# Caculus f'(x)

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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!

• Caculus f'(x) - ,

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.

• Calculus f'(x) (Misspelled Subject Name) - ,

Oh, and by the way, it is actually Calculus instead of Caculus for the subject name of your problem. I just noticed it.

• Caculus f'(x) - ,

For a if it is 4, you will get the next equation of:

4 + 4∙8 = b

Finally, you solve, and you get:

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.

• Caculus f'(x) - ,

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

• Caculus f'(x) - typo - ,

typo near the end

y = 2x + 8/3