Proof. Since \(f\) is continuous on \([a,b]\), it attains a maximum value \(M\) and a minimum value \(m\) there. Let \(K = f(a) = f(b)\), so \(m \leq K \leq M\).

If \(m = M\), then \(f\) is constant on \([a,b]\), and \(f'(c) = 0\) for every \(c\in (a,b)\).

Suppose now that \(m < M\). Then \(K < M\) or \(m < K\). Suppose that \(K < M\). Then the maximum is attained at some point \(c\) with \(a< c < b\), since \(f(a) = f(b) = K<M\). By (ii), \(f'(c)\) exists. Since \(f(x) \leq M\) for all \(x\in [a,b]\), if \(a< x < c\) then \[\frac {f(x) - f(c)}{x - c} = \frac {f(x) - M}{x - c}\geq 0 .\] Letting \(x \rightarrow c^-\) gives \begin {equation} \label {eq:rolleleft} f'(c) = \lim _{x \rightarrow c^-} \frac {f(x) - f(c)}{x - c} \geq 0 . \end {equation} On the other hand, if \(c < x < b\), then \[\frac {f(x) - f(c)}{x - c} = \frac {f(x) - M}{x - c}\leq 0,\] so that \begin {equation} \label {eq:rolleright} f'(c) = \lim _{x \rightarrow c^+} \frac {f(x) - f(c)}{x - c} \leq 0 . \end {equation} Together, (4.3) and (4.4) give \(0\leq f'(c)\leq 0\), so \(f'(c) = 0\).

If instead \(m< K\), the same argument applied at a point where \(f\) attains its minimum \(m\) gives \(f'(c) = 0\). □

Proof. Let \(K = \frac {f(b) - f(a)}{b - a}\) be the slope of the chord joining \((a, f(a))\) and \((b, f(b))\), whose equation is \(y = f(a) + K (x - a)\). Define \(G: [a,b] \longrightarrow \mathbb {R}\) by \[G(x) = f(x) - f(a) - K (x - a),\] the vertical distance from the chord to the graph. Then \(G\) is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(G(a) = 0 = G(b)\). By Rolle’s Theorem there is \(c\in (a,b)\) with \(0 = G'(c) = f'(c) - K\), that is, \(f'(c) = \frac {f(b) - f(a)}{b - a}\). □

Solution. Let \(a < x \leq b\). Then \(f\) is continuous on \([a,x]\) and differentiable on \((a,x)\), so by the Mean Value Theorem there is \(c\in (a,x)\) with \(f(x) - f(a) = f'(c)(x - a) = 0\). Hence \(f(x) = f(a)\) for every \(x\in [a,b]\).

Solution. Let \(a \leq x_1 < x_2\leq b\). By the Mean Value Theorem on \([x_1,x_2]\) there is \(c\) with \(x_1 < c < x_2\) and \(f(x_2) - f(x_1) = f'(c)(x_2 - x_1)\). Both factors are positive, so \(f(x_2) > f(x_1)\). Similarly, \(f'\geq 0\) gives an increasing function, \(f'<0\) a strictly decreasing one, and \(f'\leq 0\) a decreasing one.

Solution. The function \(h = f - g\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(h' = f' - g' = 0\) there. By the first example, \(h\) is constant, say \(h = K\). Hence \(f = g + K\). This is the fact behind the “\(+C\)” in indefinite integration.

Proof. If \(g(b) = g(a)\), Rolle’s Theorem would give a point of \((a,b)\) where \(g' = 0\), contrary to hypothesis. So \(g(b)\neq g(a)\). Define \(H: [a,b]\longrightarrow \mathbb {R}\) by \[H(x) = f(x) - f(a) - \frac {f(b) - f(a)}{g(b) - g(a)}\, \bigl (g(x) - g(a)\bigr ).\] Then \(H\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(H(a) = 0 = H(b)\). By Rolle’s Theorem there is a \(c\in (a,b)\) with \[0 = H'(c) = f'(c) - \frac {f(b) - f(a)}{g(b) - g(a)}\, g'(c) .\] Dividing by \(g'(c)\neq 0\) gives the result. Taking \(g(x) = x\) recovers the Mean Value Theorem. □

Solution. For \(x\neq y\), the Mean Value Theorem gives \(c\) between \(x\) and \(y\) with \(\sin x - \sin y = \cos c\,(x-y)\), and \(|\cos c|\leq 1\).

For the second inequality, the case \(x = 0\) is equality. If \(x\neq 0\), the Mean Value Theorem on the interval between \(0\) and \(x\) gives \(e^x - 1 = e^c x\) for some \(c\) between \(0\) and \(x\). If \(x>0\) then \(c>0\) and \(e^c>1\), so \(e^x - 1>x\). If \(x<0\) then \(c<0\) and \(e^c<1\), and multiplying by the negative number \(x\) gives \(e^cx>x\); so again \(e^x - 1>x\).

Solution. Apply the Mean Value Theorem to \(f(x) = \sqrt {x}\) on \([100,101]\): for some \(c\in (100,101)\), \[\sqrt {101} - 10 = \frac {1}{2\sqrt {c}} .\] Since \(10<\sqrt {c}<\sqrt {101}<11\), we get \(\frac {1}{22}<\sqrt {101} - 10<\frac {1}{20}\), that is, \[10.045 < \sqrt {101} < 10.05 .\] Using the upper bound once more, \(\sqrt {c}<10.05\) gives \(\sqrt {101} - 10 > \frac {1}{20.1}>0.04975\), so \(\sqrt {101}\in (10.04975, 10.05)\). (In fact \(\sqrt {101} = 10.04988\ldots \).)

Solution. Let \(p(x) = x^3+3x-1\). Since \(p(0) = -1<0<3 = p(1)\), there is a root in \((0,1)\). If there were two roots, Rolle’s Theorem would give a zero of \(p'(x) = 3x^2+3\) between them, but \(p'(x)\geq 3>0\). So the root is unique.

Let \(q(x) = x^4 - 4x + 1\). Then \(q'(x) = 4(x^3-1)\) vanishes only at \(x = 1\), so by Rolle’s Theorem \(q\) has at most two real roots. Since \(q(0) = 1>0\), \(q(1) = -2<0\) and \(q(2) = 9>0\), the Intermediate Value Theorem gives one root in \((0,1)\) and one in \((1,2)\). So there are exactly two.

Solution. Let \(s(t)\) be the distance travelled after \(t\) hours, a continuous and differentiable function on \([0,2]\) with \(s(0) = 0\) and \(s(2) = 150\). By the Mean Value Theorem there is \(c\in (0,2)\) with \[s'(c) = \frac {s(2) - s(0)}{2 - 0} = 75 .\] The speed \(s'(c)\) at that instant equals the average speed.