Video: Finding Where the Local Maximum and Minimum Values for a Polynomial Function Occur

Find where (if at all) the function 𝑓(π‘₯) = βˆ’2π‘₯Β³ βˆ’ 9π‘₯Β² βˆ’ 12π‘₯ βˆ’ 15 has its local maxima and minima.

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Video Transcript

Find where, if at all, the function 𝑓 of π‘₯ equals negative two π‘₯ cubed minus nine π‘₯ squared minus 12π‘₯ minus 15 has its local maxima and minima.

So we have a polynomial function, in fact a cubic function, and we want to find its local maxima and minima. Now we have a theorem which says that if 𝑐 is a local maximum or minimum of 𝑓, then 𝑐 is a critical number of 𝑓. Okay, but what does it mean for 𝑐 to be a critical number of 𝑓?

A number 𝑐 in the domain of 𝑓 is a critical number of 𝑓 if the derivative 𝑓 prime of 𝑐 is zero or if 𝑓 prime of 𝑐 does not exist. So to find the local maxima and minima of the function, we have to find its critical numbers. And we find the critical numbers of 𝑓 by differentiating to find 𝑓 prime and seeing where 𝑓 prime is zero or undefined. So let’s differentiate our function.

We can differentiate term by term. And because our function is a polynomial function, all we need to know is how to differentiate a monomial. In other words, we need to know how to differentiate π‘Ž times π‘₯ to the 𝑛 with respect to π‘₯, where π‘Ž and 𝑛 are some real numbers. And hopefully, you know that that is π‘Ž times 𝑛 times π‘₯ to the 𝑛 minus one.

For the first term of our function, negative two π‘₯ cubed, π‘Ž is negative two and 𝑛 is three. And so differentiating, we get negative two times three times π‘₯ to the three minus one, which simplifies to negative six π‘₯ squared. From this, we subtract the derivative of the next term nine π‘₯ squared. Nine times two is 18, and π‘₯ to the two minus one is just π‘₯. And so this is 18π‘₯.

From this, we subtract the derivative of 12π‘₯, which is 12. And of course as the derivative of a constant is zero, this constant term doesn’t affect our derivative at all. And so this is our derivative: negative six π‘₯ squared minus 18π‘₯ minus 12. Remember now that we’re looking for the critical numbers of 𝑓, where the derivative 𝑓 prime either has a root or does not exist.

As our derivative 𝑓 prime is a polynomial, in fact it’s a quadratic polynomial, its domain is the set of real numbers. And so 𝑓 prime of 𝑐 exists for all 𝑐 in the set of real numbers. There is no value of 𝑐 for which 𝑓 prime of 𝑐 does not exist. And so the critical numbers, if there are any, are values of 𝑐 for which 𝑓 prime of 𝑐 is zero.

So we need to find when this quadratic polynomial is zero. And we do that by factoring the polynomial. First, we see that we can take out a common factor of negative six. And then we can factor inside the parentheses by inspection, getting negative six times π‘₯ plus two times π‘₯ plus one. So if 𝑓 prime of 𝑐 is zero, then substituting 𝑐 for π‘₯ in the expression for 𝑓 prime of π‘₯ above, we see that negative six times 𝑐 plus two times 𝑐 plus one must be zero.

And so by the zero product property, one of the expressions inside the parentheses must be zero. So either 𝑐 plus two is zero, in which case 𝑐 is negative two, or 𝑐 plus one is zero, in which case 𝑐 is negative one. These are then our two critical numbers: negative two and negative one. And we know that if 𝑓 has a local maximum or minimum anywhere, then it must be at one of these critical numbers.

However, we don’t yet know for each critical number whether there is a local maximum or a local minimum or indeed neither. To find out, we need to use another test. For a critical number 𝑐 of a function 𝑓, if 𝑓 prime changes sign from positive to negative at 𝑐, then 𝑓 has a local maximum at 𝑐. If 𝑓 prime changes sign from negative to positive at 𝑐, then 𝑓 has a local minimum at 𝑐. And otherwise if 𝑓 prime doesn’t change sign, then 𝑓 has neither local minimum nor local maximum at 𝑐.

This is called the first derivative test because it uses the first derivative 𝑓 prime of a function 𝑓 to determine whether there is a local maximum or minimum at a critical number of 𝑓, as opposed to the second derivative test which uses the second derivative of 𝑓, 𝑓 double prime, for this purpose. Now we need to check the sign of 𝑓 prime on both sides of each critical number to see whether this sign changes and hence whether there is a local maximum or local minimum of 𝑓 at the critical number in question.

