A roller coaster is one of the first things that might catch your attention in an amusement park. It is outstanding to see people scream in excitement as the cart goes from one place to another! Have you ever ridden a roller coaster?
A typical roller coaster
There are moments in which the cart is going up and moments where it is going down. In between, there is a small moment where the cart will be aligned horizontally, where you can use to catch your breath.
The points where this happens have a special names and are an important object of study in calculus. Who said that calculus and fun don't match?
Meaning of the First Derivative Test
Closely related to The First Derivative test is the concept of a stationary point. Let's begin by taking a look at its definition.
Let \( f \) be a differentiable function. A stationary point or critical point \( c \) is an x-value for which the derivative of \( f \) is equal to 0. That is,
\[f'(c)=0.\]
Since the derivative of a function at a critical point is equal to 0 a line tangent to the function at that point will have a slope of 0, that is, it will be a horizontal line.
Lines tangent to a cubic function at its critical points
A function can have more than one critical point, and The First Derivative Test is a method for finding them. We will now take a look at how to perform The First Derivative Test.
The First Derivative Test is a method for finding the critical points of a differentiable function \( f.\) It consists of the following:
- Find the derivative of the function.
- Evaluate the derivative of the function at a critical point and set it equal to 0, that is, write the equation stating that \( f'(c)=0. \)
- Solve the equation of the above step to find the value(s) of \( c \) inside the domain of \( f \) that are critical points.
The first derivative test receives its name from the fact that you need to find the derivative of the function you are inspecting.
Let's follow these steps in an example.
Consider the function\[ f(x)=x^2+6x+10. \]
Use the first derivative test to find its critical points.
Answer:
Begin by noticing that, since nothing else is stated, you can assume that the domain of the function consists of all real numbers. Now, for the steps:
1. Find the derivative of the function.
The given function is a polynomial function, so you can find its derivative using the Power Rule, that is
\[f'(x)=2x+6.\]
2. Evaluate the derivative at a critical point and set it equal to 0.
You first evaluate the derivative at a critical point \( c, \) so
\[f'(c)=2c+6,\]
and then you set it equal to 0, resulting in the equation
\[2c+6=0.\]
3. Solve the obtained equation for \( c.\)
The resulting equation is a linear equation which can be solved by isolating \( c, \) that is
\[\begin{align}2c &= -6 \\ c &= -3.\end{align}\]
Therefore, the function has one critical point at \( x=-3.\)
It is also possible for a function to not have critical points.
Consider the function
\[g(x)=e^x.\]
Use the first derivative test to find its critical points.
Answer:
Once again, you can assume that the domain of the function consists of all real numbers since nothing else is stated above.
1. Find the derivative of the function.
The given function is an exponential function, whose derivative is itself, so
\[g'(x)=e^x.\]
2. Evaluate the derivative at a critical point and set it equal to 0.
This step is rather straightforward, so first evaluate
\[g'(c)=e^c\]
and then set equal to 0 to write the equation
\[e^c=0.\]
3. Solve the obtained equation for \( c.\)
Unfortunately, since there is no function that makes the above equation true, the equation has no solution. This means that the function has no critical points.
It is important to always check the domain of the function. Let's see why.
Consider the function
\[h(x)=x^2+2x+2, \quad \text{for}\quad x\geq 0.\]
Use the first derivative test to find its critical points.
Answer:
Now the domain of the function consists of all non-negative numbers. You have to keep track of this when solving the equation in the third step!
1. Find the derivative of the function.
This polynomial function can be differentiated with the Power Rule, so
\[h'(x)=2x+2.\]
2. Evaluate the derivative at a critical point and set it equal to 0.
Evaluate \(h'(x)\) at \(x=c,\)
\[h'(c)=2c+2,\]
and set it equal to 0
\[2c+2=0.\]
3. Solve the obtained equation for \( c.\)
You can solve the equation obtained in the above step by isolating \(c,\) so
\[\begin{align} 2c &= -2 \\ c &= -1.\end{align}\]
This value is not inside the domain of the function, hence the function has no critical points!
