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.