Caculus f'(x)
posted by Ryan B. on .
For what values of a and b is the line –2x+y=b tangent to the curve y=ax^3 when x=–2?
a= ?
b= ?
Thanks guys!

First, you rewrite the equation as:
2∙2 + a∙2^3
Next, you solve some of it to get:
4 + a∙8
Finally, if you can't figure out what a and b are, you can do what I did above, and try figuring it out. I did not know how to solve to get a and b, but at least you have the things above to help. 
Oh, and by the way, it is actually Calculus instead of Caculus for the subject name of your problem. I just noticed it.

For a if it is 4, you will get the next equation of:
4 + 4∙8 = b
Finally, you solve, and you get:
36 as your answer for b.
So, a equals 4 and b equals 36.
Hint: you can always estimate values for a and then you solve, and whatever you get as your answer will be b. 
y = ax^3
y' = 3ax^2
at x=2, y' = 12a
The line y=2x+b has slope 2, so
12a=2
a = 1/6
at x = 2, y = 8a = 4/3 so the line must go through (2,4/3)
So, the tangent line has slope 2 and goes through (2,4/3)
y = 2x+b
4/3 = 4 + b
b = 8/3
So, now we have
y = 2x  8/3
is tangent to
y = 1/6 x^3
at (2,4/3)
To confirm this, visit
http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F6+x^3+and+y+%3D+2x%2B8%2F3+for+x+%3D+4+..+4 
typo near the end
y = 2x + 8/3