http://i42.tinypic.com/scv191.jpg
The problem, answer, and my questions are all in there ^^
Thanks in advance for any insight!
http://i42.tinypic.com/scv191.jpg
The problem, answer, and my questions are all in there ^^
Thanks in advance for any insight!
Uh, Soroban, that picture is copied from the answer key and is what janedoe is asking about!
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.