# Video: Using the Product Rule

The function β(π₯) = (π₯Β³ + ππ₯Β²)π^(βπ₯) has a critical number at π₯ = 1. Find π and list all the critical numbers.

06:20

### Video Transcript

The function β of π₯ is equal to π₯ cubed plus ππ₯ squared multiplied by π to the power of negative π₯ has a critical number at π₯ is equal to one. Find π and list all of the critical numbers.

The question gives us a function β of π₯, which is the product of a polynomial and an exponential function. And weβre told this function β of π₯ has a critical number when π₯ is equal to one. We need to use this information to find the value of the constant π and to list all of the critical numbers of our function β of π₯. Letβs start by recalling what it means for some value π₯ is equal to π to be a critical number of a function β of π₯. We say that this is a critical number of the function β of π₯ if either of the following conditions are met. Either the derivative of β evaluated at π is equal to zero or the derivative of β evaluated at π does not exist.

In our case, β of π₯ is the product of a polynomial and an exponential function. And polynomials are differentiable for all real numbers and exponential functions are differentiable for all real numbers. So their product will be differentiable for all real numbers. So in our case, β prime of π₯ will exist for all real values of π₯. So we only need to find the values of π such that β prime of π is equal to zero. And remember, weβre told that our function has a critical number when π₯ is equal to one, so we must have the derivative of β evaluated at one is equal to zero. So to find our critical numbers, weβre going to need to find an expression for β prime of π₯. And β of π₯ is the product of two functions, so weβll differentiate this by using the product rule for differentiation.

The product rule tells us the derivative of the product of two functions π’ times π£ is equal to π£ times π’ prime plus π’ times π£ prime. So weβll set π’ to be our polynomial factor π₯ cubed plus ππ₯ squared and π£ to be our exponential factor π to the power of negative π₯. So β of π₯ is the product π’ of π₯ and π£ of π₯. To use the product rule, weβre going to need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of the cubic polynomial π₯ cubed plus ππ₯ squared. And we can do this term by term by using the power rule for differentiation.

We multiply by the exponent of π₯ and reduce this exponent by one. This gives us three π₯ squared plus two ππ₯. Next, we need to find an expression for π£ prime of π₯. Thatβs the derivative of π to the power of negative π₯ with respect to π₯. And we can differentiate this by using our derivative rules for exponential functions. For any constant π, the derivative of π to the power of ππ₯ with respect to π₯ is equal to π times π to the power of ππ₯. In our case, π is equal to negative one. So π£ prime of π₯ is equal to negative π to the power of negative π₯. Weβre now ready to find β prime of π₯ by using the product rule. Itβs equal to π£ of π₯ times π’ prime of π₯ plus π’ of π₯ times π£ prime of π₯.

Substituting in our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯, we get that β prime of π₯ is equal to π to the power of negative π₯ times three π₯ squared plus two ππ₯ plus π₯ cubed plus ππ₯ squared times negative π to the power of negative π₯. And we can simplify this expression. Weβll take out a factor of π to the power of negative π₯ and a factor of π₯. This gives us β prime of π₯ is equal to π₯ times π to the power of negative π₯ times three π₯ plus two π minus π₯ squared minus ππ₯. Now, remember, β prime of π₯ will be equal to zero for our critical numbers π₯. So for these critical numbers, we have the product of three factors is equal to zero, which tells us one of these factors must be equal to zero.

Since our first factor is just π₯, if π₯ is equal to zero, then β prime of zero is equal to zero. So zero is a critical number of our function. We also know that π is a positive number. So π raised to any power will be positive. This tells us π to the power of negative π₯ is never equal to zero for any value of π₯. So any remaining critical numbers of our function β of π₯ must be roots of our quadratic factor. In particular, the question tells us π₯ is equal to one is a critical number. So π₯ is equal to one is a factor of this quadratic. So weβll substitute π₯ is equal to one into this quadratic, and we know this is equal to zero. This tells us zero is equal to three times one plus two π minus one squared minus π times one.

Simplifying this expression, we get zero is equal to three plus two π minus one minus π. And then weβll just simplify this and solve for π, we get that π is equal to negative two. So weβve answered the first part of this question. We find our value for π. We just need to find the remaining critical numbers. Remember, we showed that our remaining critical numbers must be roots of our quadratic factor. So letβs substitute π is equal to negative two into this quadratic factor and then solve this equal to zero. We get three π₯ plus two times negative two minus π₯ squared minus negative two times π₯ is equal to zero. Simplifying this, we get three π₯ minus four minus π₯ squared plus two π₯ is equal to zero. Since three π₯ plus two π₯ is equal to five π₯, we can simplify this to five π₯ minus four minus π₯ squared is equal to zero.

Next, weβll multiply both sides of this equation by negative one. We get π₯ squared minus five π₯ plus four is equal to zero. And thereβs a few different ways of solving this quadratic. For example, we could use the quadratic formula. However, we already know that π₯ is equal to one is a root of this quadratic. So by the factor theorem, π₯ minus one must be a root of this quadratic. We can then see by inspection or by algebraic division that our other factor will be π₯ minus four, since negative one times negative four is equal to positive four and negative one plus negative four is equal to negative five. And since the roots of this quadratic would be critical numbers of our function, we found our third and final critical number, π₯ is equal to four.

Therefore, if the function β of π₯ is equal to π₯ cubed plus ππ₯ squared times π to the power of negative π₯ has a critical number at π₯ is equal to one. Then weβve shown π must be equal to negative two. And π₯ is equal to zero, π₯ is equal to one, and π₯ is equal to four are all of the critical numbers of our function β of π₯.