A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

uselesscookies

  • one year ago

The following limits represent f'(a) for some function f and some real number a. a. Find a possible function f and number a. b. Find f'(a) by evaluating the limit. 1) lim h --> 0 (sqrt(9 + h) - sqrt(9))/h 2) lim h --> 0 ((1+h)^8 + (1 + h)^3 - 2)/h

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

    \[\large\rm \lim_{x\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}=f'(a)\]So we have to figure out our function :) hmm

  2. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So our definition tells us,\[\large\rm f'(x)=\lim_{x\to0}\frac{f(x+h)-f(h)}{h}\]If I want to evaluate this at some specific x, let's say 9, it would look like this:\[\large\rm f'(9)=\lim_{x\to0}\frac{f(9+h)-f(9)}{h}\]

  3. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you see how this relates to our problem that we're given? :d

  4. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sort of. But I don't know what to do with the square root in this case.

  5. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah silly type :3 ty

  6. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If you match up the pieces in the equations you can see this:\[\large\rm f(9+h)=\sqrt{9+h}\]\[\large\rm f(9)=\sqrt{9}\]

  7. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So it looks like our function f is simply \(\large\rm f(x)=\sqrt{x}\) and we're trying to find the derivative and evaluate it at x=a, a being 9 in this case.

  8. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So now we have to do the nasty algebra part, ya? :)

  9. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yup yup

  10. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It might not seem immediately obvious what to do, just try to remember some of these clever algebra tricks. For this problem, we're going to multiply the top and bottom by the `conjugate` of the numerator.\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\left(\frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}}\right)\]

  11. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is there a way to use derivatives to answer this problem? Because that's the lesson I'm currently on and I'm kinda confused about both limits and derivatives at the moment

  12. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use derivatives? like by using the derivative shortcuts? yes, but that's not what the problem is asking for :D they want you to do figure out the limit.

  13. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm

  14. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The chapter suggested that I should use derivatives to solve for the limit.

  15. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    They probably mean: Use the `limit definition of the derivative` to find a derivative function. That's what part B is asking for at least :\ If you look back at the original limit,\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\]Do you understand why we have a problem when we try to plug h=0 into this?

  16. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, it's 0/0

  17. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, good. The goal with limits is to try and get it to a point where you can plug the limit value directly in without running into trouble. Step 1: Plug h=0 directly in, oops we run into a problem, back up. Step 2: Do some algebra, try to cancel stuff out. Step 3: Plug h=0 directly in again, see if things have improved.

  18. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So we're on step 2, gotta do a fancy math trick. The conjugate of (a+b) is (a-b), they look the same, but with a different sign between. When we multiply them, we get the difference of squares. \(\large\rm (a-b)(a+b)=a^2-b^2\) We're going to do this to our problem because it will `get rid of the square roots in the numerator when we do`.

  19. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm \left(\sqrt{9+h}-\sqrt9\right)\left(\sqrt{9+h}+\sqrt9\right)=\sqrt{9+h}^2-\sqrt9^2=(9+h)-9\]

  20. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\left(\frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}}\right)=\lim_{h\to0}\frac{(9+h)-9}{h\left(\sqrt{9+h)}+\sqrt9\right)}\]We don't multiply out the bottom, we just leave it like that.

  21. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The numerator simplifies down a little further,\[\large\rm \lim_{h\to0}\frac{h}{h\left(\sqrt{9+h)}+\sqrt9\right)}\]Now we can make a cancellation!\[\large\rm \lim_{h\to0}\frac{\cancel h}{\cancel h\left(\sqrt{9+h)}+\sqrt9\right)}\]\[\large\rm \lim_{h\to0}\frac{1}{\left(\sqrt{9+h)}+\sqrt9\right)}\]

  22. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We found a way to cancel out some stuff. So we go to step 3, try to plug h=0 directly in again.

  23. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do we still run into a problem?

  24. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This process too confusing? What do you think? :3

  25. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Nope, it's not confusing ;_; thank you. Can you please help me with the other problem too?

