Question

How do you determine all values of $c$ that satisfy the conclusion of mean value theorem on the interval $\left[ {0,2} \right]$ for $f\left( x \right) = 2{x^2} - 5x + 1$?

Answer 1

Hint: In other the determine all the values of $c$ for the given function after applying mean value theorem, first determine the value of variable $a,b$ by comparing the given interval with $\left[ {a,b} \right]$.Now find the value for $f\left( a \right)\,and\,f\left( b \right)$ by substituting the value of $a\,and\,b$ respectively in the function. Derive the function with respect to $x$ , to obtain $f'\left( c \right)$. Now According to Mean value theorem , put all values in $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ and solve the equation for $c$ to get the required answer.

Formula used:
Differentiation Rules:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$,
$\dfrac{d}{{dx}}\left( 1 \right) = 0$,
$\dfrac{d}{{dx}}\left( x \right) = 1$

Complete step by step answer:
We are given a quadratic function in variable $x$as
$f\left( x \right) = 2{x^2} - 5x + 1$
According to the Mean Value Theorem, MVT has two hypotheses:
$f\left( x \right)$is continuous on the closed interval $\left[ {a,b} \right]$
$f\left( x \right)$is differentiable on the open interval $\left[ {a,b} \right]$
Then there is a number $c$such that $a < c < b$which satisfies
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
In this question, we are given $f\left( x \right) = 2{x^2} - 5x + 1$over the interval $\left[ {0,2} \right]$.
Comparing given interval with $\left[ {a,b} \right]$, we get value of $a = 0$and $b = 2$
Let’s find out the value for $f\left( b \right)$by substituting $x$with the value of $b$in the function $f\left( x \right)$, we get
$
  f\left( b \right) = f\left( 2 \right) = 2{\left( 2 \right)^2} - 5\left( 2 \right) + 1 \\
  f\left( b \right) = 8 - 10 + 1 \\
 $
$f\left( b \right) = - 1$----------------(1)
Similarly, finding the value for $f\left( a \right)$by substituting$x$with the value of $a$in the function $f\left( x \right)$, we obtain
$
  f\left( a \right) = f\left( 0 \right) = 2{\left( 0 \right)^2} - 5\left( 0 \right) + 1 \\
  f\left( a \right) = 0 - 0 + 1 \\
 $
$f\left( a \right) = 1$---------------(2)
Now find the first order derivative of the function $f\left( x \right)$ with respect to variable $x$ using the rules and properties of derivative
Differentiating given function with respect to $x$, we get
$\dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{d}{{dx}}\left( {2{x^2} - 5x + 1} \right)$
As we know the derivative gets separated into all the terms and the constant part of any term is pulled out from the derivative , so we can rewrite our derivative as
$f'\left( x \right) = 2\dfrac{d}{{dx}}\left( {{x^2}} \right) - 5\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 1 \right)$
Now using the rules of derivative tha t$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, $\dfrac{d}{{dx}}\left( 1 \right) = 0$, $\dfrac{d}{{dx}}\left( x \right) = 1$ we get our equation as
\[
  f'\left( x \right) = 2\left( {2x} \right) - 5\left( 1 \right) + 0 \\
  f'\left( x \right) = 4x - 5 \\
 \]
To find $f'\left( c \right)$replace all the occurrences of $x$ in the above result with $c$
\[f'\left( c \right) = 4c - 5\]-----------(3)
Hence we have obtained $f\left( a \right),f\left( b \right)\,and\,f'\left( c \right)$
AS we can clearly see that the function given is continuous on the interval $\left[ {0,2} \right]$ and also it is differentiable as \[f'\left( x \right) = 4x - 5\] which exists all $x \in R$.Thus, function is differentiable on $\left[ {0,2} \right]$.
Now According to Mean Value Theorem(MVT), there exists at least on real number $c$ such that $c \in \left( {0,2} \right)$,
We have
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Put all the values obtained in the equations 1,2 and 3 in above, we have
$4c - 5 = \dfrac{{ - 1 - 1}}{{2 - 0}}$
Simplifying further above expression for the value of $c$, we get
$
  4c - 5 = \dfrac{{ - 2}}{2} \\
  4c - 5 = - 1 \\
  4c = - 1 + 5 \\
  4c = 4 \\
 $
Dividing both sides of the equation with the coefficient of $c$i.e.$4$, we get the value of $c$ as
$
  \dfrac{{4c}}{4} = \dfrac{4}{4} \\
  c = 1 \\
 $
Therefore, the value of $c$is equal to $1$ which lies in the interval $\left( {0,2} \right)$

Note: 1. Don’t forget to cross verify the result in the end as error may come while solving.
2. Remember the derivative of a variable is equal to one and derivative of any constant term is equal to zero.
3. To apply Mean value theorem, both the conditions should be verified then only you can use MVT.
4.A student should differentiate and put the values carefully.
5.Mean value theorem is also known as LMVT (Lagrange’s Mean Value Theorem) which states for a given planar arc between two endpoints, there is at least one real point at which the tangent to arc is parallel to the secant through the endpoints.