The First Derivative Test and Local Extrema
Critical points are closely related to relative extrema, so before proceeding, here is a reminder on what is the meaning of extrema.
An extremum of a function refers to either a maximum or a minimum. Extrema is the plural of extremum.
For a more detailed review on this subject please check out our Maxima and Minima article!
The previous example about a quadratic function is excellent for studying the relationship between critical points and local extrema. In it, you found that the function \( f(x) = x^2 + 6x + 10\) has a critical point at \( -3.\) Take now a look into its graph.
Graph of a line tangent to a function at its critical point
It looks like the critical points are also the points where a function has local extrema. This is not always the case, but there is a theorem that makes this connection.
Fermat's Theorem states that, if a function \( f \) has a local maximum or local minimum at \(x=c,\) and the function is differentiable at that point, then \( f'(c)=0.\)
In other words, Fermat's Theorem tells you that if a function has a local maximum or a local minimum at a point where it is differentiable, then it is a critical point. Be careful to not misinterpret this theorem, as there are some common mistakes when making the bridge between local extrema and critical points.
Common mistakes when relating Critical Points and Local Extrema
Fermat's Theorem tells us that if a function has a local extremum at a certain point then it is a critical point. It does not always work the other way around. This is better understood with an example.
Consider the function\[f(x)=x^3-2.\]
Use the first derivative test to find its critical points.
Answer:
The domain of the function can be assumed to be all real numbers, so you should proceed normally.
1. Find the derivative of the function.
You can find the derivative of this polynomial function using the Power Rule, so
\[f'(x)=3x^2.\]
2. Evaluate the derivative at a critical point and set it equal to 0.
Next, you evaluate the derivative obtained in the previous step at \(x=c\) to obtain
\[f'(c)=3c^2,\]
and then you set it equal to 0 to write an equation for \( c\)
\[3c^2=0.\]
3. Solve the obtained equation for \( c.\)
The equation obtained in the previous step is only true when \(c=0,\) which means that there is only one critical point at \( x=0.\)
Now, you might be tempted to assume that the function has a local extremum at 0, but this is not the case. Take a look at its graph.
Graph of a cubic function with no local extrema
This cubic function has no local extrema, but still, the critical point is at \(x=0.\) Note how the slope at the critical point is 0.
Graph of the cubic function with a line tangent to its critical point
You found that not all critical points are local extrema in the above example. Let's now take a look at another common mistake.
Consider the function
\[g(x)=|x-2|+1.\]
Now, take a look at its graph.
Graph of an absolute value function
The function has a relative minimum at \(x=2,\) so you might be tempted to assume that it is also a critical point. However, the function is not differentiable at \(x=2,\) therefore it is not a critical point.
Now you found that not all local extrema are critical points. You need the function to be differentiable at its relative extrema to be a critical point.
The First Derivative Test for Multivariable Functions
When looking for information about the First Derivative Test you might come across multivariable functions. This subject is usually reserved for higher levels, so if you are curious enough dive into it!
When you are given a function of many variables the First Derivative Test is slightly modified. You need to do the following:
- Find each Partial Derivative of the function with respect to all of its variables.
- Evaluate each Partial Derivative at a critical point and set them all equal to 0. That is, write a System of Equations stating that each Partial Derivative evaluated at a critical point is equal to 0.
- Solve the System of Equations to find the critical points.
Let's see an example of the First Derivative Test applied to a function of two variables!
Consider the function
\[f(x,y)=x^2-xy+2y.\]
Find its critical points.
