Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

Let f(x)=5·x2−2·x. Find (f(x+h)−f(x))/h if h≠0. Simplify your answer.

OpenStudy (baldymcgee6):

\[f(x) = 5x^2 -2x\]

OpenStudy (baldymcgee6):

Is that what you mean?

OpenStudy (anonymous):

I don't think so. It didn't take that answer.

OpenStudy (baldymcgee6):

\[f'(x) =\frac{ f(x+h) - f(x) }{ h }\] \[f'(x) = \frac{ [5(x+h)^2 - 2(x+h)] - (5x^2-2x) }{ h }\]

OpenStudy (baldymcgee6):

Thats because thats not the answer. Thats the orignal question.

OpenStudy (baldymcgee6):

\[f'(x) = \lim_{h \rightarrow 0} \frac{ [5(x+h)^2 - 2(x+h)] - (5x^2-2x) }{ h }\]

OpenStudy (baldymcgee6):

Now simplify

OpenStudy (anonymous):

\[(5h^{2}-4x+2h) /2\] is this right?

OpenStudy (baldymcgee6):

then if you took the limit as h->0 it would be: \[f'(x) = 10x-2\] but i dont think u need that.

OpenStudy (anonymous):

How did you get 10x+5h+2 ?

OpenStudy (anonymous):

It didn't take either of those answers either /:

OpenStudy (baldymcgee6):

\[f'(x) = \lim_{h \rightarrow 0} \frac{ [5(x+h)^2 - 2(x+h)] - (5x^2-2x) }{ h }\] \[f'(x) = \lim_{h \rightarrow 0} \frac{ 5(x^2+2hx+h^2) - 2x-h- 5x^2+2x) }{ h }\]

OpenStudy (baldymcgee6):

make sure you foil properly.

OpenStudy (baldymcgee6):

10x+5h-2 is correct.

OpenStudy (baldymcgee6):

I think i put +2 instead of -2 earlier

OpenStudy (anonymous):

That was right. Thank you so much!

OpenStudy (anonymous):

How do you do a testimony?

OpenStudy (baldymcgee6):

Go into my profile and write a testimony.. you might have to be a fan first..

OpenStudy (anonymous):

Alright!

OpenStudy (anonymous):

I don't see where to write a testimony. But I became a fan!

OpenStudy (baldymcgee6):

hmm i dont see where either.. oh well

OpenStudy (baldymcgee6):

oh i see it!

OpenStudy (baldymcgee6):

along the top of ur screen, where it shows ppl online.. click that.. hover over my profile until the little box pops up and clcik write fan testimony

OpenStudy (baldymcgee6):

sweet thanks, let me know if u have more questions!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!