For what values of a and b is the line –2x+y=b tangent to the curve y=ax^3 when x=–2?
Caculus f'(x) - Brady, Friday, February 14, 2014 at 8:54pm
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) - Brady, Friday, February 14, 2014 at 9:01pm
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) - Brady, Friday, February 14, 2014 at 9:10pm
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.
Caculus f'(x) - Steve, Friday, February 14, 2014 at 10:58pm
y = ax^3
y' = 3ax^2
at x=-2, y' = 12a
The line y=2x+b has slope 2, so
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
To confirm this, visit
Caculus f'(x) - typo - Steve, Friday, February 14, 2014 at 10:59pm
typo near the end
y = 2x + 8/3