# Video: Finding the Critical Points of a Function If They Exist by Using the Product Rule of Differentiation

Find the critical points of 𝑓(𝑥) = 𝑥²(𝑥 − 1)³.

04:20

### Video Transcript

Find the critical points of the function 𝑓 of 𝑥 equals 𝑥 squared multiplied by 𝑥 minus one all cubed.

The critical points of a function are those points where its first derivative, 𝑓 prime of 𝑥, is equal to zero or is undefined. Now, looking at our function 𝑓 of 𝑥, we can see that it is just the product of polynomial terms. And so, we know that its derivative will exist for all values of 𝑥. And so, we don’t need to be concerned about points where 𝑓 prime of 𝑥 is undefined. We do need to consider, though, how we’re going to find the derivative of this function, which is the product of 𝑥 squared and 𝑥 minus one cubed. It would be possible to distribute the parentheses to give a polynomial in 𝑥. But this could be a lengthy process, and we run the risk of making a mistake in our algebra.

Instead, we could label the two factors of our function 𝑓 of 𝑥 as 𝑢 and 𝑣 and then use the product rule to find its derivative. The product rule tells us that for two differentiable functions 𝑢 and 𝑣, the derivative of their product 𝑢𝑣 is equal to 𝑢𝑣 prime plus 𝑣𝑢 prime. We multiply each function by the derivative of the other and add them together. So we can let 𝑢 equal 𝑥 squared and 𝑣 equal 𝑥 minus one all cubed. And then, we need to find each of their derivatives with respect to 𝑥. 𝑢 prime or d 𝑢 by d𝑥 is straight forward. We can apply the power rule of differentiation to give two 𝑥. But what about d𝑣 by d𝑥?

Well, we have a polynomial 𝑥 minus one raised to the power of three. We can differentiate this using the general power rule, which is an application of the chain rule. And it tells us that the derivative of some function 𝑔 of 𝑥 to the power of 𝑛 is equal to 𝑛 multiplied by 𝑔 prime of 𝑥 multiplied by 𝑔 of 𝑥 to the power of 𝑛 minus one. This is similar to the power rule of differentiation. But we have an extra factor of 𝑔 prime of 𝑥, the derivative of what’s inside the parentheses. Applying the general power rule then, we have three multiplied by the derivative of 𝑥 minus one, which is just one, multiplied by 𝑥 minus one to the power of two. Which simplifies to three 𝑥 minus one all squared. And now, we’re ready to substitute into our formula for the product rule.

𝑓 prime of 𝑥 is equal to 𝑢𝑣 prime, that’s 𝑥 squared multiplied by three 𝑥 minus one all squared, plus 𝑣𝑢 prime, that’s 𝑥 minus one all cubed multiplied by two 𝑥. Now, we can factor this expression. Each term has a shared factor of 𝑥. And they also have a shared factor of 𝑥 minus one all squared. Remaining for the first term, we have a factor of three and another factor of 𝑥. And for the second term, we have a factor of two and another factor of 𝑥 minus one. So the factored form of our expression for 𝑓 prime of 𝑥 is 𝑥 multiplied by 𝑥 minus one squared multiplied by three 𝑥 plus two multiplied by 𝑥 minus one. Simplifying within the final set of parentheses gives that 𝑓 prime of 𝑥 is equal to 𝑥 multiplied by 𝑥 minus one squared multiplied by five 𝑥 minus two.

We said that critical points occurred when the first derivative is equal to zero. So we take our expression for 𝑓 prime of 𝑥, set it equal to zero, and solve the resulting equation for 𝑥. Setting each factor in turn equal to zero gives 𝑥 equals zero, 𝑥 minus one squared equals zero, and five 𝑥 minus two equals zero. Solving the second two equations gives 𝑥 equals one as a repeated solution and 𝑥 equals two-fifths. We can answer then that our function 𝑓 of 𝑥 has critical points at 𝑥 equals zero, 𝑥 equals two-fifths, and 𝑥 equals one. There isn’t a requirement to evaluate the function itself at each critical point in this question. But if we needed to do that, we could return to the function 𝑓 of 𝑥 and substitute each 𝑥-value in turn to find the value of the function at each critical point.