Question

How do you find the value of $ c $ that satisfy the equation $ \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right) $ in the conclusion of the mean value theorem for the function $ f\left( x \right) = 4{x^2} + 4x - 3 $ on the interval $ \left[ { - 1,0} \right] $ ?

Answer 1

Hint: The mean value theorem states that if a given function $ f $ is continuous in $ \left[ {a,b} \right] $ and differentiable in $ \left( {a,b} \right) $ , then there exists a point $ c $ in the interval $ \left( {a,b} \right) $ such that the derivative of the function at the point $ c $ can be given as $ f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} $ . We can use this theorem to find the value of $ c $ in the interval $ \left( {a,b} \right) $ .

Complete step by step solution:
We have been given that the function $ f\left( x \right) = 4{x^2} + 4x - 3 $ satisfies the mean value theorem in the interval $ \left[ { - 1,0} \right] $ .
 We have to find the value of $ c $ in the interval $ \left( { - 1,0} \right) $ such that $ \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right) $ .
Here $ a = - 1 $ and $ b = 0 $ .
We can find $ f\left( a \right) $ and $ f\left( b \right) $ using $ f\left( x \right) $ as,
 $
  f\left( a \right) = 4{a^2} + 4a - 3 \\
   \Rightarrow f\left( { - 1} \right) = 4{\left( { - 1} \right)^2} + \left( {4 \times - 1} \right) - 3 = 4 - 4 - 3 = - 3 \\
  f\left( b \right) = 4{b^2} + 4b - 3 \\
   \Rightarrow f\left( 0 \right) = 4{\left( 0 \right)^2} + \left( {4 \times 0} \right) - 3 = 0 + 0 - 3 = - 3 \;
  $
Also we can find the derivative of the function $ f\left( x \right) $ as,
 $ f'\left( x \right) = \dfrac{{d\left( {4{x^2} + 4x - 3} \right)}}{{dx}} = 8x + 4 $
Thus, for any point $ c $ we have,
 $ f'\left( c \right) = 8c + 4 $
We put all the values in the formula for mean value theorem.
 $
  f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \\
   \Rightarrow 8c + 4 = \dfrac{{\left( { - 3} \right) - \left( { - 3} \right)}}{{0 - \left( { - 1} \right)}} = \dfrac{{ - 3 + 3}}{{0 + 1}} = 0 \\
   \Rightarrow 8c = - 4 \\
   \Rightarrow c = - \dfrac{4}{8} = - \dfrac{1}{2} \;
  $
Thus, we get the value of $ c = - \dfrac{1}{2} $ .
Hence, the required value is $ c = - \dfrac{1}{2} $ .
So, the correct answer is “$ c = - \dfrac{1}{2} $ ”.

Note: We used the conclusion of the mean value theorem to get the value of $ c $ . We can see that the final value of $ c $ lies in the open interval $ \left( { - 1,0} \right) $ . While solving the problem we have to be careful that the value of $ a $ is the lower end of the given interval and the value of $ b $ is the upper end of the given interval. In this question we got $ f\left( a \right) = f\left( b \right) = - 3 $ which may not always be the case in mean value theorem. When $ f\left( a \right) = f\left( b \right) $ , this becomes a special case of mean value theorem, known as Rolle’s theorem.