Let f(x)=5·x2−2·x. Find (f(x+h)−f(x))/h if h≠0. Simplify your answer.
\[f(x) = 5x^2 -2x\]
Is that what you mean?
I don't think so. It didn't take that answer.
\[f'(x) =\frac{ f(x+h) - f(x) }{ h }\] \[f'(x) = \frac{ [5(x+h)^2 - 2(x+h)] - (5x^2-2x) }{ h }\]
Thats because thats not the answer. Thats the orignal question.
\[f'(x) = \lim_{h \rightarrow 0} \frac{ [5(x+h)^2 - 2(x+h)] - (5x^2-2x) }{ h }\]
Now simplify
\[(5h^{2}-4x+2h) /2\] is this right?
then if you took the limit as h->0 it would be: \[f'(x) = 10x-2\] but i dont think u need that.
How did you get 10x+5h+2 ?
It didn't take either of those answers either /:
\[f'(x) = \lim_{h \rightarrow 0} \frac{ [5(x+h)^2 - 2(x+h)] - (5x^2-2x) }{ h }\] \[f'(x) = \lim_{h \rightarrow 0} \frac{ 5(x^2+2hx+h^2) - 2x-h- 5x^2+2x) }{ h }\]
make sure you foil properly.
10x+5h-2 is correct.
I think i put +2 instead of -2 earlier
That was right. Thank you so much!
How do you do a testimony?
Go into my profile and write a testimony.. you might have to be a fan first..
Alright!
I don't see where to write a testimony. But I became a fan!
hmm i dont see where either.. oh well
oh i see it!
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
sweet thanks, let me know if u have more questions!
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