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Prove the second derivative of a function with a given property must have a zero

Consider a function f which is continuous everywhere on an interval [a,b] and has a second derivative f'' everywhere on the open interval (a,b). Assume the chord joining two points (a, f(a)) and (b, f(b)) on the graph of the function intersects the graph of the function at a point (c, f(c)) with c \in (a,b). Prove there exists a point t \in (a,b) such that f''(t) = 0.


Proof. Let g(x) be the equation of the line joining (a, f(a)) and (b, f(b)). Then define a function

    \[ h(x) = f(x) - g(x). \]

Since f and g intersect at the values a,b, and c this means

    \[ f(a) = g(a), \qquad f(b) = g(b), \qquad f(c) = g(c). \]

By our definition of h then we have

    \[ h(a) = h(b) = h(c) = 0. \]

Further, since f' and g' are continuous and differentiable on (a,b), we apply Rolle’s theorem twice: first on the interval [a,c] and then on the interval [c,b]. These two applications of Rolle’s theorem tells us there exist points c_1 and c_2 such that

    \[ h'(c_1) = h'(c_2) = 0, \qquad \text{with } a < c_1 < c < c_2 < b. \]

Then, we apply Rolle’s theorem for a third time, this time to the function h' on the interval [c_1, c_2] to conclude that there exists a t \in (c_1, c_2) such that h''(t) = 0. Then, since t \in (c_1, c_2) we know t \in (a,b) (since a < c_1 < c_2 < b). So,

    \begin{align*}  h(x) = f(x) - g(x) && \implies && h'(x) &= f'(x) - g'(x) \\  &&  \implies && h''(x) &= f''(x) &(g''(x) = 0 \text{ since } g \text{ is linear}). \end{align*}

Thus, f''(t) = 0 for some t \in (a,b). \qquad \blacksquare

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