Video Transcript
Does the mean value theorem apply for the function 𝑦 is equal to 𝑥 divided by the cos of 𝜋 by two 𝑥 plus one over the closed interval from zero to one?
The question gives us a function and it wants us to determine whether we can use the mean value theorem on this function over the closed interval from zero to one. So, let’s start by recalling what the mean value theorem for a function over an interval tells us. The mean value theorem tells us if we have a function 𝑓 which is continuous on a closed interval from 𝑎 to 𝑏 and that function 𝑓 is differentiable on the open interval from 𝑎 to 𝑏. Then there must exist a value 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 prime of 𝑐 is equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at 𝑎 divided by 𝑏 minus 𝑎.
So, we can see to use the mean value theorem, we have two conditions which must be true. First, our function 𝑓 must be continuous on a closed interval from 𝑎 to 𝑏 and our function 𝑓 must also be differentiable on the open interval from 𝑎 to 𝑏. In our case, we want to determine whether we can use the mean value theorem on the function 𝑦 is equal to 𝑥 divided by the cos of 𝜋 by two 𝑥 plus one on the closed interval from zero to one. So, we’ll set our value of 𝑎 equal to zero, our value of 𝑏 equal to one, and our function 𝑓 of 𝑥 to be 𝑥 divided by the cos of 𝜋 by two 𝑥 plus one.
So, to use the mean value theorem for this function on this interval, we need to show the following two conditions are true. We need to show our function 𝑓 of 𝑥 is continuous on the closed interval from zero to one. And we need to show our function 𝑓 of 𝑥 is differentiable on the open interval from zero to one. Let’s start by checking our function is continuous on the closed interval from zero to one.
The first thing we notice by looking at our function is it’s the quotient of two functions. It’s the quotient of 𝑥 and the trigonometric function the cos of 𝜋 by two 𝑥 plus one. And what’s more, we know both of these functions are continuous. So, our function 𝑓 of 𝑥 is the quotient between two continuous functions. Once more, we know the quotient between two continuous functions will always be continuous on its domain. So, let’s take a closer look at the domain of our function 𝑓 of 𝑥.
We can see our numerator is just 𝑥. This will be defined for all real values of 𝑥. The same is true in our denominator, so the only way our function will not be defined as if we’re dividing by zero. So, our function 𝑓 of 𝑥 is continuous everywhere, except when the cos of 𝜋 by two 𝑥 plus one is equal to zero. So, we want to check if any of these values of 𝑥 are within our closed interval from zero to one. So, let’s solve this equation.
We’ll start by subtracting one from both sides of the equation. This gives us the cos of 𝜋 by two 𝑥 is equal to negative one. At this point, there’s a few different ways of solving this. We’re going to use the fact that the cos of 𝜃 being equal to negative one means that 𝜃 must be equal to 𝜋 plus two 𝜋𝑛, where 𝑛 is any integer. In other words, 𝜃 must be an odd integer multiple of 𝜋. However, in our case, instead of 𝜃, we have 𝜋 by two 𝑥. So, for 𝑥 to be a solution of this equation, 𝜋 by two 𝑥 must be an odd integer multiple of 𝜋. So, we must have 𝜋 by two 𝑥 is equal to two 𝑛 plus one multiplied by 𝜋, where 𝑛 is any integer.
And we can simplify this equation. We’ll start by dividing both sides of our equation through by 𝜋 . Next, we’ll multiply both sides of this equation through by two. Distributing two over our parentheses, we get 𝑥 is equal to four 𝑛 plus two. So, our function 𝑓 of 𝑥 is not continuous whenever 𝑥 is equal to four 𝑛 plus two where 𝑛 is allowed to be in the integer. Remember, we want to check if there are any discontinuities in our closed interval from zero to one. So, let’s find out where our discontinuities lie by substituting in different values of 𝑛.
If 𝑛 is equal to zero, we see that 𝑥 is equal to two. If 𝑛 is equal to one, we see that 𝑥 is equal to six. If 𝑛 is equal to negative one, then we see that 𝑥 is equal the negative two. And all we can do from this point on is add and subtract multiples of four. We can see this will never be within our closed interval from zero to one. So, the denominator of our function 𝑓 of 𝑥 is never equal to zero in the closed interval from zero to one. And this means our function 𝑓 of 𝑥 is continuous on this interval. So, we’ve shown that our first condition is true. Let’s now clear some space and work on our second condition.
We need to show our function 𝑓 of 𝑥 is differentiable on the open interval from zero to one. We’ll do this by finding an expression for 𝑓 prime of 𝑥. Since 𝑓 of 𝑥 is the quotient to two functions, we’ll do this by using the quotient rule. We recall the quotient rule tells us the derivative of the quotient of two functions 𝑢 over 𝑣 is equal to 𝑣 times 𝑢 prime minus 𝑢 times 𝑣 prime all divided by 𝑣 squared. In our case, the function in our numerator 𝑢 is equal to 𝑥 and the function in our denominator 𝑣 is equal to the cos of 𝜋 by two 𝑥 plus one.
To use the quotient rule, we need to find expressions for 𝑢 prime and 𝑣 prime. Let’s start with 𝑢 prime. That’s the derivative of 𝑥 with respect to 𝑥. We know this is just equal to one. To find an expression for 𝑣 prime, we need to recall for any constant 𝑘 the derivative of the cos of 𝑘𝑥 with respect to 𝑥 is equal to negative 𝑘 times the sin of 𝑘𝑥. In our case, our value of 𝑘 is 𝜋 by two. This gives us that 𝑣 prime is equal to negative 𝜋 by two times the sin of 𝜋 by two 𝑥.
We’re now ready to find an expression for 𝑓 prime of 𝑥 by using the quotient rule. We just need to substitute in our expressions for 𝑢, 𝑣, 𝑢 prime, and 𝑣 prime. This gives us the following expression for 𝑓 prime of 𝑥. And this is a very complicated-looking expression. We might be tempted to start simplifying. However, this is not necessary. Remember, we’re only looking to see if our function 𝑓 is differentiable on the open interval from zero to one. So, we just need to find the domain of this function. We don’t actually need to simplify our expression to do this.
For example, we can see that our numerator is defined for all real values of 𝑥 . This is because it’s the product, sum, and difference of functions which we know are defined for all real values of 𝑥. So, in fact, we only need to worry about our denominator. In fact, our denominator is also defined for all real values of 𝑥. So, the only way our derivative function could not be defined is if we divide by zero. But let’s take a closer at our denominator. It’s the cos of 𝜋 by two 𝑥 plus one all squared.
One of the entire expressions squared is equal to zero. We just must have the cos of 𝜋 by two 𝑥 plus one is equal to zero. But this is actually the denominator of our function 𝑓 of 𝑥. And when checking our first condition, we already showed this is not equal to zero on the closed interval from zero to one. This means it can’t be equal to zero on the open interval from zero to one. Therefore, our derivative function is defined for all values of 𝑥 on the open interval from zero to one.
Therefore, we were able to show that we can use the mean value theorem on the function 𝑦 is equal to 𝑥 divided by the cos of 𝜋 by two 𝑥 plus one on the closed interval from zero to one.
