MHB Solving Functions Problems: Lagrange and Rolle's Theorem

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The discussion revolves around a problem involving two differentiable functions, f and g, that share the same values at two points. The author of the problem suggests that statement b is correct, indicating that there exist points c1 and c2 where the derivatives of f and g are equal. However, another participant argues that by applying Rolle's Theorem to the function h(x) = f(x) - g(x), it can be shown that there exists a point c where the derivatives of f and g are equal, supporting statement a as the correct answer. Ultimately, it is concluded that both statements can be true under the given conditions. The conversation highlights the application of Lagrange's and Rolle's Theorems in determining the relationships between the derivatives of the functions.
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Hi, in a book I have found this problem:
"Let be $f,g:\mathbb{R}\rightarrow\mathbb{R}$ two derivable functions such that $f(0)=g(0)$ and $f(6)=g(6)$. Which of the following statements is necessarily true?:
a) $\exists\ c\in]0;6[ : f'(c)=g'(c)$;
b) $\exists\ c_1,c_2\in]0;6[ : f'(c_1)=g'(c_2)$.
"
The author of this question indicates the answer b because, for Lagrange's theorem $\exists\ c_1\in ]0;6[ : f'(c_1)=\dfrac{f(6)-f(0)}{6-0}$ and $\exists\ c_2\in ]0;6[ : g'(c_2)=\dfrac{g(6)-g(0)}{6-0}$, therefore $f'(c_1)=g'(c_2)$ but you can't be sure that $c_1=c_2$...
I think the author misses in fact if you call $h(x)=f(x)-g(x)$ then, the hypothesis of Rolle's theorem are true ($h(0)=h(6)$) therefore $\exists\ c \in ]0;6[: h'(c)=0$ then $f'(c)=g'(c)$. For me the correct answer is a.
Am I right?
 
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laura123 said:
Hi, in a book I have found this problem:
"Let be $f,g:\mathbb{R}\rightarrow\mathbb{R}$ two derivable functions such that $f(0)=g(0)$ and $f(6)=g(6)$. Which of the following statements is necessarily true?:
a) $\exists\ c\in]0;6[ : f'(c)=g'(c)$;
b) $\exists\ c_1,c_2\in]0;6[ : f'(c_1)=g'(c_2)$.
"
The author of this question indicates the answer b because, for Lagrange's theorem $\exists\ c_1\in ]0;6[ : f'(c_1)=\dfrac{f(6)-f(0)}{6-0}$ and $\exists\ c_2\in ]0;6[ : g'(c_2)=\dfrac{g(6)-g(0)}{6-0}$, therefore $f'(c_1)=g'(c_2)$ but you can't be sure that $c_1=c_2$...
I think the author misses in fact if you call $h(x)=f(x)-g(x)$ then, the hypothesis of Rolle's theorem are true ($h(0)=h(6)$) therefore $\exists\ c \in ]0;6[: h'(c)=0$ then $f'(c)=g'(c)$. For me the correct answer is a.
Am I right?

Hi laura! ;)

Yep. I believe you are right.
Therefore both statements are true.
 
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