Like what you saw?
Create FREE Account and:
- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics
The First Derivative Test for Relative Maximum and Minimum - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
The relative maxima and minima are critical points of a function. So, as we did to find critical points, to find the relative maxima and minima, first calculate the derivative of the function. Then, find the points at which the derivative is either 0 or undefined. Then, find where the function is increasing and decreasing. A critical point is a maxima if the function changes from increasing to decreasing at that point, and a minima if the function changes from decreasing to increasing at that point.
Let’s look at an example where we find the relative maxima and the relative minima of a function. Here we have to look at the relative maxima in the function; F(x) equals x cubed minus 6x² minus 63x plus 42.
Now the first step is to find the derivative. And that’s pretty easy because this is a polynomial function. So the derivative is going to be 3x² minus 12x minus 63.
And you also want to completely factor the derivative, because in a moment, we are going to find critical points. So let’s factor the 3, which is common to each of these terms out. We get x² minus 4 x minus 21. And this also factors 3 and I’m going to have two linear factors starting with x. And I think about the factors of 21. I’m thinking -7 and plus 3 will work here. Well first of all, multiplying those will give me -21 and also I’ll get minus 7x plus 3x is -4x. That works perfectly.
So this is my derivative nicely factored, so that we can go on to the next step which is identify critical points of f. So let me write the derivative; 3x minus 7, x plus 3. The critical points are points where the derivative equals 0 or is undefined.
This is never going to be undefined, so we just need to find out where it equals 0. So set it equal to 0 and solve and since it's factored, it's very easy to solve. So we get x equals 7 or x equals -3. These are our critical points.
The third step is to make a sign chart for f', the derivative. So the first thing I need to do on the sign chart, is identify the critical points. I'll put with -3 here 7 here. And remember, the sign chart tells me where the derivative is positive, negative or 0. And the derivative is 0 at each of these points.
Now I’m going to have to test in each of these three intervals to see whether f' is positive or negative. Remember the reason I’m doing this, is I want to apply the first derivative test. And the first derivative test tells us whether we are going to have a relative max or min, in each of these points, depending on where the sign is of the derivative to the left and right.
So let’s test a point here, to the left of -3. So let’s try like -4. When I plug -4 into the derivative, I get 3 times -11, a negative number, times -1. So when I get negative times negative, a positive value here. And then I’ll test a point in here like 0. If I plug in 0, I get a negative number here and a positive here. Negative times positive is negative.
I’ll test a point over here like 8. 8 minus 7, positive, 8 plus 3, positive. Positive times positive is positive. So now I know that the derivative goes from positive to negative at x equals -3. And from negative to positive at x equals +7.
Sometimes I find it really helpful to draw little lines to indicate the direction that the graph is going. If its derivative is positive, that means f is going to be increasing. If its derivative is negative, it will be decreasing, and again positive increasing.
It becomes really obvious if you draw curves like this. What’s happening at x equals -3? It looks like we have a relative maximum. So we have a relative maximum at x equals -3. And this by the first derivative test. And we have a relative minimum at x equals 7. You can see that it's decreasing and then increasing. So relative minimum at x equals 7. You can see it all if you make a really good sign chart for f'.
But the justification here would be that we are using the derivative test. The relative max happens at x equals -3, because the derivative goes from positive to negative. And the relative min happens at x equals 7, because the derivative goes from negative to positive.
Stuck on a Math Problem?
Ask Genie for a step-by-step solution
Please enter your name.
Are you sure you want to delete this comment?
Sample Problems (3)
Need help with a problem?
Watch expert teachers solve similar problems.