Monotonicity, Maxima and Minima
2.0 Lagrange's Mean Value Theorem
2.0 Lagrange's Mean Value Theorem
Statement: If function $f(x)$ is continuous in the closed interval $[a,b]$ and is differentiable in the open interval $(a,b)$, then 
there exists atleast one value of $'c'$ such that $$f'(c) = \frac{{f(b) - f(a)}}{{b - a}}$$
there exists atleast one value of $'c'$ such that $$f'(c) = \frac{{f(b) - f(a)}}{{b - a}}$$Let us assume a curve $y=f(x)$ between points $A(a,f(a))$ and $B(b,f(b))$. The slope of the chord $AB$ is written as $${\text{Slop}}{{\text{e}}_{AB}} = \frac{{f(b) - f(a)}}{{b - a}}$$ Consider any point $'c'$ between $'a'$ and $'b'$ and draw a vertical line as shown in figure. The slope of the tangent at point $'C'$ is written as $f'(c)$. According to LMVT, $$f'(c) = \frac{{f(b) - f(a)}}{{b - a}}$$ We conclude that, if a curve $AB$ has a tangent at each of its point then there is a point $'C'$ on this curve between $'A'$ and $'B'$, the tangent at which is parallel to the chord $AB$.
Question 4. Using LMVT, prove that if $b>a>0$, then $$\frac{{b - a}}{{1 + {b^2}}} < {\tan ^{ - 1}}b - {\tan ^{ - 1}}a < \frac{{b - a}}{{1 + {a^2}}}$$
Solution: Let us assume the function be $$f(x) = {\tan ^{ - 1}}x\;;\quad x \in [a,b]$$ Applying LMVT, we get $$f'(c) = \frac{{{{\tan }^{ - 1}}b - \;{{\tan }^{ - 1}}a}}{{b - a}}\;;\quad {\text{for }}a < c < b$$ On differentiating the function $f(x)$, we get $$f'(x) = \frac{1}{{1 + {x^2}}}$$ Since, it is a monotonically decreasing function, we can say that $$\begin{equation} \begin{aligned} a < c < b \Rightarrow f'(b) < f'(c) < f'(a) \\ \Rightarrow \frac{1}{{1 + {b^2}}} < \frac{{{{\tan }^{ - 1}}b - {{\tan }^{ - 1}}a}}{{b - a}} < \frac{1}{{1 + {a^2}}} \\\end{aligned} \end{equation} $$
