Proof. Since \(f\) is continuous on the closed interval \([a,b],\, f\) is bounded on \([a,b]\). Let \(\hspace {0.2cm} \displaystyle {M = \sup _{a\leq x \leq b}\, f(x)}\, \) \(\hspace {0.2cm} \displaystyle {m = \inf _{a\leq x \leq b}\, f(x)}\). Also let \(f(a) = K = f(b)\). Then we must have \(m \leq K \leq M\).
If \(m = K = M\), then \(f(x) = K\) for all \(x \in [a,b]\) and \(f\) is a constant. We must have \(f'(c) = 0\) for any \(c\) such that \(a\leq c\leq b\).
Suppose now that \(m \neq M\). Then either \(m < K\) or \(K < M\). Suppose that \(K < M\). Then there is some point \(c\) with \(a< c < b\) at which \(f(c) = M\), since a continuous function on a closed interval assume its maximum and its minimum values. By condition (ii) \(f'(c)\) exists because \(a< c < b\). But \(f(x) \leq M\) for all \(x\in [a,b]\), if \(a< x < c\) then we have \[\dfrac {f(x) - f(c)}{x - c} = \dfrac {f(x) - M}{x - c}\geq 0\] Thus \(0\leq \displaystyle {\lim _{x \rightarrow c^-} \frac {f(x) - f(c)}{x - c} = f'(c)}\hspace {0.3cm}\) i.e \(\hspace {0.3cm} f'(c) \geq 0\hspace {0.3cm}\cdots \cdots \hspace {0.3cm}(I)\)
On the other hand, if \(c < x < b\), then we have \[\dfrac {f(x) - f(c)}{x - c} = \dfrac {f(x) - M}{x - c}\leq 0\] So that \(\hspace {0.3cm}\displaystyle {f'(c) = \lim _{x \rightarrow c^+} \, \dfrac {f(x) - f(c)}{x - c} \leq 0}\hspace {0.3cm}\) i.e \(\hspace {0.2cm} f'(c) \leq 0 \hspace {0.3cm}\cdots \cdots \hspace {0.3cm} (II)\)
But now \(f'(c)\) must satisfy both condition \((I)\) and condition \((II)\). That is \(0\leq f'(c)\leq 0\).
Therefore we must have \(f'(c) = 0\).

A similar argument can be used in the case \(m< K\) to prove that at the point \(c\), \(\hspace {0.2cm} f(c) = m\) then \(f'(c) = 0\).

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Proof. Let \(\, K = \dfrac {f(b) - f(a)}{b - a}\,\) be the slope of the line joining \((a, f(a))\) and \((b, f(b))\). Then the equation of this line is \(y = f(a) + K \, (x - a)\). Define a function \(G:\, [a,b] \longrightarrow \mathbb {R}\) \begin {align*} G(x) & = f(x) - \big (f(a) + K (x - a)\big )\\ G(x) & = f(x) - f(a) - K\, (x - a) \end {align*}

Then \(G(x)\) is continuous on \([a,b]\) and is differentiable on \((a,b)\). \(\, G(a) = 0\)
\(G(a) = f(a) - f(a) - K\, (a-a) = 0\,\) and \(G(b) = 0\). Thus \(G\) satisfies the hypothesis of Rolle’s Theorem given in the Remark. So there is \(c\in (a,b)\) such that \(\hspace {0.3cm} 0 = G'(c) = f'(c) - K\) \begin {align*} \implies & f'(c) - K = 0\\ \implies & f'(c) = K = \dfrac {f(b) - f(a)}{b - a}\\ \end {align*} □

Proof. We shall use the MVT. Let \(x_1\), be such that \( a < x_1 \leq b\). Then \(f\) is continuous on \([a,x_1]\) and differentiable on \((a,x_1)\). By the Mean Value Theorem, there is \(c\in (a,x_1)\) such that \[\dfrac {f(x_1) - f(a)}{x_1 - a} = f'(c)\] That is, \(f(x_1) - f(a) = (x_1 - a)\, f'(c)\). But \(f'(x) = 0\) for all \(x\in (a,b)\). Therefore \(f'(c) = 0\). This implies that \(f(x_1) - f(a) = 0\implies f(x_1) = f(a)\). This is true for all \(x_1\) with \(a < x_1 \leq b\). Therefore we conclude that \(f\) is a constant on \([a,b]\).

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Proof. Let \(x_1,\, x_2\) be two numbers such that \(a < x_1 < x_2\leq b\). Then \(f\) is continuous on \([x_1,x_2]\) and differentiable on \((x_1, x_2)\). By the mean value theorem, there is \(c\) with \(x_1 < c < x_2\) such that \(f(x_2) - f(x_1) = (x_2 - x_1)\, f'(c)\). But \(f'(c) > 0\) and \(x_2 -x_1>0\), therefore we must have \(f(x_2) - f(x_1) > 0\) giving \(f(x_2) > f(x_1)\). Since this is true for all \(x_1, \, x_2\) with \(a < x_1 \leq b\). We conclude that \(f\) is strictly increasing on \([a,b]\).

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Proof. Note that since \(g'(x) \neq 0\) for all \(x\in (a,b)\) then \(g(b) - g(a)\neq 0\). Now, consider as in the proof of the mean value theorem the function \(\, H:\, [a,b]\longrightarrow \mathbb {R}\) defined by \[H(x) = f(x) - f(a) = \dfrac {f(b) - f(a)}{g(b) - g(a)}\, \big (g(x) - g(a)\big )\] Then \(H\) is continuous on \([a,b]\) is differentiable on \((a,b)\). Also \(H(a) = 0 = H(b)\). The conditions of the Rolle’s theorem are satisfied. Therefore, there is a \(c\) with \(a < c < b\) such that \(H'(c) = 0\). That is \(\hspace {0.3cm}0 = H'(c) = f'(c) = \dfrac {f(b) - f(a)}{g(b) - g(a)}\, g'(c)\) \begin {align*} \implies & \hspace {0.3cm}\dfrac {f(b) - f(a)}{g(b) - g(a)} \,\, g'(c) = f'(c)\\\\ \implies & \hspace {0.3cm}\dfrac {f(b) - f(a)}{g(b) - g(a)} = \dfrac {f'(c)}{g'(c)}\hspace {0.4cm}\text {as required}\\\\ \end {align*} □