Question

if g(x) =x^2+3x-1, find g(x+\[\Delta\]x)-g(x)/\[\Delta\]x

Answer 1

hahaha definition of derivative

Answer 2

umm whatt?

Answer 3

are you in calculus?

Answer 4

yes......

Answer 5

that is the definition of a derivative of a function the answer will be 2x+3

Answer 6

you will find a much more simpler way to solve this but uuntil then you must use this

Answer 7

not quite, it will involve terms of delta as well

Answer 8

but wouldnt there have to be a delta x in the middle of that?

Answer 9

no the deltas will cancel i believe if i remember right

Answer 10

you're thinking of evaluating the limit as delta --> 0

Answer 11

well why the hell would they have you do this in calculus without the limit to 0?

Answer 12

it's just the difference quotient, not the actual derivative


it's a lead up to derivatives

Answer 13

so there is a delta x in it? can one of u show how u guys r getting two different answers?

Answer 14

o wow i never did it without it

jim_thompson is right. You will have deltas however the whole point of this is to use it in this formula

\[\lim_{\Delta x \rightarrow 0}\frac{ f(c + \Delta x)-f(c)}{\Delta x}\]

Answer 15

ohh ok and i ended up with 2x+deltax+3 and thanksss

Answer 16

not always...sometimes you are interested in the average rate of change and not the instantaneous rate of change

Answer 17

\[\large g(x) = x^2+3x-1\]


\[\large g(x+\Delta) = (x+\Delta)^2+3(x+\Delta)-1\]


\[\large g(x+\Delta) = x^2+2x\Delta+\Delta^2+3x+3\Delta-1\]


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\[\large \frac{g(x+\Delta)-g(x)}{\Delta}\]


\[\large \frac{(x^2+2x\Delta+\Delta^2+3x+3\Delta-1)-(x^2+3x-1)}{\Delta}\]


\[\large \frac{x^2+2x\Delta+\Delta^2+3x+3\Delta-1-x^2-3x+1}{\Delta}\]


\[\large \frac{2x\Delta+\Delta^2+3\Delta}{\Delta}\]


\[\large \frac{\Delta(2x+\Delta+3)}{\Delta}\]


\[\large \frac{\cancel{\Delta}(2x+\Delta+3)}{\cancel{\Delta}}\]


\[\large 2x+\Delta+3\]


\[\large 2x+3+\Delta\]


So \[\large \frac{g(x+\Delta)-g(x)}{\Delta}=2x+3+\Delta\]

Answer 18

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