# Question Video: Finding the Local Maximum and Minimum Values of a Function Involving Solving Exponential Equations Mathematics • Higher Education

Find, if they exist, the local maximum and/or minimum values of the function 𝑓(𝑥) = 9𝑒^(9𝑥) + 9𝑒^(−9𝑥). Also specify what type of value they are.

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### Video Transcript

Find, if they exist, the local maximum and/or minimum values of the function 𝑓 of 𝑥 equals nine 𝑒 to the nine 𝑥 plus nine 𝑒 to the negative nine 𝑥. Also specify what type of value they are.

We have a function, and we want to find its local maxima and minima. The local maxima and minima will occur at critical numbers of the function; that is, where the derivative of the function is zero or does not exist. So let’s differentiate. We differentiate term by term. What is the derivative of nine 𝑒 to the nine 𝑥? It’s nine times the derivative of 𝑒 to the nine 𝑥, which is nine 𝑒 to the nine 𝑥, and so we get 81𝑒 to the nine 𝑥.

Here we’ve used a special case of the chain rule, the derivative of a function of a number 𝑎 times 𝑥 is that number 𝑎 times the derivative of 𝑓 evaluated at 𝑎𝑥. In our application, 𝑎 was nine and the function 𝑓 was the exponential function 𝑒 to the 𝑥. Using the same rule, we can differentiate the second term nine 𝑒 to the negative nine 𝑥. That’s nine times negative nine 𝑒 to the negative nine 𝑥, because the derivative of 𝑒 to the negative nine 𝑥 is negative nine 𝑒 to the negative nine 𝑥.

Simplifying, we get 81𝑒 to nine 𝑥 minus 81𝑒 to the negative nine 𝑥. Remember, we’re looking for local maxima and minima of our function, which occur at the critical numbers where the derivative is zero or doesn’t exist. We can see that our derivative exists for all real values of 𝑥. There’s no value of 𝑥 for which this is undefined. And so the local maxima and minima, if they exist, occur when the derivative is zero. So let’s solve this equation. We can divide through by 81, add 𝑒 to the negative nine 𝑥 to both sides.

We divide through by 𝑒 to the nine 𝑥 using laws of exponents to write 𝑒 to the negative nine 𝑥 divided by 𝑒 to the nine 𝑥 as 𝑒 to the negative 18𝑥. We take the natural logarithm on both sides and see that 𝑥 must be zero. So we have a critical number of zero. And this is the only place where the function can have a local maximum or a local minimum. But we don’t know which happens at zero; it could be a local maximum or a local minimum. And in fact it could be neither! So how do we tell?

One way to do this, though not the only way, is to use the second derivative test. Suppose that 𝑓 prime of 𝑐, the derivative of 𝑓 at 𝑐, is zero and 𝑓 double prime, the second derivative of 𝑓, is continuous near 𝑐. In other words, it’s continuous in an open interval containing 𝑐. If the value of the second derivative of 𝑓 at 𝑐 is greater than zero, then 𝑓 has a local minimum at 𝑐. But if the second derivative of 𝑓 is less than zero at 𝑐, then 𝑓 has a local maximum at 𝑐.

If, however, the second derivative of 𝑓 at 𝑐 is zero, then we can’t say anything. They could be a local maximum or a local minimum or neither. Let’s clear some room so we can apply our test. The second derivative test involves the second derivative of 𝑓, hence the name. So we need to differentiate 𝑓 prime to get 𝑓 double prime, the second derivative of 𝑓. And this is a very similar process to differentiating 𝑓.

The derivative of 81𝑒 to the nine 𝑥 is 81 times nine 𝑒 to the nine 𝑥. And from this, we subtract the derivative of 81𝑒 to the negative nine 𝑥, which is 81 times negative nine times 𝑒 to the negative nine 𝑥. And the two minus signs make a plus, so we get 729𝑒 to the nine 𝑥 plus 729𝑒 to the negative nine 𝑥. Okay, now that we have the second derivative of 𝑓, let’s apply the second derivative test.

Suppose that 𝑓 prime of 𝑐 is zero. Well that’s true when 𝑐 is the critical number zero and 𝑓 double prime is continuous near 𝑐. Is our second derivative 𝑓 double prime continuous near zero? Yes, it’s continuous, not just around the critical number zero, but everywhere. The conditions for the second derivative test are satisfied then. The test itself relies on the sign of 𝑓 double prime at 𝑐.

Now what is the sign of 𝑓 double prime at zero? We substitute zero for 𝑥 and see that we can simplify the exponents: nine times zero is zero as is negative nine times zero; that’s also zero. So 𝑓 double prime of zero is 729𝑒 to the zero plus 729𝑒 to the zero. And anything to the power of zero is one. So we have 729 times one plus 729 times one. And that’s just 729 plus 729, which is 1458. The precise value of 𝑓 double prime of zero isn’t important. What is important is the sign of this value.

And in our case, we see that it’s positive. Looking across at the second derivative test, we see that if 𝑓 double prime of 𝑐 is greater than zero, and we’ve just seen that it is, then 𝑓 has a local minimum at 𝑐. Now what is this local minimum value? Well it’s 𝑓 of zero. So we substitute zero for 𝑥 in the expression for 𝑓 of 𝑥 that we have. It’s nine 𝑒 to the nine times zero plus nine 𝑒 to the negative nine times zero.

And as we saw before, both 𝑒 to the nine times zero and 𝑒 to the negative nine times zero are one. And so the value of our function at zero is nine plus nine equals 18. What’s our final answer: 𝑓 of zero equals 18, and this is a local minimum value of our function 𝑓 of 𝑥 equals nine 𝑒 to the nine 𝑥 plus nine 𝑒 to the negative nine 𝑥. This is the only local minimum of our function, and there are no local maxima.

Let’s recap! We differentiated 𝑓 to find the critical number zero, which is the only place at which a local maximum or minimum can occur. And then we used the second derivative test to test this critical value, showing that there was a local minimum there. The second derivative test depends on the sign of the second derivative at 𝑐. You can understand the second derivative test intuitively by considering the second derivative 𝑓 double prime as being the rate of change of the slope function 𝑓 prime.

At a local minimum of a function 𝑓, the slope 𝑓 prime is zero because the slope of the tangent to the curve at this point is zero. To the left of this point, the slope is negative. And to the right, it is positive. The slope, and therefore the slope function 𝑓 prime, is increasing as we pass 𝑐. So the rate of change of the slope function 𝑓 prime is positive, and the rate of change of the slope function 𝑓 prime is just the second derivative: 𝑓 double prime. So 𝑓 double prime is positive at 𝑐.

Similarly, near a local maximum, the slope function 𝑓 prime decreases as 𝑥 increases, starting positive but decreasing to zero as we reach the maximum and then decreasing further to become a negative as we pass this local maximum point. Hence at a local maximum, 𝑓 double prime, which is the rate of change of the slope function 𝑓 prime, is negative.