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Question
If from Lagrange's mean value theorem, we have \[f' \left( x_1 \right) = \frac{f' \left( b \right) - f \left( a \right)}{b - a}, \text { then }\]
Options
a < x1 ≤ b
a ≤ x1 < b
a < x1 < b
a ≤ x1 ≤ b
Solution
a < x1 < b
In the Lagrange's mean value theorem,\[c \in \left( a, b \right)\] such that \[f'\left( c \right) = \frac{f\left( b \right) - f\left( a \right)}{b - a}\].
So, if there is \[x_1\] such that \[f'\left( x_1 \right) = \frac{f\left( b \right) - f\left( a \right)}{b - a}\] then \[x_1 \in \left( a, b \right)\].
\[\Rightarrow a < x_1 < b\]
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