# First Derivative Test

Hi, and welcome to this video on the First Derivative Test! In this video we’ll go over some important definitions, talk about how the test works, and go over some practice problems. Grab some paper, a pencil, and a graphing calculator, and let’s get started!

First, let’s review the definitions of a few terms that will be important to this test. The first thing we need to remember is what a critical point is. A critical point is any point on a graph where the derivative equals zero. This graph labels each critical point of this function.

The other terms we need to review are maximums and minimums. A maximum of a function occurs at a point where the derivative equals zero and the slope is changing from positive to negative. A minimum of a function occurs at a point where the derivative equals zero and the slope changes from negative to positive.

A maximum and minimum can further be described as a relative max or min, or an absolute max or min. Any maximum or minimum is considered a relative max or min, and a graph can have multiple of each. An absolute max or min is the highest or lowest point on a graph, which means any function can only have one absolute max and one absolute min. Look at our same graph as before, this time with the relative and absolute max and mins labeled. Notice that one of our critical points did not change to a maximum or a minimum. This is because the slope was negative, flattened out to zero, which is why we have a critical point, but then continued to be negative. Since the slope did not change signs, it is neither a maximum nor a minimum.

The reason these definitions are so important is because the First Derivative Test tells you where relative maximums and minimums are on a function without having to graph it. The First Derivative Tests says:

Suppose there is a critical point for the function fx at x=c:

- If f’x>0 when x
c, then x=c is a relative maximum. - If f’x<0 when x
0 when x>c, then x=c is a relative minimum. - If f'(x) has the same sign on either side of x=c, then x=c is neither a maximum nor a minimum.

In other words, if you have a critical point and the derivative to the left of that point is positive but to the right of it is negative, then you have a relative maximum. Similarly, if you have a critical point and the derivative to the left of that point is negative but to the right of it is positive, then you have a relative minimum.

Let’s try an example.

For the following function, find the critical points and determine if each is a relative max, relative min, or neither.

\(f(x)=x^3+3x^2-2x+4\)To use the First Derivative Test, we need to take the derivative.

Remember, when taking the derivative of a function, we multiply the exponent by the coefficient and then decrease the exponent by 1.

3 times 1 is 3, so our first term is \(3x^2\).

\(f(x)=3x^2\)2 times 3 is 6, so we then have 6x.

\(f'(x)=3x^2+6x\)And finally, 1 times negative 2 is negative 2, so the final derivative is

\(f'(x)=3x^2+6x-2\)Next, we need to find the critical points. Critical points occur when the derivative is equal to 0. So we are going to set \(3x^2+6x-2\) equal to 0 and solve.

\(0=3x^2+6x-2\)This function doesn’t factor easily, so we will have to use the quadratic equation.

\(x=\frac{-6±\sqrt{(6)^2-4(3)(-2)}}{2(3)}=\frac{-6±\sqrt{36+24}}{6}=\frac{-6±\sqrt{60}}{6}=\frac{-6±2\sqrt{15}}{6}=-1±\frac{1}{3}\sqrt{15}\)

x=0.291, -2.291

Now that we’ve found that the derivative is equal to 0 at the points x=0.291 and x=-2.291, we can use these to test our intervals.

I’m going to set up a chart that looks like this:

This allows me to check the three intervals we are interested in: when x < -2.291, when x is between -2.291 and 0.291, and when x > 0.291. Notice that our intervals are based on the critical points that we found.

Next, I’m going to choose a number less than -2.291. Since we are just testing to see if the derivative in this interval is positive or negative, this number can be any number less than -2.291, so -157 would work just as well as -3. But, to avoid making things unnecessarily complicated, I’m going to use -3. We then plug this number into our f’(x) equation and solve.

\(f'(-3)=3(-3)^2+6(-3)-2=3(9)+6(-3)-2=27-18-2=7\)

7 is positive, which means the slope when x < -2.291 is positive, so I’m going to write a plus sign right here:

Now we do this for our middle interval. Choose any number between -2.291 and 0.291. I’m going to use 0.

\(f'(0)=3(0)^2+6(0)-2=0+0-2=-2\)

The slope for this interval is negative, so I’m going to write a negative sign right here:

We’ve got one more interval to go. I’m going to choose 1 and plug it in because it is greater than 0.291.

