# Video: Finding the Local Maximum and Minimum Values of a Function Involving Exponential Functions If They Exist

Find (if any) the local maxima and local minima of 𝑓(𝑥) = 𝑒^(6𝑥) + 𝑒^(−𝑥).

05:07

### Video Transcript

Find, if any, the local maxima and local minima of 𝑓 of 𝑥 equals 𝑒 to the power of six 𝑥 plus 𝑒 to the power of negative 𝑥.

In order to find local minima and local maxima of a function, we begin by recalling how we find any critical points. The critical points of a function occur when the first derivative 𝑓 prime of 𝑥 is either equal to zero or does not exist. And there are two ways we can establish whether these critical points are local minimums or local maximums. The first is to use the second derivative test. We differentiate again; that gives us 𝑓 double prime of 𝑥. And if, at the critical points, the second derivative is greater than zero, we know we have a local minimum. If it’s less than zero, we know we have a local maximum.

Now, if the second derivative is equal to zero, that could indicate we have a point of inflection. But we would need to perform some second test to be sure. Alternatively, we could use the first derivative test. That involves substituting values just either side of the critical point into the expression for the first derivative and looking to see what the value of that expression tells us about the slope of the graph of that point. We’re going to use the second derivative test in this question though. And we can see quite clearly that, to find the critical points, we’re going to need to evaluate the first derivative.

So next, we recall that the derivative of 𝑒 to the power of 𝑎𝑥 with respect to 𝑥, where 𝑎 is some constant, is 𝑎𝑒 to the power of 𝑎𝑥. This means the derivative of 𝑒 to the power of six 𝑥 is six 𝑒 to the power of six 𝑥. And the derivative of 𝑒 to the power of negative 𝑥 is negative 𝑒 to the power of negative 𝑥. So we see that 𝑓 prime of 𝑥 is equal to six 𝑒 to the power of six 𝑥 minus 𝑒 to the power of negative 𝑥. To find our critical points, we’re going to set this equal to zero and solve for 𝑥.

Now, what we’re not going to do is attempt to divide through by some 𝑒 to the power of 𝑥. Instead, we’re going to factor the expression on the right-hand side of our equation. There are a couple of ways we can do this, but let’s factor by removing the common factor of 𝑒 to the power of negative 𝑥. And so we get 𝑒 to the power of negative 𝑥 times six 𝑒 to the power of seven 𝑥 minus one. Now, of course, six 𝑒 to the power of seven 𝑥 is there because we know that when we redistribute these parentheses, we add the exponents. So we get six 𝑒 to power seven 𝑥 times 𝑒 to the power of negative 𝑥 equals six 𝑒 to the power of six 𝑥.

Now, for this entire expression to be equal to zero, either 𝑒 to the power of negative 𝑥 is equal to zero or six 𝑒 to the power of seven 𝑥 minus one equals zero. But there’s no way the 𝑒 to the power of negative 𝑥 can be equal to zero. So this means that six 𝑒 to the power of seven 𝑥 minus one must be equal to zero. We’ll solve for 𝑥 by adding one to both sides. We then divide through by six. And then, for the next step, we recall that the natural log of 𝑒 to the power of 𝑥 is simply 𝑥. And so we’re going to take the natural log of both sides of this equation. When we do, we find that the natural log of 𝑒 to the power of seven 𝑥 is simply seven 𝑥. So seven 𝑥 is equal to the natural log of one-sixth.

But then we also know that the natural log of 𝑎 divided by 𝑏 can also be written as the natural log of 𝑎 minus the natural log of 𝑏. And so the natural log of one-sixth must be equal to the natural log of one minus the natural log of six. But then the natural log of one is zero. And so our equation becomes seven 𝑥 equals the negative natural log of six. And if we divide through by seven, we found our value of 𝑥, where the critical points occur. They occur at 𝑥 equals negative the natural log of six over seven. So we found the location of the critical point. Our next job is to determine whether this is a local maximum or local minimum. Let’s clear some space.

We said that one way we can test for local minimums and local maximums is by considering the second derivative. Well, if we differentiate 𝑓 prime of 𝑥 to get 𝑓 double prime of 𝑥, we get six times six 𝑒 to the power of six 𝑥, which is just 36𝑒 to the power of six 𝑥 minus negative 𝑒 to the power of negative 𝑥, which is, of course, plus 𝑒 to the power of negative 𝑥. We’re going to substitute 𝑥 equals negative the natural log of six over seven into this equation. Typing this into our calculator and we get 9.041. Now, this is of course greater than zero, so we have a local minimum. We now know we have a local minimum at 𝑥 equals negative the natural log of six over seven. All we need to do now is find the value of the function at that point.

So we’re going to substitute 𝑥 equals negative the natural log of six over seven into our original function. And that gives us 𝑓 of 𝑥 equals 𝑒 to the power of negative six times the natural log of six over seven plus 𝑒 to the power of the natural log of six over seven. Let’s see if we can simplify this somewhat. By using our laws of exponents, we can write 𝑒 to the power of negative six the natural log of six over seven as 𝑒 to the power of the natural log of six all to the power of negative six-sevenths. And of course, 𝑒 to the power of the natural log of six is simply six, whereas 𝑒 to the power of the natural log of six over seven is equal to 𝑒 to the power of the natural log of six all to the power of one-seventh.

And since 𝑒 to the power of the natural log of six is simply six, we find the value of the function at our local minimum to be six to the power of negative six over seven plus six to the power of one-seventh. And so the function 𝑓 of 𝑥 equals 𝑒 to the power of six 𝑥 plus 𝑒 to the power of negative 𝑥 has a local minimum at 𝑥 equals negative the natural log of six over seven. And this has a value of six to the power of negative six-sevenths plus six to the power of one-seventh.