Questions & Answers

Question

Answers

Answer

Verified

94.2k+ views

Formula used:

Lagrange’s mean value theorem given by,

$ f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} $

Such that $ c $ will lie in the interval of $ \left[ {a,b} \right] $ .

So we have the function given as $ f\left( x \right) = {x^3} - 5x - 3 $ .

Since we know that the polynomial function is everywhere continuous and differentiable. Therefore, $ f\left( x \right) $ will be continuous on the interval $ \left[ {1,3} \right] $ and also will be differentiable on $ \left( {1,3} \right) $ . And in both conditions, Lagrange’s mean value theorem will be satisfied.

So by using Lagrange’s mean value theorem,

$ \Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}} $

And on solving the denominator we get

$ \Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2} $

Now we have $ f\left( x \right) = {x^3} - 5x - 3 $

On differentiating it with respect to $ x $ , we get the equation as

$ \Rightarrow f'\left( x \right) = 3{x^2} - 5 $

Now on solving for the value of $ x = 1 $ in $ f\left( x \right) = {x^3} - 5x - 3 $ , we get

$ \Rightarrow f\left( 1 \right) = - 7 $

Now on solving for the value of $ x = 3 $ in $ f\left( x \right) = {x^3} - 5x - 3 $ , we get

$ \Rightarrow f\left( 3 \right) = 9 $

Therefore, $ f'\left( x \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2} $

On substituting the values, we get

$ \Rightarrow 3{x^2} - 5 = 8 $

Now taking the constant term to the right side, we get

$ \Rightarrow 3{x^2} = 8 - 5 $

And on solving it, we get

$ \Rightarrow {x^2} = 1 $

And therefore, $ x = 1 $ which comes in the interval such that $ f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2} $

Hence, Lagrange’s theorem is verified.