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?
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