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If from mean value theorem, $f^{\prime}\left(x_1\right)=\frac{f(b)-f(a)}{b-a}$, then
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The correct answer is:
$a \lt x_1 \lt b$
According to mean value theorem, In interval $[a, b]$ for $f(x)$
$\begin{aligned}& \frac{f(b)-f(a)}{b-a}=f^{\prime}(c) \\& \therefore a \lt x_1 \lt b .\end{aligned}, \text { where } a \lt c \lt b$
$\begin{aligned}& \frac{f(b)-f(a)}{b-a}=f^{\prime}(c) \\& \therefore a \lt x_1 \lt b .\end{aligned}, \text { where } a \lt c \lt b$
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