Question

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

Short answer

It does not contradict the Mean Value Theorem.

Step-by-step solution

Given data

Let\(f\)be a function that satisfies the following hypothesis:

1.\(f\)is continuous on the closed interval\(\left( {a\;,{\rm{ }}b} \right)\).

2.\(f\)is differentiable on the open interval\((a\;,{\rm{ }}b)\).

Then, there is a number\(c\)in\(\left( {a\;,{\rm{ }}b} \right)\)such that\({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently,\(f(b) - f(a) = {f^\prime }(c)(b - a)\).

Calculation of the derivative of the function \(f(x)\)

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{l}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{{(x - 3)}^{ - 2}}} \right)\\{f^\prime }(x) &= - 2{(x - 3)^{ - 3}}\\{f^\prime }(x) &= \frac{{ - 2}}{{{{(x - 3)}^3}}}\end{aligned}\)

Replace \(x\) by \(c\) in \({f^\prime }(x)\).

\({f^\prime }(c) = \frac{{ - 2}}{{{{(c - 3)}^3}}}\)

Find the functional values at the end points of the given interval, \((1,4)\).

Substitute 1 in \(f(x)\) and obtain the value of \(f(1)\).

\(\begin{aligned}{l}f(1) &= {(1 - 3)^{ - 2}}\\f(1) &= \frac{1}{{{{( - 2)}^2}}}\\f(1) &= \frac{1}{4}\end{aligned}\)

Substitute \(4\) in \(f(x)\) and obtain the value of \(f(4)\).

\(\begin{aligned}{l}f(4) &= {(4 - 3)^{ - 2}}\\f(4) &= \frac{1}{{{{(1)}^2}}}\\f(4) &= 1\end{aligned}\)

Calculation of the tangent function

It is given that \(f(4) - f(1) = {f^\prime }(c)(4 - 1)\).

Substitute the respective values in \(f(4) - f(1) = {f^\prime }(c)(4 - 1)\) and find the values of \(c\).

\(\begin{aligned}{c}1 - \frac{1}{4} &= \frac{{ - 2}}{{{{(c - 3)}^3}}}(4 - 1)\\\frac{3}{4} &= \frac{{ - 6}}{{{{(c - 3)}^3}}}\\4 &= \frac{{ - {{(c - 3)}^3}}}{2}\\ - 8 &= {(c - 3)^3}\end{aligned}\)

Simplify further and the find the value of \(c\).

\(\begin{aligned}{c}c - 3 &= \sqrt(3){{ - 8}}\\c - 3 &= - 2\\c &= 1\end{aligned}\)

Notice that \(c = 1 \notin (1,4)\).

Therefore, there is no existence of point c on the interval, \((1,4)\).

To check the continuity, set the denominator of the given function \(f(x) = \frac{1}{{{{(x - 3)}^2}}}\) to zero and obtain the value of \(x\).

Thus, the function \(f\) fails to satisfy the Mean Value Theorem conditions as \(f\) is not continuous at \(x = 3\).

Logically, if hypotheses (conditions) are false and the conclusion is either true or false, then the statement is true.

Thus, it does not contradict the Mean Value Theorem.