# Question Video: Finding the Average Rate of Change Expression of a Rational Function Mathematics • Higher Education

Determine the average rate of change function 𝐴(ℎ) for 𝑓(𝑥) = (𝑥² + 2)/8𝑥 when 𝑥 changes from 𝑥₁ to 𝑥₁ + ℎ.

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### Video Transcript

Determine the average rate of change function 𝐴 of ℎ for 𝑓 of 𝑥 equals 𝑥 squared plus two over eight 𝑥 when 𝑥 changes from 𝑥 one to 𝑥 one plus ℎ.

The average rate of change of a function between 𝑎 and 𝑏 is 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. Looking at the graph of some function, suppose you want to see how the function changes as 𝑥 varies from 𝑎 to 𝑏, we can read off the values of 𝑓 of 𝑎 and 𝑓 of 𝑏 from the graph. The average rate of change of the function between 𝑎 and 𝑏 is the slope of the line segment connecting the two points on the graph. Now that we’ve seen the formula for the average rate of change of a function and we’ve seen how that formula relates to the graph of the function, we can apply this formula to the question we have.

So we want to find 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. Comparing our question to the definition we have, we see that 𝑎 is 𝑥 one and 𝑏 is 𝑥 one plus ℎ. So we replace 𝑏 by 𝑥 one plus ℎ and 𝑎 by 𝑥 one in our formula. What is 𝑓 of 𝑥 one plus ℎ? Well, it’s what you get when you replace 𝑥 by 𝑥 one plus ℎ in the definition of our function. 𝑥 squared becomes 𝑥 one plus ℎ squared and eight 𝑥 becomes eight 𝑥 one plus ℎ. And the same thing happens for 𝑓 of 𝑥 one, 𝑥 squared becomes 𝑥 one squared and eight 𝑥 becomes eight 𝑥 one. And finally, we can simplify the denominator, 𝑥 one plus ℎ minus 𝑥 one is just ℎ.

And having used the average rate of change formula and the definition of 𝑓 of 𝑥, the rest is just algebra. We can move the ℎ from the denominator in the big fraction to the denominators of the fraction in the numerator. This works because division is distributive over subtraction. Now we have the difference of two fractions, and we can combine those two fractions by finding a common denominator. Looking at these denominators, we have factors of eight, 𝑥 one plus ℎ, ℎ, and 𝑥 one. And the lowest common denominator is the product of those, eight times 𝑥 one plus ℎ times ℎ times 𝑥 one. To write this first fraction with this new denominator, we need to multiply the numerator by 𝑥 one. And the numerator of the second fraction, we need to multiply it by 𝑥 one plus ℎ before subtracting.

Now we have a single fraction where the numerator is quite complicated. So perhaps we can simplify by expanding the terms, but first I think we need to clear some space. Now we have some more room, we can expand the terms in the numerator. We expand 𝑥 one plus ℎ squared to 𝑥 one squared plus two ℎ𝑥 one plus ℎ squared. And we also expand 𝑥 one plus ℎ times 𝑥 one squared plus two to get 𝑥 one cubed plus two 𝑥 one plus ℎ𝑥 one squared plus two ℎ. The denominator of the fraction remains the same. We can expand further, multiplying 𝑥 one by everything in the first set of parentheses. And we can distribute the minus sign over the terms in the other set of parentheses.

Now that we’ve expanded as far as possible in the numerator, we can see that some terms cancel out. The 𝑥 one cubed cancels out with the minus 𝑥 one cubed. We can combine the next term, two ℎ𝑥 one squared, with the penultimate term, minus ℎ 𝑥 one squared. Two ℎ𝑥 one squared minus ℎ𝑥 one squared is just ℎ𝑥 one squared. And we can also cancel out plus two 𝑥 one with minus two 𝑥 one. We have a bit of a mess in the numerator, so let’s tidy it up. Having done this, we see that we have simplified greatly, and we can also see that all the terms in the numerator have a factor of ℎ which will cancel out with the factor of ℎ in the denominator.

Having done this, we are pretty much done. The only thing that we could do is expand the parentheses in the denominator as well. Doing this and swapping some of the terms, both in the numerator and the denominator, we get ℎ𝑥 one plus 𝑥 one squared minus two over eight ℎ𝑥 one plus eight 𝑥 one squared.

This is the average rate of change of the function 𝑓 of 𝑥 equals 𝑥 squared plus two over eight 𝑥, as 𝑥 changes from 𝑥 one to 𝑥 one plus ℎ. You might just like to think about what happens as ℎ gets smaller and smaller, closer and closer to zero.

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