Question Video: Finding the Rate of Change of a Reciprocal Function at a Given 𝑥-Value Mathematics

Video Transcript

Determine the rate of change of the function 𝑓 of 𝑥 is equal to five divided by nine 𝑥 at 𝑥 is equal to the square root of two.

The question wants us to find the rate of change of the function 𝑓 of 𝑥 when 𝑥 is equal to the square root of two. The rate of change of our function 𝑓 of 𝑥 when 𝑥 is equal to 𝑎 is given by 𝑓 prime of 𝑎. And we say this is equal to the limit as ℎ approaches zero of 𝑓 evaluated at 𝑎 plus ℎ minus 𝑓 evaluated at 𝑎 divided by ℎ if this limit exists. Since we want to find the rate of change when 𝑥 is equal to the square root of two, we’ll set 𝑎 equal to the square root of two and 𝑓 of 𝑥 to be five divided by nine 𝑥.

We can do this in steps. Let’s start by simplifying the numerator inside of our limit. With 𝑎 equal to the square root of two, we get 𝑓 evaluated at the square root of two plus ℎ minus 𝑓 evaluated at the square root of two. So we want to evaluate 𝑓 at the square root of two plus ℎ and then subtract 𝑓 evaluated at the square root of two. And, remember, 𝑓 of 𝑥 is five divided by nine 𝑥. So 𝑓 of 𝑥 evaluated at the square root of two plus ℎ is equal to five divided by nine times the square root of two plus ℎ. Similarly, 𝑓 evaluated at the square root of two is five divided by nine root two.

We now want to simplify this expression. We can do this by using cross multiplication. When we cross multiply these two fractions, in our denominator, we’ll get the product of the two denominators. That’s nine times root two plus ℎ multiplied by nine root two. And we get the first term in our numerator by multiplying our first numerator by our second denominator. That’s five multiplied by nine root two. Then we want to subtract the second term, which will be the numerator of our second fraction multiplied by the denominator in our first fraction. That’s five times nine times the square root of two plus ℎ.

And we can simplify the numerator in this expression. Five times nine is equal to 45. Multiplying out and simplifying our second term, we get 45 root two plus 45ℎ. So this gives us a numerator of 45 root two minus 45 root two plus 45ℎ. And we can simplify this further, since 45 root two minus 45 root two is equal to zero. This means our numerator is just negative 45ℎ. And we can simplify this even further. 45 divided by nine is equal to five.

So we’ve found the following expression for 𝑓 evaluated at the square root of two plus ℎ minus 𝑓 evaluated at the square root of two. So we can substitute this into our limit for 𝑓 prime of root two. That’s the limit as ℎ approaches zero of 𝑓 evaluated at root two plus ℎ minus 𝑓 evaluated at root two divided by ℎ. Substituting in our expression for the numerator, we get the limit as ℎ approaches zero of negative five ℎ divided by the square root of two plus ℎ times nine times the square root of two all divided by ℎ.

Now, we’re ready to simplify this expression. First, instead of dividing through by ℎ, we’re going to multiply through by one divided by ℎ. Next, we’ll cancel the shared factor of ℎ in our numerator and our denominator. This gives us the limit as ℎ approaches zero of negative five divided by the square root of two plus ℎ times nine times root two. And this is just the limit of a rational function. We can attempt to do this by direct substitution.

Substituting ℎ is equal to zero, we get negative five divided by root two plus zero multiplied by nine root two. And we can then just calculate this expression to get negative five divided by 18. Therefore, we’ve shown from first principles the rate of change of the function 𝑓 of 𝑥 is equal to five divided by nine 𝑥 where 𝑥 is equal to root two is given by negative five divided by 18.