  26. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We're not quite done with this one yet!! >.< What happens when we plug in h=0 from this point?

  27. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I can plug in 0 and get 1/6

  28. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yay good job \c:/

  29. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm \lim_{h\to0}\frac{(1+h)^8 + (1 + h)^3 - 2}{h}\]Hmm

  30. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm \lim_{h\to0}\frac{\color{orangered}{(1+h)^8 + (1 + h)^3} - 2}{h}\]This part represents f(x+h) being evaluated at some x value.\[\large\rm \color{orangered}{f(?+h)=(1+h)^8+(1+h)^3}\]What x value is being plugged in?

  31. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1?

  32. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm f(\color{royalblue}{1+h})=(\color{royalblue}{1+h})^8+(\color{royalblue}{1+h})^3\]Good, that seems to make sense here. So if we want to know the original function, let's evaluate this at x, not 1+h.\[\large\rm f(\color{royalblue}{x})=(\color{royalblue}{x})^8+(\color{royalblue}{x})^3\]Ok good, this is the original function we're taking the derivative of. If we evaluate this at x=1,\[\large\rm f(\color{royalblue}{1})=(\color{royalblue}{1})^8+(\color{royalblue}{1})^3\]\[\large\rm f(1)=2\]Which further agrees with our limit.

  33. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok good so we've found our function f(x)=x^8+x^3 and our x=a value a=1.

  34. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh boy this one would be pretty intense to do algebraically since we have an 8th power :\ hmmm

  35. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm a little confused. Is f(x) = 2?

  36. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm \lim_{h\to0}\frac{\color{orangered}{(1+h)^8 + (1 + h)^3} - 2}{h}=\lim_{h\to0}\frac{\color{orangered}{f(1+h)}-f(1)}{h}\] \(\large\rm f(1+h)=(1+h)^8+(1+h)^3\) This information led us to realize that \(\large\rm f(x)=(x)^8+(x)^3\)

  37. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    They really want us to evaluate the limit for this one? +_+ That's so mean lol. Have you learned any of your derivative shortcuts at this point? Power rule? :d

  38. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes. At least I'm supposed to have. My teacher kinda goes too fast I only get like brief understanding of it.

  39. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess we can use the shorter method if you prefer for this one :P The instructions were pretty vague... hmm

  40. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's just assumed that you would use derivatives since this section's about derivatives and the different methods for it. This particular set of questions has the title "Derivatives from Limits" if that helps

  41. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Recall our power rule for differentiation:\[\large\rm f(x)=x^n\qquad\to\qquad f'(x)=nx^{n-1}\]The power comes down in front, then we subtract 1 from the power. Example: \(\large\rm f(x)=x^4\qquad\to\qquad f'(x)=4x^3\)

  42. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Since we've determined the function for this particular problem:\[\large\rm f(x)=x^8+x^3\]\[\large\rm f'(x)=?\]Do you understand how we can apply our `power rule` to this?

  43. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I would assume that it would be f'x=8x^7 + 3x^2?

  44. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Good :) Now remember, we determined that we want to evaluate this derivative at x=1. \(\large\rm f'(1)=?\)

  45. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Uh...is my answer supposed to be in the 2 million range?

  46. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmmm no

  47. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large\rm f'(x)=8(x)^7+3(x)\]We're evaluating this derivative function at x=1,\[\large\rm f'(x)=8(1)^7+3(1)^2\]You're only raising the 1 to the 7th power, not the 8 as well.

  48. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh you're supposed to raise the exponent first?

  49. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So it'll just be f'(x) = 11

  50. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yay good job \c:/ woops I made a typo there, should be f'(1)=11

  51. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So I can just plug in 1 again for the rest of the problem after that?

  52. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, you're done. I just made a typo on the notation. \[\large\rm f'(x)=8(x)^7+3(x)^2\]I plugged 1 into the x's but forgot about the x on the left side. So the next line should have read\[\large\rm f'(1)=8(1)^7+3(1)^2\]But nothing to worry about :) done.

  53. uselesscookies
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Uh what about the -2 and the h on the bottom?

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.