. This happens when the value of the function decreases as its independent variable increases.
Figure 1. Graph of a linear function with a negative growth rate
In the above graph you can note that the values of \(f(x)\) decrease as \( x \) increases.
Growth Rate of Functions From a Table
Sometimes you will be asked to find the growth rate of a function given a table. Consider the following table.
| \[ x \] | \[ f(x)\] |
| \[ 0 \] | \[ 1 \] |
| \[ 1 \] | \[ 2 \] |
| \[ 2\] | \[ 5\] |
| \[ 3\] | \[ 10\] |
| \[4\] | \[ 17\] |
Suppose you are asked to find the growth rate of the function in the interval from \(x=2\) to \(x=4\). You will need to use the growth rate formula
\[ G = \frac{f(x_2)-f(x_1)}{x_2-x_1}\]
with the values from the table. In this case \( x_2=4\) and \(x_1=2\), so
\[ \begin{align} G &= \frac{f(4)-f(2)}{4-2} \\ &= \frac{17-5}{4-2} \\ &= \frac{12}{2} \\ &= 6. \end{align}\]
The downside of using a table for finding the growth rate of a function is that the task of finding the instant growth rate becomes harder. Since you are not given an expression of the function, you cannot find its derivative in standard ways.
Luckily, whenever you are asked to find the growth rate of a function by means of a table, you will usually be asked to do it in a given interval.
Comparing Growth Rates of Functions
As mentioned previously in this article, speaking of the growth rate of functions usually refers to the comparison between two functions. In other words, you would be looking at which of two functions grows faster, and usually, this is within the context of functions whose domain extends to infinity.
The following definition allows you to compare the growth rate between two functions as \(x \rightarrow \infty\). For two increasing functions \(f(x)\) and \(g(x)\):
\(f(x)\) grows faster than \(g(x)\) as \(x \rightarrow \infty\) if
\[\lim_{x \to \infty} \frac{f(x)}{g(x)}=\infty.\]
\(f(x)\) grows slower than \(g(x)\) as \(x \rightarrow \infty\) if\[\lim_{x \to \infty} \frac{f(x)}{g(x)}=0.\]
\(f(x)\) and \(g(x)\) grow at the same rate as \(x \rightarrow \infty\) if\[\lim_{x \to \infty} \frac{f(x)}{g(x)}=L, \quad \text{where}\quad L\neq0.\]
It does not matter which names you are using for the function. What matters is that you identify which function is in the numerator (the top) and which is in the denominator (the bottom). If the limit goes to \(\pm\infty\), then the function at the top grows faster. If the limit goes to \(0\), then the function at the bottom grows faster.
Typically, you will need to use L'Hôpital's Rule to compute these limits, as these will often lead to indeterminate forms. Check out or take on L'Hôpital's Rule if you need a refresher on the topic.
In order to compare the growth rate of the two functions, you will need to evaluate the limit
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)}\]
and then check under which case it falls. If the resulting limit has an indeterminate form, you need to use L'Hôpital's Rule as many times as needed until you can determine the limit.
Consider the functions
\[ f(x) = x^3\]
and
\[ g(x) = 100x^2.\]
Which function grows faster as \(x \rightarrow \infty\) ?
Solution:
You might be tempted to assume that, since the function \( g(x) \) has a coefficient of \(100\), it grows faster. Rather than doing this, you should evaluate the limit
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)},\]
so
\[ \begin{align} \lim_{x \to \infty} \frac{f(x)}{g(x)} &= \lim_{x \to \infty} \frac{x^3}{100x^2} \\ &= \lim_{x \to \infty} \frac{x}{100} \\ &= \infty. \end{align}\]
Since the limit goes to infinity, you can assume that \( f(x)=x^3\) grows faster than \(g(x)\) as \( x\rightarrow\infty\).
Here is an example that involves L'Hôpital's Rule.
Consider the functions
\[ f(x)=x^2-10\]
and
\[ g(x) = 2x^2+6.\]
Which function grows faster as \( x\rightarrow\infty\)?
Solution:
Once again, begin by evaluating the limit
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)}\]
so
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)}= \lim_{x \to \infty} \frac{x^2-10}{2x^2+6}.\]
This time you cannot factor or cancel anything, and if you substitute \( x \) for \( \infty \) you will have an indeterminate form of
\[ \frac{\infty}{\infty}\]
which suggests the use of L'Hôpital's Rule. This means that you will need to find the derivatives of \(f\) and \(g,\) so
\[ f'(x)=2x\]
and
\[g'(x)=4x.\]
so you can now evaluate
\[ \begin{align} \lim_{x \to \infty} \frac{f(x)}{g(x)} &= \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \\ &= \lim_{x \to \infty} \frac{2x}{4x} \\ &= \frac{1}{2}. \end{align}\]
Since the limit is a number other than \(0\), you can conclude that both functions grow at the same rate as \(x \to \infty\).
You could have also divided by \( x^2 \) on the numerator and the denominator. The point of this exercise was to illustrate the use of L'Hôpital's Rule. You can use the method that works better for you. Suit yourself!
Finding the Growth Rate of an Exponential Function
The exponential function has a peculiar characteristic that stands out from other functions: it is its own derivative, that is
\[\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x.\]
This means that this function can be differentiated over and over again, and every time you will be getting yet another exponential function!
This fact becomes particularly outstanding when comparing the growth rate of the exponential function to another function.
