Monday

March 2, 2015

March 2, 2015

Posted by **Catherine** on Tuesday, March 15, 2011 at 6:09pm.

dy/dx=________

B) Using your answer for dy/dx, fill in the following table of approximate y-values of points on the curve near x=1 y=2.

0.96 ______

0.98 ______

1.02 ______

1.04 ______

C) Finally, find the y-value for x=0.96 by substituting x=0.96 in the original equation and solving for y using a computer or calculator.

y(0.96)= ________

D) How large (in magnitude) is the difference between your estimate for y(0.96) using dy/dx and your solution with a computer or calculator?

___________

- Calculus Please Help -
**bobpursley**, Tuesday, March 15, 2011 at 6:12pmI am not going to do this for you. What are you stuck on, or dont understand?

- Calculus Please Help -
**Catherine**, Tuesday, March 15, 2011 at 6:27pmIn the first part... I got (-2x+y^2)/(3y^2+x2y) but apparently that's wrong which will make everything else wrong.... I just need to know how to do that part right...

- Calculus Please Help -
**Catherine**, Tuesday, March 15, 2011 at 7:01pmI see what is wrong with my equation... now I have (-2x+y^2)/(3y^2-x2y) and for the second part I tried to plug 1 for x and all the other values for y and then in the third part I plugged 2 for y and 0.96 for x and I got them all wrong:

my values are:

0.96 = -1.2765

0.98= -1.1285

1.02= -0.8875

1.04= -0.7885

and for part C) 0.255

and For part D) -1.5315

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