Q: Find all points on the graph of y = (1/3)X^3 + X^2 - X where the tangent line has slope 1

im not sure how to solve this problem or where to start. can some one help?

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- Oct 18th 2006, 02:36 PMvietFind points on the graph...
Q: Find all points on the graph of y = (1/3)X^3 + X^2 - X where the tangent line has slope 1

im not sure how to solve this problem or where to start. can some one help? - Oct 18th 2006, 02:53 PMCaptainBlack
Points on the graph of:

y = (1/3)x^3 + x^2 - x

where the tangen has slope 1, are the points where dy/dx=1.

So you need to differentiate (1/3)x^3 + x^2 - x with respect to x, and then

set the derivative equal to zero and solve the resulting equation. This will

give the x values where the slope is 1, the y values are obtained by

substituting these x's back into y = (1/3)x^3 + x^2 - x.

RonL - Oct 18th 2006, 03:14 PMviet
ok so the derivative is x^2+2x-1, x = -2, x = +sqrt(8), x = -sqrt(8)

then plug x into y = (1/3)X^3 + X^2 - X ? - Oct 18th 2006, 03:29 PMCaptainBlack
- Oct 18th 2006, 03:41 PMviet
ahh ok, thank you so much CaptainBlack.

-tom