Question

If from Lagrange's mean value theorem, we have

$f\text{'}\left(x_1\right)=\frac{f\text{'}\left(b\right)-f\left(a\right)}{b-a},\mathrm{then}$
(a) a < x₁ ≤ b
(b) a ≤ x₁ < b
(c) a < x₁ < b
(d) a ≤ x₁ ≤ b

Answer 1

(c) a < x₁ < b

In the Lagrange's mean value theorem, $c\in \left(a,b\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

So, if there is $x_1$ such that $f\text{'}\left(x_1\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$, then $x_1\in \left(a,b\right)$.
$\Rightarrow a<x_1<b$