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Differentaitin help watch

1. Please can someone help, ive tried to do this so many times

1)find the x coordinate of the point on the curve y= (x^3)/3 - 3x^2 + 2/3 where the tangent is parallel to the tangent drawn at point (1,2)

thank you
2. (Original post by slothgal)
Please can someone help, ive tried to do this so many times

1)find the x coordinate of the point on the curve y= (x^3)/3 - 3x^2 + 2/3 where the tangent is parallel to the tangent drawn at point (1,2)

thank you
The point (1,2) isn't on the curve. Is that intentional, or should it be (1,-2) ?
3. yes sorry it should be (1,-2), thanks
5. (Original post by slothgal)
unpick the question

You want 2 parallel lines

How will you find the gradient of the tangent to the curve at (1, -2)
6. parallel lines have the same gradient, so it has the same gradient as the curve?
7. (Original post by slothgal)
parallel lines have the same gradient, so it has the same gradient as the curve?
yes
8. thank you, ill try to work it out now
9. sorry how can i differentiate (x^3)/3
10. (Original post by slothgal)
sorry how can i differentiate (x^3)/3
How would you usually differentiate
11. x^3 is 3x^2
12. (Original post by slothgal)
x^3 is 3x^2

What you're trying to say is that the first derivative of with respect to is .
13. (Original post by slothgal)
x^3 is 3x^2
yep

So if you divide that by 3 you will get ... ...
14. X^2! Thank you, im a fool
15. (Original post by slothgal)
X^2! Thank you, im a fool
less of a fool if you use the quote facility so that I can see you have responded

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