Video: Finding the Average Rate of Change of Root Functions between Two Points

Evaluate the average rate of change of 𝑓(π‘₯) = √(2π‘₯ βˆ’ 1) when π‘₯ varies from 5 to 5.62.

01:50

Video Transcript

Evaluate the average rate of change of the function 𝑓 of π‘₯ is equal to the square root of two π‘₯ minus one when π‘₯ varies from five to 5.62.

The question wants us to calculate the average rate of change of our function 𝑓 of π‘₯. And we recall the average rate of change of the function 𝑓 of π‘₯ from π‘₯ is equal to π‘Ž to π‘₯ is equal to 𝑏 is given by 𝑓 average is equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at π‘Ž divided by 𝑏 minus π‘Ž. We want to calculate the average rate of change when π‘₯ is varying from five to 5.62.

So, we’ll set π‘Ž equal to five and 𝑏 equal to 5.62. We have that 𝑓 evaluated at five is equal to the square root of two times five minus one, which is the square root of nine which is equal to three. And we have that 𝑓 evaluated at 5.62 is equal to the square root of two times 5.62 minus one, which we can calculate to give us 3.2.

We’re now ready to calculate the average rate of change of our function 𝑓 of π‘₯, when π‘₯ is varying from 5 to 5.62. Using our formula, we have that 𝑓 average is equal to 𝑓 evaluated at 5.62 minus 𝑓 evaluated at five divided by 5.62 minus five. We already calculated that 𝑓 evaluated at 5.62 is 3.2, and 𝑓 evaluated at five is equal to three.

In our denominator, we have 5.62 minus five is equal to 0.62. We can then evaluate this expression to give us 10 divided by 31. So, what we’ve shown is the average rate of change of our function 𝑓 of π‘₯ is equal to the square root of two π‘₯ minus one when π‘₯ varies from 5 to 5.62 is 10 divided by 31.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.