Rolle's theorem?

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1 Answer
Jul 30, 2017

Apply Rolle's Theorem to D = f-gD=fg on [-1,2][1,2] (and on [a,b][a,b]). Then solve D'(x) = 0for part (a).

Explanation:

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.