And the easiest way to check the sign of 𝑓 prime is using the factored form, negative six times π‘₯ plus two times π‘₯ plus one. It’s straightforward to check the sign of each factor. And then using the fact that a negative times a negative is a positive and so on, we can find the sign of the entire expression. So let’s use a table to find the sign of 𝑓 prime. And as we said before, we’re going to use the signs of each factor, negative six π‘₯ plus two and π‘₯ plus one to help us.

So we put them in our table too. We want to know the sign of 𝑓 prime when π‘₯ is below the critical number negative two. At negative two itself, we know that 𝑓 prime is zero. We also need to know the sign of 𝑓 prime when π‘₯ is greater than the critical number negative two, but less than the other critical number negative one. Of course at negative one, we know that 𝑓 prime is zero. But to see if 𝑓 primer changes sign at negative one, we need to know the sign of 𝑓 prime when π‘₯ is greater than negative one.

Let’s start filling in this table. We want to know the sign of 𝑓 prime when π‘₯ is less than negative two. Well the sign of negative six is negative, as it always is. π‘₯ plus two is also negative in this region. Why? Well in this region, π‘₯ is less than negative two. And adding two to both sides, we see that π‘₯ plus two is less than zero. In other words, π‘₯ plus two is negative. And it’s a similar story for π‘₯ plus one. Adding one to both sides, we see that π‘₯ plus one is less than negative one. And so π‘₯ plus one is negative.

Having found the sign of the factors, we can find the sign of 𝑓 prime of six, which is negative six times π‘₯ plus two times π‘₯ plus one. So it’s a negative number times a negative number times another negative number. Well a negative number times a negative number is a positive number. And multiplying out by another negative number, we get back to negative again. So 𝑓 prime is negative in this region.

Moving on to the second region when π‘₯ is between negative two and negative one, negative six is still negative. As π‘₯ is between negative two and negative one, π‘₯ plus two is between zero and one. We just add two to the sides. And so π‘₯ plus two is suddenly positive. What about π‘₯ plus one? Well, adding one to all the parts of the inequality, negative two is less than π‘₯ is less than negative one, we see that π‘₯ plus one is between negative one and zero. And so it is negative.

In this region, 𝑓 prime is a product of a negative number, a positive number, and another negative number. And so it is positive. Finally, in the region π‘₯ is greater than negative one, negative six is negative. π‘₯ plus two is greater than one, and so it is positive. And π‘₯ plus one is greater than zero, and so it too is positive. And so 𝑓 prime of π‘₯ is the product of a negative number, a positive number, and another positive number. And so it is negative.

Now having found the sign of 𝑓 prime in each region, we can get rid of our working. This gives us some space to continue solving this problem. We can see then at negative two that 𝑓 prime changes sign from negative to positive. Looking again at the first derivative test, we see that if 𝑓 prime changes sign from negative to positive at 𝑐, then 𝑓 has a local minimum at 𝑐. And so 𝑓 has a local minimum at π‘₯ equals negative two.

We also see that 𝑓 prime changes sign from positive to negative at the critical number negative one. And so according to the first derivative test, 𝑓 has a local maximum at π‘₯ equals negative one. These two critical numbers were the only places where the function could have a local maximum or minimum. And so this is our final answer: the function 𝑓 of π‘₯ equals negative two π‘₯ cubed minus nine π‘₯ squared minus 12π‘₯ minus 15 as a local maximum at π‘₯ equals negative one and a local minimum at π‘₯ equals negative two.

Having found the answer, you might now like to check it using a graphing calculator or graphing software. Let’s recap how we found our answer. We first found the critical members of 𝑓; that is, the values of 𝑐 for which 𝑓 prime of 𝑐 is zero or 𝑓 prime of 𝑐 does not exist. These critical numbers are the only places where a local maximum or minimum can occur. And to find them, we need to differentiate our function. Having found these critical numbers, we checked the sign on either side of each of them.

In both cases, we found that the sign of that 𝑓 prime changed. And the direction of that change told us for each critical number whether it was a local maximum or a local minimum. Had the sign of 𝑓 prime have been the same on both sides of one of our critical numbers, then that’s critical number would have been neither a local minimum nor a local maximum of our function. Instead, it would have been a point of inflection of the graph of the function. But that didn’t happen in this case. We found a local maximum at π‘₯ equals negative one and a local minimum at π‘₯ equals negative two.

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