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Chapter 4: Problem 45

Let \(f:[a, b] \rightarrow \mathbb{R}\) be such that \(f^{\prime}\) is continuous on \([a, b]\) and \(f^{\prime \prime}\) exists on \((a, b)\). Given any \(\xi \in[a, b]\), show that there is \(c \in(a, b)\) such that $$ f(\xi)-f(a)=\frac{f(b)-f(a)}{b-a}(\xi-a)+\frac{f^{\prime \prime}(c)}{2}(\xi-a)(\xi-b) $$

Short Answer

Expert verified
By constructing an auxiliary function \(g(x)\) and applying Rolle's theorem twice, we showed that there exists \(c \in (a, b)\) such that the equation \[ f(\xi) - f(a) = \frac{f(b) - f(a)}{b - a}(\xi - a) + \frac{f^{\prime \prime}(c)}{2}(\xi - a)(\xi - b) \] holds for any \(\xi \in [a, b]\).

Step by step solution

01

Construction of the Auxiliary Function

Construct the auxiliary function \(g(x)\) as: \[g(x) = \frac{f(b) - f(a)}{b - a}(x - a) - f(x) + f(a)\] Notice that \(g(a) = g(b) = 0\). Now we will proceed to find the derivatives of \(g(x)\) on the given interval.
02

Derivatives of the Auxiliary Function

To differentiate \(g(x)\), let's find the first and second derivatives: \(g'(x) = \frac{f(b) - f(a)}{b - a} - f'(x)\) and \(g''(x) = -f''(x)\)
03

Application of Rolle's Theorem

Since \(g(a) = g(b) = 0\), and \(g'(x)\) exists and is continuous on \([a, b]\), we can apply Rolle's Theorem, which states that there exists at least one value \(c_1 \in (a, b)\) such that \(g'(c_1) = 0\). Therefore, we have: \(g'(c_1) = \frac{f(b) - f(a)}{b - a} - f'(c_1) = 0\)
04

Another Application of Rolle's Theorem

Now consider \(\xi \in [a, b]\). Let's find \(g'(a)\) and \(g'(\xi)\): \(g'(a) = \frac{f(b) - f(a)}{b - a} - f'(a)\) and \(g'(\xi) = \frac{f(b) - f(a)}{b - a} - f'(\xi)\) Since \(g'(c_1) = 0\) and \(g''(x)\) exists and is continuous on \((a, b)\), we can apply Rolle's theorem again to find at least one value \(c \in (a, b)\) such that: \(g''(c) = -f''(c) = \frac{g'(\xi) - g'(a)}{\xi - a}\)
05

Isolating the Target Expression

Isolating \(-f''(c)\) in the last equation, we get: \(-f''(c) = \frac{g'(\xi) - g'(a)}{\xi - a}\) \(-f''(c)(\xi - a) = g'(\xi) - g'(a)\) \(-f''(c)(\xi - a) = \frac{f(b) - f(a)}{b - a} - f'(\xi) - \frac{f(b) - f(a)}{b - a} + f'(a)\) \(-f''(c)(\xi - a) = f'(a) - f'(\xi)\) Now, let's rewrite this expression for \(g(x)\) (from Step 1): \(g(\xi) = \frac{f(b) - f(a)}{b - a}(\xi - a) - f(\xi) + f(a)\) Substitute the expression we derived in the previous step: \(\frac{f^{\prime \prime}(c)}{2}(\xi - a)(\xi - b) = \frac{f(b) - f(a)}{b - a}(\xi - a) - f(\xi) + f(a)\) We have successfully shown that there exists \(c \in (a, b)\) such that \[ f(\xi) - f(a) = \frac{f(b) - f(a)}{b - a}(\xi - a) + \frac{f^{\prime \prime}(c)}{2}(\xi - a)(\xi - b) \]

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Most popular questions from this chapter

Show that the \(x\) -axis is a normal to the curve \(y^{2}=x\) at \((0,0)\). If three normals can be drawn to this curve from a point \((a, 0)\), show that \(a\) must be greater than \(\frac{1}{2}\). Find the value of \(a\) such that the two normals, other than the \(x\) -axis, are perpendicular to each other.

Let \(m \in \mathbb{N}\) and \(f, g:[a, b] \rightarrow \mathbb{R}\) be such that \(f, f^{\prime}, \ldots, f^{(m-1)}\) as well as \(g, g^{\prime}, \ldots, g^{(m-1)}\) are continuous on \([a, b]\) and \(f^{(m)}, g^{(m)}\) exist on \((a, b)\). Suppose \(f^{\prime}(a)=f^{\prime \prime}(a)=\cdots=f^{(m-1)}(a)=0\) and \(g^{\prime}(a)=g^{\prime \prime}(a)=\cdots=\) \(g^{(m-1)}(a)=0\), but \(g^{(m)}(x) \neq 0\) for all \(x \in(a, b)\). Prove that there exist \(c_{1}, \ldots, c_{m} \in(a, b)\) such that $$ \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^{\prime}\left(c_{1}\right)}{g^{\prime}\left(c_{1}\right)}=\frac{f^{\prime \prime}\left(c_{2}\right)}{g^{\prime \prime}\left(c_{2}\right)}=\cdots=\frac{f^{(m)}\left(c_{m}\right)}{g^{(m)}\left(c_{m}\right)} $$

Let \(f:(a, b) \rightarrow \mathbb{R}\) be a function such that $$ |f(x+h)-f(x)| \leq C|h|^{r} \quad \text { for all } x, x+h \in(a, b), $$ where \(C\) is a constant and \(r \in \mathbb{Q}\) with \(r \geq 1\). Show that \(f\) is differentiable on \((a, b)\) and compute \(f^{\prime}(x)\) for \(x \in(a, b)\).

Give an alternative proof of Taylor's Theorem with a single application of Rolle's Theorem by proceeding as follows. Let the notation and hypothesis be as in the statement of Taylor Theorem (Proposition \(4.23\) ). Also, as in the proof of Taylor's Theorem, for \(x \in[a, b]\), let $$ P(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^{n} . $$ Define \(g:[a, b] \rightarrow \mathbb{R}\) by $$ g(x)=f(x)+f^{\prime}(x)(b-x)+\frac{f^{\prime \prime}(x)}{2 !}(b-x)^{2}+\cdots+\frac{f^{(n)}(x)}{n !}(b-x)^{n}+s(b-x)^{n+1}, $$ where \(s=[f(b)-P(b)] /(b-a)^{n+1} .\) Show that \(g(a)=g(b)=f(b) .\) Apply Rolle's Theorem to \(g\) to deduce Taylor's Theorem.

Let \(f:\left[-\frac{1}{2}, \frac{1}{2}\right] \rightarrow \mathbb{R}\) be given by $$ f(x)=\left\\{\begin{array}{ll} \sqrt{2 x-x^{2}} & \text { if } 0 \leq x \leq \frac{1}{2} \\ \sqrt{-2 x-x^{2}} & \text { if }-\frac{1}{2} \leq x \leq 0 . \end{array}\right. $$ Show that \(f\left(\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)\) but \(f^{\prime}(x) \neq 0\) for all \(x\) with \(0<|x|<\frac{1}{2}\). Does this contradict Rolle's Theorem? Justify your answer.
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