Answer:
Since its inputs are points in the plane, its critical points will be ordered pairs of the form \( (a,b). \)
1. Find each Partial Derivative of the function with respect to all of its variables.
Both of the partial derivatives of this function can be found with the Power Rule, so
\[\frac{\partial f}{\partial x}=2x-y,\]
and
\[\frac{\partial f}{\partial y}=-x+2.\]
2. Evaluate each Partial Derivative at a critical point and set them all equal to 0.
Since the critical points have the form \( (a,b), \) evaluate both partial derivatives using \( x=a \) and \( y=b,\) so
\[\left. \frac{\partial f}{\partial x}\right|_{(a,b)}=2a-b,\]
and
\[\left. \frac{\partial f}{\partial y}\right|_{(a,b)}=-a+2.\]
By setting above expressions equal to 0 you get the system of equations
\[\begin{cases}2a-b &= 0 \\ -a+2 &= 0.\end{cases}\]
3. Solve the System of Equations to find the critical points.
You can now solve the above system of equations using a method of your choice, obtaining
\[\begin{cases}a &= 2 \\ b &= 4.\end{cases}\]
This means that there is only one stationary point, which is at \( (2,4).\)
Please note that if the function has three variables you will have to evaluate all three partial derivatives and write three equations. In general, if the function has \(n\) variables you will need \(n\) partial derivatives and \(n\) equations.
Examples of the First Derivative Test
Here we will take a look at more examples of The First Derivative Test. Let's dig in!
Consider the function
\[f(x)=x^4-2x^2+1.\]
Use the First Derivative Test to find its critical points.
Answer:
1. Find the derivative of the function.
You can find the derivative of this polynomial function using the Power Rule, so
\[f'(x)=4x^3-4x.\]
2. Evaluate the derivative at a critical point and set it equal to 0.
Next, you evaluate the derivative obtained in the previous step at \(x=c\) to obtain
\[f'(c)=4c^3-4c,\]
and then you set it equal to 0 to write an equation for \( c\)
\[4c^3-4c=0.\]
3. Solve the obtained equation for \( c.\)
To solve this equation begin by noting that you can factor out \(4c\) in the left-hand side of the equation, that is
\[4c^3-4c=4c(c^2-1).\]
Furthermore, you can factor the difference of squares as well, obtaining
\[4c^3-4c=4c(c-1)(c+1).\]
Substituting this back into the equation gives you
\[4c(c-1)(c+1)=0,\]
which means that
\[4c=0,\]
\[c-1=0,\]
and
\[c+1=0.\]
By solving the above equations you can find that \( c=0,\) \(c=1,\) and \(c=-1.\) Therefore, this function has three critical points!
Let's now see an example involving a trigonometric function.
Consider the function
\[g(x)=\sin{x}.\]
Use the First Derivative Test to find its critical points.
Answer:
1. Find the derivative of the function.
The derivative of the sine function is the cosine function, that is
\[g'(x)=\cos{x}.\]
2.Evaluate the derivative at a critical point and set it equal to 0.
By evaluating the above function at \(x=c\)
\[g'(c)=\cos{c},\]
and setting it equal to 0, you obtain an equation for \(c,\)
\[\cos{c}=0.\]
3.Solve the obtained equation for \( c.\)
The cosine function has its zeros at positive and negative odd multiples of \( ^\pi / _2, \) that is,
\[\pm\frac{\pi}{2},\, \pm\frac{3\pi}{2},\, \pm\frac{5\pi}{2},\]
and so on. You can express these odd multiples by using a factor of \( 2n-1\) because subtracting 1 from an even number results in an odd number, so
\[c = \left( 2n-1 \right)\frac{\pi}{2}, \quad \text{ for }\, n=1, 2, 3, \dots .\]
The sine function has an infinite amount of critical points!
The first derivative test and concavity
Despite being related to the shape of a graph, the First Derivative Test is a test for finding critical points, it is not a test for finding the concavity of a graph. In that case you should take a look at our article on the Second Derivative Test instead.
The First Derivative Test - Key takeaways
- A stationary point or critical point is an x-value for which the derivative of a function is equal to 0.
- The First Derivative Test consists in finding the critical points of a function.
- You can do The First Derivative Test by following these steps:
- Find the derivative of the function.
- Evaluate the derivative of the function at a critical point \(c\) and set it equal to 0. That is, write the equation \( f'(c)=0.\)
- Solve the above equation to find the critical points. Don't forget to only include values inside the domain of the function!
- The First Derivative Test is related to local extrema by means of the Fermat's Theorem about stationary points.
- Just because something is a critical point does not mean it is a local extremum!
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