Question Video: Finding the Local Maximum Value of a Function Involving Using the Product Rule with Exponential Functions | Nagwa Question Video: Finding the Local Maximum Value of a Function Involving Using the Product Rule with Exponential Functions | Nagwa

# Question Video: Finding the Local Maximum Value of a Function Involving Using the Product Rule with Exponential Functions Mathematics • Third Year of Secondary School

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Determine where π(π₯) = 3π₯Β²π^(βπ₯) has a local maximum, and give the value there.

08:09

### Video Transcript

Determine where π of π₯ equals three π₯ squared π to the negative π₯ has a local maximum and give the value there.

We should recall first that a local maximum is a type of stationary or critical point. And the critical points of a function occur when its first derivative, in this case π prime of π₯, is equal to zero. More specifically, in order to be a local maximum, the first derivative of the function must change from positive immediately to the left of the critical point to negative immediately to the right of the critical point. We therefore need to find the value of π₯ and then the value of the function at which both of these things are true.

Our first step is going to be to find an expression for the first derivative or slope function of π of π₯, which we call π prime of π₯. Now, looking carefully at the function π of π₯, we see that it is a product. Itβs three π₯ squared, which is a polynomial term, multiplied by π to the power of negative π₯, an exponential term. In order to differentiate π of π₯, we need to recall the product rule, which tells us that for two differentiable functions π’ and π£, the derivative with respect to π₯ of their product π’π£ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Weβll therefore let π’ equal three π₯ squared and π£ equal π to the power of negative π₯. And then we need to find their individual derivatives with respect to π₯.

To differentiate three π₯ squared, first of all, we recall the general power rule of differentiation, which tells us that for real values of π and π, the derivative with respect to π₯ of π multiplied by π₯ to the πth power is equal to π multiplied by π multiplied by π₯ to the π minus first power. We multiply by the power and then we reduce the power by one. So the derivative with respect to π₯ of three π₯ squared is three multiplied by two multiplied by π₯ to the first power. Of course, three multiplied by two is six and π₯ to the first power is simply π₯. So dπ’ by dπ₯ is equal to six π₯.

To differentiate π to the power of negative π₯, we recall the exponential rule of differentiation. If we have π to the power of some function π of π₯, then the derivative of this with respect to π₯ is equal to π prime of π₯ multiplied by π to the power of π of π₯. We multiply by the derivative of the power or exponent. In this case, the exponent is negative π₯. And the derivative of negative π₯ with respect to π₯ is negative one. So we have that dπ£ by dπ₯ is equal to negative π to the power of negative π₯.

Now we can substitute into the product rule. π prime of π₯ is equal to π’ times dπ£ by dπ₯. Thatβs three π₯ squared multiplied by negative π to the negative π₯. Then, we add π£ times dπ’ by dπ₯. Thatβs π to the power of negative π₯ multiplied by six π₯. We can factor this expression by three π₯π to the power of negative π₯. And it gives three π₯π to the power of negative π₯ multiplied by negative π₯ plus two. So we found our expression for the first derivative of this function.

Now, to find the critical points of this function, which could be local maxima but they could also be local minima or points of inflection, weβre looking for the point at which π prime of π₯ is equal to zero. In general, π prime of π₯ can also be undefined at critical points. But that isnβt the case at a local maximum. So we arenβt concerned about this here. We can therefore take the factored expression we have for π prime of π₯, and set it equal to zero.

To find where a product is equal to zero, we can consider where each part of the product is equal to zero. So we have either three π₯ equals zero, π to the power of negative π₯ equals zero, or negative π₯ plus two equals zero. If three π₯ is equal to zero, then it follows that π₯ is equal to zero. And if negative π₯ plus two is equal to zero, it follows that π₯ is equal to two. On the other hand, there are no solutions to the equation π to the power of negative π₯ equals zero because π to the power of negative π₯ is strictly positive.

So we found that this function has two critical points. They occur when π₯ equals zero and π₯ equals two. But we donβt yet know whether they are local maxima. To determine this, we need to consider the first derivative of the function a little either side of each critical point. So we can see what the shape of the curve is around this point. This is called the first derivative test. So letβs clear some room to perform this test.

Letβs begin with the critical point at π₯ equals zero. We know that the slope at π₯ equals zero is equal to zero. And weβll now consider the slope a little either side. So weβll choose the integer values of negative one and positive one. We then evaluate the slope at each of these values of π₯ by substituting them into our slope function. Thatβs this function here. At π₯ equals negative one, π prime of negative one is equal to three multiplied by negative one multiplied by π to the power of negative negative one. And this is all multiplied by negative negative one plus two. That gives negative three π to the first power multiplied by one plus two, which simplifies to negative nine π to the first power, or simply negative nine π.

Now, weβre not actually concerned exactly what the slope is, but more whether it is positive or negative. So the slope here at π₯ equals negative one is less than zero. At positive one, the slope is equal to three multiplied by one multiplied by π to the negative one multiplied by negative one plus two. And that simplifies to three multiplied by π to the negative one, or three over π.

Again, weβre not overly concerned exactly what the value is. But we do observe that the slope is positive when π₯ equals one. If we consider then the slope of the curve at this critical point, it changes from negative to zero to positive. And this tells us that the critical point at π₯ equals zero is in fact a local minimum. We may assume then that our other critical point at π₯ equals two is a local maximum, but we need to check this.

Weβll perform the first derivative test again, this time using integer values of π₯ immediately either side of two. So weβre using the values one and three. We already found that the slope of the curve when π₯ is equal to one is three over π, which we found to be positive. At π₯ equals three, the slope is equal to three multiplied by three multiplied by π to the power of negative three multiplied by negative three plus two. And that simplifies to negative nine π to the power of negative three, or negative nine over π cubed. Again, we arenβt overly concerned with the exact value. But whatβs important is that the slope here is negative.

Weβve now found that, at this critical point, the slope changes from positive when π₯ equals one to zero when π₯ equals two to negative when π₯ equals three. And so the shape of the curve here is indeed the correct shape for a local maximum.

So weβve answered the first part of this question. We have determined where π of π₯ has a local maximum. Itβs when π₯ is equal to two.

The final part of the question is to give the value there, which means we need to evaluate the function itself when π₯ equals two. Substituting π₯ equals two into the function itself, remember, thatβs three π₯ squared multiplied by π to the power of negative π₯, gives π of two equals three times two squared multiplied by π to the power of negative two. That simplifies to 12π to the power of negative two. Or equivalently, we can write this as 12 over π squared.

And so we have our answer to the problem. This function π of π₯ has a local maximum when π₯ equals two. And the value of the function there is 12 over π squared.

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