Chapter 4, Problem 37RE

(a)

To determine

To show: The given function satisfies the mean value theorem within given interval.

Answer to Problem 37RE

The given function satisfies mean value theorem.

Explanation of Solution

Given:

The function is $f(x)=x\ln x$ and the interval is $\left[0.5,3\right]$

Calculation:

According to the Mean value theorem, if any function is continuous on any closed interval $\left[a,b\right]$ and differential on any open interval $\left(a,b\right)$ then, there exist a point (say c) in that interval in such a way that the first derivative of function at point c is equal to the average change rate of function over interval $\left[a,b\right]$

Hence, he given function f(x) is continuous on $\left[0.5,3\right]$ and differentiable on $\left(0.5,3\right)$

Conclusion:

The given function satisfies mean value theorem.

(b)

To determine

The value of c in (a,b).

Answer to Problem 37RE

The value of c is $1.579$ .

Explanation of Solution

Given:

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

Calculation:

  $f\text{'}(x)=x.\frac{1}{x}+\ln x$

Now, use the value of a and b to solve:

  $\begin{array}{c}f\text{'}(c)=\frac{f(3)-f(0.5)}{3-0.5}\\ 1+\ln c=\frac{3\ln 3-0.5\ln (0.5)}{3-0.5}\\ \ln c=\frac{\ln \left(\frac{3^3}{{0.5}^{0.5}}\right)}{2.5}-1\\ \ln c=0.4\ln {\left(27\sqrt{2}\right)}^{0.4}\\ c=e^{-1}{\left(27\sqrt{2}\right)}^{0.4}\\ c=e^{-1}\sqrt[5]{1458}\\ c=1.579\end{array}$

Conclusion:

The value of c is $1.579$

(c)

To determine

The equation for the secant line AB.

Answer to Problem 37RE

The equation for secant line is $y=1.457x-1.075$

Explanation of Solution

Given:

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

Calculation:

The slope of a line is:

  $\begin{array}{l}m=\frac{f(b)-f(a)}{b-a}\\ m=0.4\ln \left(27\sqrt{2}\right)\\ m=0.2\ln 1458\end{array}$

Now, the line passes through $(3,3\ln 3)$ .

Hence, the equation becomes:

  $\begin{array}{l}y=0.2(\ln 1458)(x-3)+3\ln 3\\ y=1.457x-1.075\end{array}$

Conclusion:

The equation for secant line is $y=1.457x-1.075$

(d)

To determine

The equation for the tangent line which is parallel to secant line AB.

Answer to Problem 37RE

The equation for tangent line is $y=1.457x-1.579$ .

Explanation of Solution

Given:

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

Calculation:

The slope of a line is:

  $\begin{array}{l}m=\frac{f(b)-f(a)}{b-a}\\ m=0.4\ln \left(27\sqrt{2}\right)\\ m=0.2\ln 1458\end{array}$

Now, the line passes through:

  $\begin{array}{l}\left(c,f(c)\right)=\left(e^{-1}\sqrt[5]{1458},e^{-1}\sqrt[5]{1458}(-1+0.2\ln 1458)\right)\\ \text{ = }\left(1.579,0.722\right)\end{array}$

  $\begin{array}{l}y=0.2(\ln 1458)(x-c)+f(c)\\ y=0.2\ln 1458(x-e^{-1}\sqrt[5]{1458})+e^{-1}\sqrt[5]{1458}\left(-1+0.2\ln 1458\right)\\ y=0.2\left(\ln 1458\right)x-e^{-1}\sqrt[5]{1458}\\ y=1.457x-1.579\end{array}$

Conclusion:

The equation for tangent line is $y=1.457x-1.579$ .