http://i42.tinypic.com/scv191.jpg

The problem, answer, and my questions are all in there ^^

Thanks in advance for any insight! (Happy)

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- Feb 26th 2009, 05:25 AMjanedoeCalculus...graphs/derivatives
http://i42.tinypic.com/scv191.jpg

The problem, answer, and my questions are all in there ^^

Thanks in advance for any insight! (Happy) - Feb 26th 2009, 06:11 AMSoroban
Hello, janedoe!

Looks good! . . . Nice work!

- Feb 26th 2009, 06:56 AMjanedoe
I think you misunderstood me...that's the book's answer, but I don't know how they got it. My questions are in blue at the bottom.

- Feb 26th 2009, 07:03 AMHallsofIvy
Uh, Soroban, that picture is copied from the answer key and is what janedoe is asking about!(Evilgrin)

janedoe, exactly how you would put it into a calculator depends on the calculator.

You certainly should be able to put in $\displaystyle y1= (2(x+ 0.01)- (x+0.01)^2- 2(x)+ x^2)/0.01$, for the straight line. The parabola is, of course, [tex]y2= 2x- x^2[/itex].

Finally, they know it is "similar to graph of the derivative of f(x)" because the derivative is given by $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)- f(x)}{h}$ and 0.01 is not very far from 0.