Question:
If from Lagrange's mean value theorem, we have
$f^{\prime}\left(x_{1}\right)=\frac{f^{\prime}(b)-f(a)}{b-a}$, then
(a) $a
(b) $a \leq x_{1}
(c) $a
(d) $a \leq x_{1} \leq b$
Solution:
(c) $a In the Lagrange's mean value theorem, $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$. So, if there is $x_{1}$ such that $f^{\prime}\left(x_{1}\right)=\frac{f(b)-f(a)}{b-a}$, then $x_{1} \in(a, b)$. $\Rightarrow a
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