Consider the functions
\[ f(x) = e^x\]
and
\[ g(x) = x^2.\]
If you were to compare their growth rate as \( x \to \infty \) you would first find an indeterminate limit, that is
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{e^x}{x^2}.\]
This suggests the use of L'Hôpital's Rule. In fact, you have to use it twice! This means that
\[ \begin{align} \lim_{x \to \infty} \frac{e^x}{x^2} &= \lim_{x \to \infty} \frac{e^x}{2x} \\ &= \lim_{x \to \infty} \frac{e^x}{2} \\ &= \infty. \end{align}\]
where you have differentiated twice each function. Since the limit goes to infinity, the function on the numerator grows faster, so \( f(x)=e^x \) is the function with the faster growth rate.
In particular, given any polynomial function \( P(x) \) you will always find out that
\[ \lim_{x \to \infty} \frac{e^x}{P(x)} = \infty,\]
which means that the exponential function grows faster than any polynomial function, no matter its degree! Dive in to see why!
Here you can see why the exponential function grows faster than any polynomial function. Without loss of generality, you can compare the growth rate of
\[ P(x) = x^n\]
to the growth rate of
\[ f(x) = e^x,\]
so the goal is the evaluation of
\[ \lim_{x \to \infty} \frac{e^x}{x^n}.\]
Direct evaluation of the limit would give you an indeterminate form of
\[ \frac{\infty}{\infty}\]
so you will be required to use L'Hôpital's Rule. The exponential function will stay the same, no matter how many times you differentiate it. However, the function \( x^n\) will start to decrease its degree with each differentiation, so
\[ P'(x)= nx^{n-1},\]
then
\[ P''(x) = n(n-1)x^{n-2},\]
and so on. If you differentiate it \( n \) times, the power will eventually reach to \( 0\), which will yield a constant function, that is
\[ \begin{align} P^{(n)}(x) &= n(n-1)(n-2)\dots(2)(1) \\ &= n!. \end{align}\]
Recall that \( n! \) reads as the factorial of \(n\), and it is equal to taking the product of all numbers starting from \( 1 \) up to \(n\).
This way you can evaluate the required limit, that is
\[ \lim_{x \to \infty} \frac{e^x}{x^n} = \lim_{x \to \infty} \frac{e^x}{n!}. \]
The number \(n!\) is usually a very large number, but this does not matter, it is still a number. This means that the above limit will tend to infinity, so
\[ \lim_{x \to \infty} \frac{e^x}{x^n} = \infty.\]
From here, you can conclude that the exponential function grows faster than any polynomial function!
Growth Rate of Functions Examples
Comparing the growth rate of functions is an excellent way to practice both derivatives and limits!
Show that
\[ m(x) = e^{3x} \]
grows faster than
\[ n(x) = x^3+2x-1.\]
Solution:
You need to evaluate the limit
\[ \lim_{x \to \infty} \frac{m(x)}{n(x)}\]
and show that it goes to infinity. When evaluating these types of limits to infinity when polynomial functions are involved, you will usually need to differentiate a number of times equal to the degree of the polynomial, so in order to evaluate
\[ \lim_{x \to \infty} \frac{m(x)}{n(x)} = \lim_{x \to \infty} \frac{e^{3x}}{x^3+2x-1}\]
you need to diferentiate three times. This will give you
\[ \begin{align} \lim_{x \to \infty} \frac{m(x)}{n(x)} &= \lim_{x \to \infty} \frac{m'(x)}{n'(x)} \\ &= \lim_{x \to \infty} \frac{3e^{3x}}{3x^2+2},\end{align}\]
then
\[ \begin{align} \lim_{x \to \infty} \frac{m(x)}{n(x)} &= \lim_{x \to \infty} \frac{m''(x)}{n''(x)} \\ &= \lim_{x \to \infty} \frac{9e^{3x}}{6x}, \end{align}\]
and finally
\[ \begin{align} \lim_{x \to \infty} \frac{m(x)}{n(x)} &= \lim_{x \to \infty} \frac{m'''(x)}{n'''(x)} \\ &= \lim_{x \to \infty} \frac{27e^{3x}}{6}. \end{align}\]
Now you can factor out the constants and evaluate the limit, that is
\[ \begin{align} \lim_{x \to \infty} \frac{m(x)}{n(x)} &= \frac{27}{6}\lim_{x \to \infty} e^{3x}, \end{align}\]
which goes all the way up to infinity, so
\[ \lim_{x \to \infty} \frac{m(x)}{n(x)} = \infty.\]
This means that the function
\[ m(x)= e^{3x}\]
grows faster than
\[ n(x) = x^3+2x-1\]
as \( x\to \infty\).
Sometimes you will not need to use L'Hôpital's Rule!
Consider the functions
\[ r(x) = -\frac{1}{x}\]
and
\[ s(x) = \ln{x}.\]
Both are slowly increasing functions. Which one grows faster as \(x \to \infty\)?
Solution:
In order to determine which function grows faster, you should evaluate the limit
\[ \lim_{x \to \infty}\frac{r(x)}{s(x)} = \lim_{x \to \infty} \frac{-\frac{1}{x}}{\ln{x}},\]
which can be written as
\[ \lim_{x \to \infty}\frac{r(x)}{s(x)} = -\lim_{x \to \infty} \frac{1}{x\ln{x}}. \]
Since the denominator of the above rational expression goes to infinity as \(x\to\infty\), the above limit is equal to \(0\), that is
\[ \lim_{x \to \infty}\frac{r(x)}{s(x)} = 0.\]
This means that
\[s(x)=\ln{x}\]
is the function that grows faster.
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