A community for students.
Here's the question you clicked on:
 0 viewing
clara1223
 one year ago
PLEASE HELP
Will award medal and fan for an explanation to this question! Question in comments (Involves simplifying (f(xh)f(x))/h
clara1223
 one year ago
PLEASE HELP Will award medal and fan for an explanation to this question! Question in comments (Involves simplifying (f(xh)f(x))/h

This Question is Closed

clara1223
 one year ago
Best ResponseYou've already chosen the best response.1Given \[f(x)=3x+\frac{ 1 }{ x }3 \] Simplify \[\frac{ f(xh)f(x) }{ h }\] When x=4

itsmichelle29
 one year ago
Best ResponseYou've already chosen the best response.0what are u looking for

clara1223
 one year ago
Best ResponseYou've already chosen the best response.1I just need to simplify (f(xh)f(x))/h when x=4

clara1223
 one year ago
Best ResponseYou've already chosen the best response.1another way to write \[3x+\frac{ 1 }{ x }3\] is \[\frac{ 3x ^{2}3x+1 }{ x }\] so then f(x+h) is \[\frac{ 3(x+h)^{2}3(x+h)+1 }{ x+h }\] and then f(x+h)f(x) is \[\frac{ 3(x ^{2}+2xh+h ^{2})3x3h+1 }{ x+h }\frac{ 3x ^{2}3x+1 }{ x }\] and then I give them the same denominator so that I can combine them\[\frac{ 3x ^{3}6x ^{2}h3xh ^{2}3x ^{2}3xh+x }{ x(x+h) }+\frac{ (x+h)(3x ^{2}+3x1) }{ x(x+h) }\]and then multiply the numerator of the second part\[\frac{ 3x ^{3}+3x ^{2}h+3x ^{2}+3xhxh }{ x(x+h) }\] and then cancel out as much as i can\[\frac{ 3x ^{2}h3xh ^{2}h }{ x(x+h) }\]and then I take an h out of the numerator\[\frac{ h(3x ^{2}3xh1) }{ x(x+h) }\] so basically my question is: how to I simplify:\[\frac{ \frac{ h(3x ^{2}3xh1) }{ x(x+h) } }{ h }\]

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.2To simplify, start by just substituting 4 in for every x. Then follow your order of operations.

clara1223
 one year ago
Best ResponseYou've already chosen the best response.1@marihelenh quick question, how to I get rid of the bottom h? would I multiply it by the denominator in the numerator? in other words, would I multiply x(x+h) by h to get a simple fraction instead of a fraction inside of a fraction?

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.2If you recall, when you divide fractions, you can multiply by the reciprocal and it will give you the same thing. So, you could multiply the top part by 1/h. dw:1437424044483:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0have you thought of doing this term by term ?! that way, the 3 drops out immediately and you therefore get closer to the answer that much more quickly. what you are trying to do here is calculate the derivative of your function \(f(x)\). and constants "drop out" :p so work out f(xh) then work out f(xh)  f (x) only then divide by h

clara1223
 one year ago
Best ResponseYou've already chosen the best response.1answer choices are: a) \[\frac{ 4912h }{ 164h }\] b) \[0\] c) \[\frac{ 3h23h+1 }{ h }\] d) \[\frac{ 59+4h }{ 4 }\] e) \[\frac{ 4912h }{ 16+4h }\]

clara1223
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 could you please explain to me how to do that? Someone else taught me how to do it that way but a faster way would be much appreciated!!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i cannot do it for you but: \( \large f(x)=−3x+ \frac{1}{x}−3\) so, \( \large f(xh)=−3(xh)+ \frac{1}{xh}−3\) and so you can see that working out \( \large f(xh) f(x)\) should already be a lot easier once you do that, you can think about dividing 'h' in
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.