To show that the tangent line to the graph of #f# at #(c,f(c))# is parallel to the tangent line to the graph of #g# at #(c,g(c))#, it suffices to show that #f'(c) = g'(c)#.
Let #D# be the difference #f-g#
So #D(x) = f(x) - g(x)#
Observe that
#D'(x) = f'(x)-g'(x)#, so #D# is differentiable and therefore continuous on #[-1,2]#. (Part (b), "on #[a,b]#.")
Furthermore, #D(-1) = 0# and #D(2) = 0# (Part (b) substitute #a# and #b#.)
By Rolle's Theorem, there is a #c# in #(-1,2)# (in #[a,b]#) such that
#D'(c) = 0#
The implies that #f'(c) - g'(c)=0# so #f'(c) = g'(c)#.
To find the #c# in part (a), find #D'(x)#, set it to #0# and solve in the interval #(-1,2)#.
#D(x) = (x^2)-(-x^3+x^2+3x+2) = x^3-3x-2#
So #D'(x) = 3x^2-3#
and #3x^2-3 = 0# at #x = +-1#.
#-1# is not in #(-1,2)# but #1# is
so the only possible value for #c# is #c = 1#.
