Question

Find the value or values of$c$that satisfy the equation \[\dfrac{{\left( {f\left( b \right) - f\left( a \right)} \right)}}{{\left( {b - a} \right)}}\] in the conclusion of the Mean Value Theorem for the function \[f\left( x \right) = {x^2} + 2x + 2\;\] on the interval \[\left[ { - 2,1} \right]\].

Answer 1

Hint:The Mean Value Theorem simply states that if a function $f$ is continuous on the closed interval \[\left[ {a,b} \right]\] and differentiable on the open interval \[\left( {a,b} \right)\], then there exists a point $c$ in the interval \[\left( {a,b} \right)\] such that $f'\left( c \right)$ is equal to the function's average rate of change over\[\left[ {a,b} \right]\].In other words, the graph has a tangent somewhere in \[\left( {a,b} \right)\] that is parallel to the secant line over \[\left[ {a,b} \right]\].Using the above definition we can solve the given question.

Complete step by step answer:
Given, \[f\left( x \right) = {x^2} + 2x + 2\;....................\left( i \right)\]
Interval \[\left[ { - 2,1} \right]\]
So our aim is to find the value of $c$. For that we have to find two equations containing the value $c$ and thereby solve it. First we have to create an equation for $f'\left( c \right)$ by simply differentiating the given $f\left( x \right)$ and then we have to take the equation for $f'\left( c \right)$ by using the Mean Value Theorem.Now differentiating (i), we can write:
$f'\left( x \right) = 2x + 2$
Now let’s find $f'\left( c \right)$ by substituting $c$ in the place of $x$.
So we get:
$f'\left( c \right) = 2c + 2........................\left( {ii} \right)$

Now let’s find the equation for $f'\left( c \right)$ by using the Mean Value Theorem:
Now using Mean Value Theorem we can write:
\[f'\left( c \right) = \dfrac{{\left( {f\left( b \right) - f\left( a \right)} \right)}}{{\left( {b - a} \right)}}.....................\left( {iii} \right)\]
Over the interval \[\left[ { - 2,1} \right]\], so here:
$a = - 2 \\
\Rightarrow b = 1 \\ $
Now we have:
\[f\left( x \right) = {x^2} + 2x + 2\; \\
\Rightarrow f\left( { - 2} \right) = {\left( { - 2} \right)^2} + 2\left( { - 2} \right) + 2 \\
\Rightarrow f\left( { - 2} \right) = 4 - 4 + 2 \\
\Rightarrow f\left( { - 2} \right) = 2 \\
\Rightarrow f\left( 1 \right) = {\left( 1 \right)^2} + 2\left( 1 \right) + 2 \\
\Rightarrow f\left( 1 \right) = 1 + 2 + 2 \\
\Rightarrow f\left( 1 \right) = 5 \\ \]
Now we can write (iii) as:
\[f'\left( c \right) = \dfrac{{\left( {f\left( 1 \right) - f\left( { - 2} \right)} \right)}}{{\left( {1 - \left( { - 2} \right)} \right)}} \\
\Rightarrow f'\left( c \right)= \dfrac{{5 - 2}}{{1 + 2}} \\
\Rightarrow f'\left( c \right)= \dfrac{3}{3} \\
\Rightarrow f'\left( c \right) = 1.......................\left( {iv} \right) \\ \]
Now we can see that the LHS of both (ii) and (iv) are the same so equating their RHS and then solving for $c$.
Such that:
$1 = 2c + 2 \\
\Rightarrow 2c + 2 = 1 \\
\Rightarrow 2c = - 1 \\
\therefore c = - \dfrac{1}{2}....................\left( v \right) \\ $
Therefore the value of $c$ that satisfy the equation \[\dfrac{{\left( {f\left( b \right) - f\left( a \right)} \right)}}{{\left( {b - a} \right)}}\] in the conclusion of the Mean Value Theorem for the function \[f\left( x \right) = {x^2} + 2x + 2\;\] on the interval \[\left[ { - 2,1} \right]\] is $\left( { - \dfrac{1}{2}} \right)$.

Note:Similar questions involving Mean Value Theorem have to be proceeded in the same way as given above since it has a high rate of efficiency and accuracy. Also care must be taken while solving equations since there are places where the human error can occur and interfere.