dy/dx = 3 ax^2
when x = 4, dy/dx= 48 a = slope
but y = 4 x + b has slope of 4
so
48 a = 4
a = 4/48 = 1/12
and
y = (1/12) x^3
when x = 4
y = 16/3
-4(4) + 16/3 = b
-16 + 16/3 = b
b = -32/3
so
For what values of a and b is the line -4x+y=b tangent to the curve y=ax^3 when x=4?
Can anyone give me the hint?
2 answers
you can verify this at
http://www.wolframalpha.com/input/?i=plot+y%3D1%2F12+x^3,+y%3D4x-32%2F3,+0%3C%3Dx%3C%3D6
http://www.wolframalpha.com/input/?i=plot+y%3D1%2F12+x^3,+y%3D4x-32%2F3,+0%3C%3Dx%3C%3D6