Question

How do you determine the values of c that satisfy the mean value theorem on the interval \[\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right] \] for \[f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right)\] ?

Answer 1

Hint: In order to solve the above question, we will use the mean value theorem which states that if a function is continuous on the closed interval \[\left[ {a,b} \right] \] and differentiable in the open interval \[\left( {a,b} \right)\] such that \[f\left( a \right) = f\left( b \right)\] then \[f'\left( x \right) = 0\] for some \[c\] in \[\left[ {a,b} \right] \] . Using this concept, we will solve the above sum.
Formula used:
To solve the above question, we will be using the mean value theorem.
 \[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\] .

Complete step by step solution:
We are given: \[f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right)\] is continuous in \[\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right] \] .
Also, \[f\left( x \right)\] is differential in \[\left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right)\] .
So, there must exist \[c\] on \[\left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right)\] such that
 \[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\] .
On equating the values of \[a\] and \[b\] , we get,
 \[ \Rightarrow f'\left( c \right) = \dfrac{{f\left( {\dfrac{{3\pi }}{2}} \right) - f\left( {\dfrac{\pi }{2}} \right)}}{{\dfrac{{3\pi }}{2} - \dfrac{\pi }{2}}}\] .
Now, \[f\left( {\dfrac{{3\pi }}{2}} \right) = \sin \left( {\dfrac{{3\pi }}{4}} \right)\]
 \[ = \dfrac{{\sqrt 2 }}{2}\] .
And,
 \[f\left( {\dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)\]
 \[ = \dfrac{{\sqrt 2 }}{2}\] .
From this, we get that,
 \[
  f'\left( c \right) = \dfrac{{f\left( {\dfrac{{3\pi }}{2}} \right) - f\left( {\dfrac{\pi }{2}} \right)}}{\pi } \\
   \Rightarrow f'\left( c \right) = \dfrac{{\dfrac{{\sqrt 2 }}{2} - \dfrac{{\sqrt 2 }}{2}}}{\pi } \\
 \] ,
Now.
 \[f'\left( c \right) = 0\]
But,
 \[f'\left( x \right) = \left( {\dfrac{1}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)\] ,
 \[f'\left( c \right) = \left( {\dfrac{1}{2}} \right)\cos \left( {\dfrac{c}{2}} \right)\]
 \[ = 0\]
From this we get,
 \[ \Rightarrow \cos \left( {\dfrac{c}{2}} \right) = 0\] ,
This can be written as \[\dfrac{c}{2} = \dfrac{\pi }{2}\] , as \[{\cos ^{ - 1}}0 = \dfrac{\pi }{2}\] .
Therefore, we get, \[c = \pi \] .
So, from this we found the value of \[c\] , that is, \[c = \pi \] .
So, the correct answer is \[c = \pi \].

Note: While solving sums similar to the one given above, you need to remember Rolle’s mean value theorem in which he stated that if a function is continuous on the closed interval \[\left[ {a,b} \right] \] and differentiable in the open interval \[\left( {a,b} \right)\] such that \[f\left( a \right) = f\left( b \right)\] then \[f'\left( x \right) = 0\] for some \[c\] in \[\left[ {a,b} \right] \] . Also, you need to remember the formula to calculate \[f'\left( c \right)\] , which is, \[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\] . In questions similar to this, you also need to remember the values of trigonometric functions, for example, in this question we used the value of \[{\cos ^{ - 1}}0\] which is equal to \[\dfrac{\pi }{2}\] .