f = 6x(3-2x)^3
f' = 6(3-8x)(3-2x)^2
f(1) = 6
f'(1) = -30
so, the tangent line is
y-6 = -30(x-1)
see at
http://www.wolframalpha.com/input/?i=plot+y%3D6x%283-2x%29^3,+y%3D-30x%2B36
Suppose that
f(x)=6x(3−2x)3.
f(x)=6x(3−2x)3.
Find an equation for the tangent line to the graph of ff at x=1x=1.
1 answer