Question

Let \(f\left( x \right) = {\left( {x - 3} \right)^{ - 2}}\). Show that there is no value of \(c\) in \(\left( {1,4} \right)\) such that \(f\left( 4 \right) - f\left( 1 \right) = f'\left( c \right)\left( {4 - 1} \right)\). Why does this not contradict the Mean Value Theorem?

Short answer

It is proved that there is no value of \(c\)in\(\left( {1,4} \right)\) such that \(f\left( 4 \right) - f\left( 1 \right) = f'\left( c \right)\left( {4 - 1} \right)\).

The function does not contradict the Mean Value Theorem as \(f\left( x \right)\) is not continuous at \(x = 3\) which is in domain \(\left( {1,4} \right)\).

Step-by-step solution

Mean Value Theorem

The Mean Value Theorem is applicable for any function \(f\left( x \right)\) if and only if

\(\begin{array}{l}f\left( x \right) \to {\rm{continuous within }}\left( {a,b} \right)\\f\left( x \right) \to {\rm{differentiable within }}\left( {a,b} \right)\\{\rm{then, for }}c \in \left( {a,b} \right) \to f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\end{array}\).

Applying the Mean Value theorem to the function:

The given function is:

\(f\left( x \right) = {\left( {x - 3} \right)^{ - 2}}\)

Now, according to theMean Value Theorem, there exists a point \(c \in \left( {1,4} \right)\)such that:

\(f'\left( x \right)\left| {_{x = c}} \right. = \frac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}}\)

On solving, we have:

\(\begin{aligned}{c}f'\left( x \right)\left| {_{x = c}} \right. &= \frac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}}\\f\left( 4 \right) - f\left( 1 \right) &= f'\left( c \right)\left( {4 - 1} \right)\\{\left( {4 - 3} \right)^{ - 2}} - {\left( {1 - 3} \right)^{ - 2}} &= \left\{ { - 2{{\left( {c - 3} \right)}^{ - 3}}} \right\}\left( 3 \right)\\{\left( {c - 3} \right)^{ - 3}} &= \frac{{\frac{1}{1} - \frac{1}{4}}}{{ - 6}} = - \frac{1}{8}\\{\left( {c - 3} \right)^3} &= - 8\\c - 3 &= - 2\\c &= 1\end{aligned}\)

Since, \(c \notin \left( {1,4} \right)\). Hence, there is no such number \(c\)exists that satisfies the theorem.

Also, the function does not contradict the Mean Value Theorem because \(f\left( x \right) = {\left( {x - 3} \right)^{ - 2}}\) can never be continuous at \(x = 3\) which is in domain \(\left( {1,4} \right)\).