Question Video: Finding the Local Maximum and Minimum Values of a Function Involving Exponential Equations | Nagwa Question Video: Finding the Local Maximum and Minimum Values of a Function Involving Exponential Equations | Nagwa

# Question Video: Finding the Local Maximum and Minimum Values of a Function Involving Exponential Equations Mathematics • Third Year of Secondary School

## Join Nagwa Classes

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

07:17

### Video Transcript

Find the local maximum or minimum values, if they exist, of the function 𝑓 of 𝑥 equal to nine 𝑒 to the power of nine 𝑥 plus nine 𝑒 to the power of negative nine 𝑥 and specify their type.

We begin by recalling that the local maximum or minimum values occur at the critical points of a function. And these critical points are found where the first derivative 𝑓 prime of 𝑥 is equal to zero or where the function does not exist. In this question, we are given the function 𝑓 of 𝑥, which is equal to nine 𝑒 to the power of nine 𝑥 plus nine 𝑒 to the power of negative nine 𝑥. We will begin by differentiating this function term by term.

The general rule for differentiating an exponential function tells us that if 𝑦 is equal to 𝑎 multiplied by 𝑒 to the power of 𝑛𝑥, then d𝑦 by d𝑥 is equal to 𝑛 multiplied by 𝑎 multiplied by 𝑒 to the power of 𝑛𝑥. This means that differentiating nine 𝑒 to the power of nine 𝑥 gives us nine multiplied by nine multiplied by 𝑒 to the power of nine 𝑥. And this is equal to 81𝑒 to the power of nine 𝑥. In the same way, differentiating nine 𝑒 to the power of negative nine 𝑥 with respect to 𝑥 gives us negative 81𝑒 to the power of negative nine 𝑥.

We now have an expression for the first derivative of our function. 𝑓 prime of 𝑥 is equal to 81𝑒 to the power of nine 𝑥 minus 81𝑒 to the power of negative nine 𝑥. To find the critical values, we set this expression equal to zero. We can then divide through by 81, giving us 𝑒 to the power of nine 𝑥 minus 𝑒 to the power of negative nine 𝑥 is equal to zero. There are several ways to solve this equation for 𝑥. One way is to multiply both sides by 𝑒 to the power of nine 𝑥. The left-hand side is equal to zero.

When distributing the parentheses on the right-hand side, we recall one of our laws of exponents. 𝑥 to the power of 𝑎 multiplied by 𝑥 to the power of 𝑏 is equal to 𝑥 to the power of 𝑎 plus 𝑏. This means that 𝑒 to the power of nine 𝑥 multiplied by 𝑒 to the power of nine 𝑥 is equal to 𝑒 to the power of nine 𝑥 plus nine 𝑥, which simplifies to 𝑒 to the power of 18𝑥. Multiplying negative 𝑒 to the power of negative nine 𝑥 by 𝑒 to the power of nine 𝑥 gives us negative 𝑒 to the power of negative nine 𝑥 plus nine 𝑥, which is equal to negative 𝑒 to the power of zero. And as 𝑒 to the power of zero is equal to one, we have 𝑒 to the power of 18𝑥 minus one equals zero.

Adding one to both sides, we have 𝑒 to the power of 18𝑥 is equal to one. This means that the power 18𝑥 must equal zero. And we could equivalently see this by taking the natural logarithm of both sides. Dividing through by 18, we have 𝑥 is equal to zero. This means that the function 𝑓 of 𝑥 has a critical point when 𝑥 is equal to zero. And we can calculate the value of this by substituting 𝑥 equals zero into our original function. 𝑓 of zero is equal to nine 𝑒 to the power of nine multiplied by zero plus nine 𝑒 to the power of negative nine multiplied by zero, which simplifies to nine 𝑒 to the power of zero plus nine 𝑒 to the power of zero. And as 𝑒 to the power of zero is equal to one, we have nine plus nine, which is equal to 18. When 𝑥 is equal to zero, the function has a critical value that is 𝑓 of zero equals 18.

Our next step is to determine whether this is a local maximum or local minimum. To do this, we will consider the second derivative of our function. If the second derivative is greater than zero, we have a local minimum, whereas if the second derivative is less than zero, we have a local maximum. And if the second derivative 𝑓 double prime of 𝑥 equals zero, we have a possible inflection point. In order to find an expression for this second derivative, we will differentiate 𝑓 prime of 𝑥 term by term. 81 multiplied by nine is equal to 729. So differentiating 81𝑒 to the power of nine 𝑥 with respect to 𝑥 gives us 729𝑒 to the power of nine 𝑥.

We differentiate the second term in the same way, giving us 𝑓 double prime of 𝑥 is equal to 729𝑒 to the power of nine 𝑥 plus 729𝑒 to the power of negative nine 𝑥. We will now substitute the value of 𝑥 at the critical point into this expression. When 𝑥 is equal to zero, we have 729𝑒 to the power of nine multiplied by zero plus 729𝑒 to the power of negative nine multiplied by zero. Once again, since 𝑒 to the power of zero is equal to one, we are left with 729 plus 729. And this is equal to 1,458.

Since this value is positive, or greater than zero, we have a local minimum. We can therefore conclude that the function 𝑓 of 𝑥 has a local minimum value 𝑓 of zero equals 18 and has no local maximums and no other local minimums.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions