A community for students.
Here's the question you clicked on:
 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
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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm \lim_{x\to0}\frac{\sqrt{9+h}\sqrt{9}}{h}=f'(a)\]So we have to figure out our function :) hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So 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:\[\large\rm f'(9)=\lim_{x\to0}\frac{f(9+h)f(9)}{h}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Do you see how this relates to our problem that we're given? :d

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Sort of. But I don't know what to do with the square root in this case.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0If you match up the pieces in the equations you can see this:\[\large\rm f(9+h)=\sqrt{9+h}\]\[\large\rm f(9)=\sqrt{9}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So now we have to do the nasty algebra part, ya? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0It 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.\[\large\rm \lim_{h\to0}\frac{\sqrt{9+h}\sqrt{9}}{h}\left(\frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}}\right)\]

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Is 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0use 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.

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0The chapter suggested that I should use derivatives to solve for the limit.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0They 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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So we're on step 2, gotta do a fancy math trick. The conjugate of (a+b) is (ab), they look the same, but with a different sign between. When we multiply them, we get the difference of squares. \(\large\rm (ab)(a+b)=a^2b^2\) We're going to do this to our problem because it will `get rid of the square roots in the numerator when we do`.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm \left(\sqrt{9+h}\sqrt9\right)\left(\sqrt{9+h}+\sqrt9\right)=\sqrt{9+h}^2\sqrt9^2=(9+h)9\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0The 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!\[\large\rm \lim_{h\to0}\frac{\cancel h}{\cancel h\left(\sqrt{9+h)}+\sqrt9\right)}\]\[\large\rm \lim_{h\to0}\frac{1}{\left(\sqrt{9+h)}+\sqrt9\right)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0We found a way to cancel out some stuff. So we go to step 3, try to plug h=0 directly in again.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Do we still run into a problem?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0This process too confusing? What do you think? :3

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Nope, it's not confusing ;_; thank you. Can you please help me with the other problem too?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0We're not quite done with this one yet!! >.< What happens when we plug in h=0 from this point?

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0I can plug in 0 and get 1/6

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm \lim_{h\to0}\frac{(1+h)^8 + (1 + h)^3  2}{h}\]Hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Ok good so we've found our function f(x)=x^8+x^3 and our x=a value a=1.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Oh boy this one would be pretty intense to do algebraically since we have an 8th power :\ hmmm

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0I'm a little confused. Is f(x) = 2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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 \(\large\rm f(x)=(x)^8+(x)^3\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0They 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

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Yes. At least I'm supposed to have. My teacher kinda goes too fast I only get like brief understanding of it.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0I guess we can use the shorter method if you prefer for this one :P The instructions were pretty vague... hmm

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0It'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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Recall our power rule for differentiation:\[\large\rm f(x)=x^n\qquad\to\qquad f'(x)=nx^{n1}\]The power comes down in front, then we subtract 1 from the power. Example: \(\large\rm f(x)=x^4\qquad\to\qquad f'(x)=4x^3\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Since 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?

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0I would assume that it would be f'x=8x^7 + 3x^2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Good :) Now remember, we determined that we want to evaluate this derivative at x=1. \(\large\rm f'(1)=?\)

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Uh...is my answer supposed to be in the 2 million range?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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.

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Oh you're supposed to raise the exponent first?

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0So it'll just be f'(x) = 11

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0yay good job \c:/ woops I made a typo there, should be f'(1)=11

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0So I can just plug in 1 again for the rest of the problem after that?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0No, 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.

uselesscookies
 one year ago
Best ResponseYou've already chosen the best response.0Uh what about the 2 and the h on the bottom?
Ask your own question
Sign UpFind 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
 Engagement 19 Mad Hatter
 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.