Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find the average rate of change of the function over the given interval. Compare the average rate of change with the instant rate of change at the endpoints of the interval. f(x)=-1/x [1,2]

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

f(x)=-1x^-1 -1(-1)x^-2 1/x^2
1/(1)^2=1 1/(2)^2=2 instantaneous rates: f'(1)=1 ; f'(2)= 1/4
how do i find the average rate?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i know the average rate is 1/2 because the back of the book tells me, but I do not see how to get that from my work. :(
think of slope between the 2 endpoints avg rate = f(2) -f(1) / 2-1 = -1/2 -(-1) / 1 = 1/2
so basically f(1)-f(2)?
or more technically using calculus \[avg rate = \frac{1}{b-a}\int\limits_{a}^{b}f'(x) dx = \frac{f(b) -f(a)}{b-a}\]
im trying to understand really
so if i take the original equation f(x)=-1/x
then i do the f(2)-f(1)
wait +f1)
im just confusing myself now lol
yeah f(2) -f(1)

Not the answer you are looking for?

Search for more explanations.

Ask your own question