Question

The value of c in Mean value theorem for the function, $f\left( x \right) = {x^2}$ in [2,4] is:
A. 4
B. 2
C. $\dfrac{7}{2}$
D. 3

Answer 1

Hint: We will be using the conditions of the Mean value theorem to find the value of c. For that, we will first find the value of the function at the endpoints of the interval. Then we will be differentiating the function, and equate it to the value of the derivative of the function at point c. After solving the obtained equation, we will get the value of c.

Complete step by step answer:

First let us note that the given function being a polynomial function, is continuous in the interval [2,4] and also it is differentiable in the interval (2,4). Hence, the given function satisfies the condition for the Mean value theorem.
Let us first consider what a Mean Value theorem really is. According to this theorem, if a function f is continuous on any closed interval, say $\left[ {a,b} \right]$ and it is differentiable on the open interval $\left( {a,b} \right)$ then there exists a point c in the open interval $\left( {a,b} \right)$, such that,
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Let us read the given question carefully and write down the unknown values in the condition for the Mean value theorem.
$
  a = 2, \\
  b = 4, \\
  f\left( x \right) = {x^2} \\
 $
Let us now find the value of the function at points 2 and 4.
$
  f\left( x \right) = {x^2} \\
   \Rightarrow f\left( a \right) = {a^2} \\
   \Rightarrow f\left( 2 \right) = {2^2} \\
   \Rightarrow f\left( 2 \right) = 4 \\
 $
And,
$
  f\left( x \right) = {x^2} \\
   \Rightarrow f\left( b \right) = {b^2} \\
   \Rightarrow f\left( 4 \right) = {4^2} \\
   \Rightarrow f\left( 4 \right) = 16 \\
 $
Now, let us find the value of the derivative of the function at the point c.
\[
   \Rightarrow f'\left( x \right) = 2x \\
   \Rightarrow f'\left( c \right) = 2c \\
 \]
Now, we will use the condition for the mean value theorem and substitute the obtained values into it.
$
   \Rightarrow f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \\
   \Rightarrow f'\left( c \right) = \dfrac{{f\left( 4 \right) - f\left( 2 \right)}}{{4 - 2}} \\
   \Rightarrow 2c = \dfrac{{16 - 4}}{{4 - 2}} \\
   \Rightarrow 2c = \dfrac{{12}}{2} \\
   \Rightarrow c = \dfrac{6}{2} \\
   \Rightarrow c = 3 \\
 $
Also, we note that 3 lies in between 2 and 4, thus this is a valid choice for the value of c, as it satisfies the Mean value theorem.
Hence, the value of c in Mean value theorem for the function, $f\left( x \right) = {x^2}$ in [2,4] is 3.

Note: You need to first check if the given function satisfies the conditions for the given function. If yes, then you will proceed with further solutions. If not, the question could not be solved. Also, make sure to check the continuity for the closed interval and differentiability for the open interval. Towards the end, when you get the value of c, verify that it lies in the interval given to you in the question.