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# 数学代写|微积分代写Calculus代考|Geometric interpretation of the proof of Lagrange’s Theorem

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## 数学代写|微积分代写Calculus代考|Geometric interpretation of the proof of Lagrange’s Theorem

In order to prove Lagrange’s Theorem (see Fig. 5.1), it suffices to find a function that ‘Rollefies’ Lagrange’s Theorem (that is, a function that reduces Lagrange’s

Theorem to Rolle’s Theorem). Namely, it is convenient to set
$$\varphi(x)=y_{\text {curve }}-y_{\text {line } \mathrm{AB}}$$
that is
$$\varphi(x)=f(x)-\left(\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\right) .$$
Function $\varphi$ fulfils the hypothesis of Rolle’s Theorem, hence there exists $c \in$ $(a, b)$ such that $\varphi^{\prime}(c)=0$, that is $\varphi^{\prime}(c)=f^{\prime}(c)-m_{A B}=0$ hence $f^{\prime}(c)=m_{A B}$. Since
$$m_{A B}=\frac{f(b)-f(a)}{b-a}$$
the proof of Lagrange’s Theorem follows.
Note that in Fig. 5.2, the point $c$ is a point which maximizes the function $\varphi$, which means that $\varphi$ reaches its maximum at $c$. However, in general the point $c$ provided by Rolle’s Theorem could be any local maximum or minimum.

## 数学代写|微积分代写Calculus代考|Geometric interpretation of the proof of Cauchy’s Theorem

For simplicity we assume that $f(a)=g(a)=0$. We consider the rectangles in Fig. 5.3 with sides depending on $x$. Their area is given by
\begin{aligned} & S_f(x)=f(x)(g(b)-g(a)) \ & S_g(x)=g(x)(f(b)-f(a)) \end{aligned}

We have that $S_f(a)=S_g(a)=0$ and $S_f(b)=S_g(b)$. Thus, we can apply Rolle’s Theorem to the function
$$S_f(x)-S_g(x)$$
hence there exists $c \in(a, b)$ such that
$$S_f^{\prime}(c)-S_g^{\prime}(c)=0$$
that is
$$f^{\prime}(c)(g(b)-g(a))=g^{\prime}(c)(f(b)-f(a))$$
Thus, keeping in mind the fact that the point $c$ provided by Rolle’s Theorem is actually a point which maximizes or minimizes the function $S_f(x)-S_g(x)$, we understand that the point $c$ from Cauchy’s Theorem is related to the gap between the size of two rectangles and ‘appears’ when the gap reaches a global maximum or a minimum but also when it reaches a local maximum or minimum.

Remark Is there a loss of generality in assuming that $f(a)=g(a)=0$ ? No. If the condition $f(a)=g(a)=0$ is not satisfied then we can simply consider the functions $g_1(x)=g(x)-g(a)$ and $f_1(x)=f(x)-f(a)$. Indeed:
$$\begin{array}{ll} g_1^{\prime}(x)=g^{\prime}(x), \quad f_1^{\prime}(x)=f(x), & g_1(b)-g_1(a)=g(b)-g(a) \ f_1(b)-f_1(a)=f(b)-f(a), \quad g_1(a)=f_1(a)=0 . \end{array}$$

## 数学代写|微积分代写Calculus代考|Geometric interpretation of the proof of Lagrange’s Theorem

$$\varphi(x)=y_{\text {curve }}-y_{\text {line } \mathrm{AB}}$$

$$\varphi(x)=f(x)-\left(\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\right) .$$

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