Chapter 5, Problem 37RE

a.

To determine

To show that $f$ satisfies the hypotheses of the mean value theorem on the interval $\left[a,b\right]=\left[0.5,3\right]$ .

Explanation of Solution

Given:

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

Calculation:

According to mean value theorem $f\left(x\right)$ must be continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ .

Below is the graph of $f\left(x\right)=x\ln x$

  

From graph of $f\left(x\right)=x\ln x$ it can be observed that in closed interval $\left[0.5,3\right]$ , because the curve does not breaks at any point in this interval and also there does not exist any vertical tangent or non-differentiable corner for open interval $\left(0.5,3\right)$ .

Therefore, $f$ satisfies the hypotheses of the mean value theorem on the interval $\left[a,b\right]=\left[0.5,3\right]$ .

b.

To determine

To find the values of $c$ in $\left(a,b\right)$ for which

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Answer to Problem 37RE

The value of $c$ in $\left(a,b\right)$ is $1.568$ .

Explanation of Solution

Given:

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

Calculation:

According to mean value theorem $f\left(x\right)$ must be continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there is at least one point $c$ in $\left(a,b\right)$ at which

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Since, from part (a) it is proved that $f$ satisfies the hypotheses of the mean value theorem on the interval $\left[a,b\right]=\left[0.5,3\right]$ .

Here $a=0.5$ and $b=3$

So , $f\left(b\right)=f\left(3\right)=3\ln 3$ , $f\left(a\right)=f\left(0.5\right)=0.5\ln \left(0.5\right)$ and

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\ln x+1\\ f^{\prime }\left(c\right)=\ln c+1\end{array}$

Since,

  $\begin{array}{l}{f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ \ln c+1=\frac{3\ln 3-0.5\ln 0.5}{3-0.5}=1.45\\ \ln c+1=1.45\\ \ln c=0.45\\ c=e^{0.45}=1.568\end{array}$

Therefore, the value of $c$ in $\left(a,b\right)$ is $1.568$ .

c.

To determine

To write an equation for the secant line $AB$ where $A=\left(a,f\left(a\right)\right)$ and $B=\left(b,f\left(b\right)\right)$

Answer to Problem 37RE

The equation of secant is $y=1.452x-1.066$

Explanation of Solution

Given:

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

Calculation:

  $a=0.5$ and $f\left(a\right)=f\left(0.5\right)=0.5\ln \left(0.5\right)=-0.34$

, $b=3$ and $f\left(b\right)=f\left(3\right)=3\ln 3=3.29$

Therefore,

  $A=\left(a,f\left(a\right)\right)=\left(0.5,-0.34\right)$ , $B=\left(b,f\left(b\right)\right)=\left(3,3.29\right)$

Equation of secant given two points $A=\left(a,f\left(a\right)\right)=\left(0.5,-0.34\right)$ , $B=\left(b,f\left(b\right)\right)=\left(3,3.29\right)$ is given by

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\\ y-3.29=\frac{-0.34-3.29}{0.5-3}\left(x-3\right)\\ y=1.452x-1.066\end{array}$

Therefore, the equation of secant is $y=1.452x-1.066$

d.

To determine

To write an equation for the tangent line that is parallel to the secant line $AB$ .

Answer to Problem 37RE

The equation of tangent is $y=1.452x-1.57$ .

Explanation of Solution

Given:

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

Calculation:

Since, it is given that tangent line is parallel to the secant line $AB$ so tangent line have same slope as the secant line.

General equation of line with slope $m$ is given by $y=mx+c$

Equation of secant line is $y=1.452x-1.066$

On comparing equation of secant line with general equation of line

Slope $m=1.452$

  $\begin{array}{l}{y}^{\prime }=\ln x+1\\ 1.452=\ln x+1\\ x=1.568\\ y=x\ln x=\left(1.568\right)\ln \left(1.568\right)=0.70\end{array}$

Equation of tangent is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-0.70=1.452\left(x-1.568\right)\\ y=1.452x-1.57\end{array}$

Therefore, the equation of tangent is $y=1.452x-1.57$ .