f(x)=6x^2+5 (f(a+h)-f(a))/h

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f(x)=6x^2+5 (f(a+h)-f(a))/h

Mathematics
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So in this example f(x)=6x^2+5, if I told you find f(2) would you know what to do?
yes its just plugging it in to the function
You would plug 2 in everyplace there was an x. In the problem above there is an (a+h) in the argument. So every place you see an x in the original replace it with (a+h). Then the second part shows f(a) so plug a in for every instance of x.

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Then continue to manipulate algebraically
\[f(a+h)=6(a+h)^2 +5 =6(a^2 +2ah +h^2)+5\]
\[6(a^2 +2ah +h^2)+5 = 6a^2 +12ah +6h^2 +5\]
and then subtract that from 6a^2+5?
yes :)
so i got 12ah+6h^2+10
so divide by h?
\[\frac{ 6a^2 +12ah +6h+5-(6a^2 +5) }{ h }\]
\[\frac{ 12ah +6h }{ h }=\frac{ h(12a+6) }{ h }=12a+6\]
Hope that helps
im sorry but it says its wrong
I see my mistake.... look up at the equation I posted just after you said "So divide by h?". In that equation I had 6h in the numerator. it should have been 6h^2. Then you would have had 12a+6h as an answer sorry about that
ohhhh okay got it, thanks!
Your welcome
It should only be \(\huge 12a\).
You're finding the derivative, right?
no his answer was right
Ok if that's what you're saying.
@Maharot there is one thing missing for it to be a derivative. This was an example of the "difference of a quotient" function which represents the slope of a line secant to two points on a curve. If you wanted a derivative you are asking for the slope of that same curve but at any point. In order to do that you need to consider the "h" in the above problem as a change in the x direction and then take the limit as delta x approaches infinity. That is to say that the two points on the curve are the same and the slope represented is at a point rather than in between two points

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