y"' = 24x-6
y" = 12x^2-6x+2a
y' = 4x^3-3x^2+2ax+b
y = x^4-x^3+ax^2+bx+c
Now what else do we know?
y(-1) = 8
y(1) = 4
y'(1) = 4
So,
1+1+a-b+c = 8
1-1+a+b+c = 4
4-3+2a+b = 4
Solve for a,b,c and you have
y = x^4-x^3+2x^2-x+3
I'll let you verify that it meets the conditions.
find the equation of the curve for which y'''=24x-6 if the curve passes through (-1,8) and is tangent to the line y=4x at (1,4).
help please thanks :-)
1 answer