\(f'(1)=3(1)^2+61-2=3+6-2=7\)

This is positive again, so I’ll write a plus sign right here:

This chart isn’t necessary to perform the First Derivative Test, but it is helpful to visualize and remember what’s going on with our graph.

Remember, we are looking for points where the intervals change signs. For this example, it happens at both points. -2.291 changes from positive to negative, so there is a relative maximum at x = -2.291. And 0.291 changes from negative to positive, so there is a relative minimum at x = 0.291.

One of the things this test is useful for is estimating what the graph might look like. Let’s try that with this function.

First, we need to find the y-coordinates for our critical points. Make sure you plug the x-values into the original equation and not the derivative equation.

\(f(-2.291)=(-2.291)^3+3(-2.291)^2-2(-2.291)+4=12.303\)

\(f(0.291)=(0.291)^3+3(0.291)2-2(0.291)+4=3.697\)

We know we have a point at (-2.291, 12.303) and another point at (0.291, 3.697). Let’s draw a rough sketch of those points.

We know that the slope is positive before -2.291 so we can draw in a segment like this:

We also know that it then switched so that the slope was negative all the way to 0.291, like this:

Then, when it gets to 0.291, it switches to positive again.

Remember, this isn’t an exact graph but it gives us a general idea. Let’s compare it to the actual graph of the function and see if we are close.

Notice that we go from positive to negative to positive again, just like the First Derivative Test told us and our sketch showed us.

Let’s try another one.

For the following function, find the critical points and determine if each is a relative max, relative min, or neither. Then, sketch a graph of the function.

\(f(x)=-8x^3-2x^2+17x+1\)Pause the video and try this one on your own, or pull out some paper and try this one step-by-step with me.

First, take the derivative of the function.

\(f'(x)=-24x^2-4x+17\)Then, set it equal to 0 and solve for x.

\(0=-24x^2-4x-17\)

\(x=\frac{4±\sqrt{(-4)^2-4(-24)(17)}}{2(-24)}=\frac{4±\sqrt{16+1,632}}{-48}=\frac{4±\sqrt{1,648}}{-48}=\frac{4±4\sqrt{103}}{-48}=\frac{1±\sqrt{103}}{-12}\)

x=-0.929, 0.762

Next, set up your interval chart.

Choose a number less than -0.929 to plug into the derivative function.

\(f'(-1)=-24(-1)^2-4(-1)+17=-24+4+17=-3\)This is negative, so write a minus sign here:

Then, choose a number between -0.929 and 0.762.

\(f'(0)=-24(0)^2-4(0)+17=17\)Since 17 is positive, write a plus sign here:

Finally, choose a number greater than 0.762.

\(f'(1)=-24(1)^2-4(1)+17=-24-4+17=-11\)Write another negative sign here:

Now we can determine that the point x = -0.929 is a relative minimum because the slope of the function changes from negative to positive, and the point x = 0.762 is a relative maximum because the slope of the function changes from positive to negative.

Now, let’s sketch a graph of the function.

First, we need to find our y-coordinates that correspond with our x-coordinates at our critical points.

\(f(-0.929)=-8(-0.929)^3-2(-0.929)^2+17(-0.929)+1=-0.105\)

\(f(0.762)=-8(0.762)^3-2(0.762)^2+17(0.762)+1=9.253\)

Then, we’ll roughly plot these on our graph.

We know that at the first point the graph changes from decreasing to increasing, so let’s sketch that.

And then at the second point it changes from increasing to decreasing.

Let’s compare this to the actual graph of the function.

It looks very similar to the one we sketched!

Remember, the First Derivative Test is used to determine at which x-values the function has a maximum or a minimum. It only tells us relative maximums and minimums and not absolute ones. First, find the critical points of the function by taking the derivative of the function, setting it equal to zero, and solving for x. Then, plug in different numbers to determine if the derivative changes signs or not. If it changes from positive to negative, you have a relative maximum. If it changes from negative to positive, you have a relative minimum. And if it doesn’t change signs, you don’t have a maximum or a minimum. Then, you can use this information to draw a rough sketch of your graph.

I hope this video on the First Derivative Test was helpful. Thanks for watching and happy studying!