Question

Given the function f(x) = 2x^3 - 5x^2. Use the expression lim h-->0 f(x) f(x+h)- f(x)/h

Please help have exams in the morning :(

Answer 1

So your expression is what's called the "difference quotient." It's a geometric way of getting at the tangent to a polynomial like f(x) here.

Answer 2

\[
\frac{f(h+x)-f(x)}{h}=\frac{2
(h+x)^3-5 (h+x)^2-2 x^3+5
x^2}{h}=\\\frac{2 h^3+6 h^2 x-5 h^2+6
h x^2-10 h x}{h}=2 h^2+6 h x-5 h+6
x^2-10 x
\]

Answer 3

What he said

Answer 4

lol^

Answer 5

the limit when h goes to zero is
\[
-10 x +6 x^2
\]

Answer 6

definition of derivative =P

Answer 7

Yes

Answer 8

ontech does this make sense?

Answer 9

what she did is for f(x+h) , she replaced all the x's with (x+h) in the function


\[f(x)=2x^3-5x^2\]

Answer 10

Pretty sure he is the guy on the right. :)

Answer 11

left

Answer 12

And he is very quick with the equation editor.

Answer 13

?

Answer 14

Thanks .. im really trying to figure it out here..

Answer 15

Stil there @ontech I have a comment if you want...

Answer 16

im still here.. sorry for being so slow.. Im trying to figure it out as much as I can.. I need to understnad this..

Answer 17

Seems like you are. I would forget the cofactors and expand the basic functions
They expand with cofactors from Pascal's Triangle

(x+h)^2 = x^2 + 2xh + h^2
(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
(x+h)^4 = .. the coefficients will be 1 4 6 4 1

Don't add the cofactors (like the 2 in 2x^3 in your problem) until you have it all multiplied out.

Answer 18

And of course the basic point here is that you will subtract the first term (that's f(x)) and divide by h, and then all the other terms disappear as h -> 0, and you have the derivative by the power rule. Which is the whole point.

Answer 19

ahhh