Question

The value of c in Lagrange's theorem for the function f(x)=∣x∣ in the interval [-1, 1] is
A.0
B.½
C.-1/2
D.None existence in the interval

Answer 1

Hint: First check the definition of Lagrange's theorem that all conditions are valid or not for this question check the function given which is differentiable or not in the given interval.

Complete step-by-step answer:
Given f(x) = ∣x∣ in the interval [-1, 1]
Since, |x| is not differentiable at x=0
So, Lagrange theorem is not applicable to f(x) in [-1, 1].
Hence, no value of 'c' exists.
There is an option in which given none exists in the interval in option D.
Hence option D is the correct answer for the question.

Additional information:
The Mean Value Theorem (MVT)
Lagrange’s mean value theorem (MVT) states that if a function f(x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there is at least one point x=c on this interval, such that
\[f\prime \left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}\]
This theorem (also known as First Mean Value Theorem) allows to express the increment of a function on an interval through the value of the derivative at an intermediate point of the segment.
Proof.
Consider the auxiliary function
\[F\left( x \right) = f\left( x \right) + \lambda x\]
We choose a number λ such that the condition F(a)=F(b) is satisfied. Then
\[f\left( a \right) + \lambda a = f\left( b \right) + \lambda b\]
\[ \Rightarrow f\left( b \right) - f\left( a \right) = \lambda \left( {a - b} \right)\]
\[f\prime \left( c \right) - \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = 0\] \[ \Rightarrow \lambda = - \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\]
As a result, we have
\[ \Rightarrow F(x) = f(x) - \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}x\]
 The function F(x) is continuous on the closed interval [a, b] differentiable on the open interval (a, b) and takes equal values at the endpoints of the interval. Therefore, it satisfies all the conditions of Rolle’s theorem. Then there is a point c in the interval (a, b) such that
\[F\prime \left( c \right) = 0\]
It follows that
\[f\prime \left( c \right) - \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = 0\]
Or
\[f\prime \left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}\]


Note: Modulus functions are not differentiable at x=0.Lagrange’s mean value theorem (MVT) states that if a function f(x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there is at least one point x=c on this interval, such that
\[f\prime \left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}\]