Question

9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers that satisfy the conclusion of the Mean Value Theorem.

11. \(f(x) = \ln x\), \((1\,,\;4)\)

Short answer

The number \(c \approx 2.1640\) satisfies the conclusion of Mean Value Theorem.

Step-by-step solution

Given data

The function is \(f(x) = \ln x\).

Concept of Mean value theorem

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

The function \(f(x) = \ln x\) is continuous on the interval \((0,1) \cup (1,\infty )\) for all \(x > 0\).

Therefore, \(f(x)\) is continuous on the interval, \(\left( {1\;,\;4} \right)\).

The function \(f(x) = \ln x\) is differentiable for all \(x > 0\).

Therefore, \(f(x)\) is differentiable on the interval, \(\left( {1\;,\;4} \right)\).

Hence, the function \(f(x)\) satisfies the conditions of the Mean Value Theorem.

Since the number \(c\) satisfies the conditions of Mean Value Theorem, it should lie on the open interval \(\left( {1\;,\;4} \right)\).

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

\(\begin{array}{l}{f^\prime }(x) = \frac{d}{{dx}}(\ln x)\\{f^\prime }(x) = \frac{1}{x}\end{array}\)

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

\({f^\prime }(c) = \frac{1}{c}\)

Calculation for the number which satisfies the mean value theorem

Obtain the functional values at the end points of the given interval.

Substitute \(1\) for \(x\) in \(f(x)\).

\(\begin{array}{c}f(a) = f(1)\\ = \ln 1\end{array}\)

Substitute \(4\) for \(x\) in \(f(x)\).

\(\begin{array}{c}f(b) = f(4)\\ = \ln 4\end{array}\)

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

\(\frac{1}{c} = \frac{{\ln 4 - \ln 1}}{{4 - 1}}\)

Since, \(\ln 1 = 0\).

\(\begin{array}{c}c = \frac{{\ln 4}}{3}\\c = \frac{3}{{\ln 4}}\\c \approx 2.1640\end{array}\)

Notice that the number \(c \approx 2.1640\) belongs to the open interval \((1,4)\).

Therefore, it can be concluded that the number \(c \approx 2.1640\) satisfies the conclusion of Mean Value Theorem.