Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Berndawg

  • 2 years ago

For session 2 dealing with derivatives as a rate of change. In the problem pdf for question 2. Taking f'(10) why do we automatically assume to take f(11) to find the value of 175 in order to validate the $40 dollar decrease?

  • This Question is Open
  1. MattBenjamins
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hello, f'(10) is the rate at which the bank balance is changing during the month October (the tenth month). f(10) is the actual balance in October and f(11) the balance in November. By subtracting f(10) from f(11) you can simply check how much the balance has changed from October to November. In this case: \[f(11)-f(10) = 175 - 220 = -45\] Which is close enough to -40 to verify the rate of change found by taking f'(10). On a side note. The solution pdf suggest trying f(10)-f(11) to check the change but that's a bad idea. The difference will still be 45 but it will be a positive number when the actual change has been a decrease. As a rule, subtract function values for smaller x's from function values for larger x's to not just get a change, but also whether that change has been positive or negative.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.