Chapter 2.1, Problem 3E

The point P(2, –1) lies on the curve y = 1/(1 – x).

(a) If Q is the point (x, 1/(1 – x)), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x :

(i) 1.5

(ii) 1.9

(iii) 1.99

(iv) 1.999

(v) 2.5

(vi) 2.1

(vii) 2.01

(viii) 2.001

(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(2, –1).

(c) Using the slope from part (b), find an equation of the tangent line to the curve at P(2, –1).

To determine

To find: The slope of the secant line PQ for $x=1.5$.

Answer to Problem 3E

The slope of the secant line PQ when $x=1.5$ is 2.

Explanation of Solution

Given:

The equation of the curve is $y=\frac{1}{\left(1-x\right)}$.

The point $P\left(2,-1\right)$ lies on the curve y.

Calculation:

The slope of the secant line between the points, $P\left(2,-1\right)$  and $Q\left(x,\frac{1}{1-x}\right)$ is

$m=\frac{\frac{1}{1-x}-\left(-1\right)}{x-2}$ (1)

Obtain the slope of the secant line PQ when $x=1.5$.

Substitute 1.5 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-1.5\right)}\\ =\frac{1}{-0.5}\\ =-2\end{array}$

Substitute 1.5 for x and −2 for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-2-\left(-1\right)}{1.5-2}\\ =\frac{-2+1}{-0.5}\\ =\frac{-1}{-0.5}\\ =2\end{array}$

Thus, the slope of the secant line PQ when $x=1.5$ is 2.

To determine

To find: The slope of the secant line PQ for $x=1.5$.

Answer to Problem 3E

The slope of the secant line PQ when $x=1.5$ is 2.

Explanation of Solution

Given:

The equation of the curve is $y=\frac{1}{\left(1-x\right)}$.

The point $P\left(2,-1\right)$ lies on the curve y.

Calculation:

The slope of the secant line between the points, $P\left(2,-1\right)$  and $Q\left(x,\frac{1}{1-x}\right)$ is

$m=\frac{\frac{1}{1-x}-\left(-1\right)}{x-2}$ (1)

Obtain the slope of the secant line PQ when $x=1.5$.

Substitute 1.5 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-1.5\right)}\\ =\frac{1}{-0.5}\\ =-2\end{array}$

Substitute 1.5 for x and −2 for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-2-\left(-1\right)}{1.5-2}\\ =\frac{-2+1}{-0.5}\\ =\frac{-1}{-0.5}\\ =2\end{array}$

Thus, the slope of the secant line PQ when $x=1.5$ is 2.

To determine

To find: The slope of the secant line PQ for $x=1.9$.

Answer to Problem 3E

The slope of the secant line PQ when $x=1.9$ is 1.111111.

Explanation of Solution

Substitute 1.9 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-1.9\right)}\\ =\frac{1}{-0.9}\\ =-1.11111\end{array}$

Substitute 1.9 for x and $-1.11111$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-1.11111-\left(-1\right)}{1.9-2}\\ =\frac{-1.11111+1}{-0.1}\\ =\frac{-0.11111}{-0.1}\\ \approx 1.111111\end{array}$

Thus, the slope of the secant line PQ when $x=1.9$ is 1.111111.

To determine

To find: The slope of the secant line PQ for $x=1.99$.

Answer to Problem 3E

The slope of the secant line PQ when $x=1.99$ is 1.010101.

Explanation of Solution

Obtain the slope of the secant line PQ when $x=1.99$.

Substitute 1.99 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-1.99\right)}\\ =\frac{1}{-0.99}\\ =-1.\overline{01}\end{array}$

Substitute 1.99 for x and $-1.\overline{01}$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-1.\overline{01}-\left(-1\right)}{1.99-2}\\ =\frac{-1.\overline{01}+1}{-0.01}\\ =\frac{-0.\overline{01}}{-0.01}\\ \approx 1.010101\end{array}$

Thus, the slope of the secant line PQ when $x=1.99$ is 1.010101.

To determine

To find: The slope of the secant line PQ for $x=1.99$.

Answer to Problem 3E

The slope of the secant line PQ when $x=1.99$ is 1.010101.

Explanation of Solution

Obtain the slope of the secant line PQ when $x=1.999$.

Substitute 1.999 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-1.999\right)}\\ =\frac{1}{-0.999}\\ =-1.\overline{001}\end{array}$

Substitute 1.999 for x and $-1.\overline{001}$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-1.\overline{001}-\left(-1\right)}{1.999-2}\\ =\frac{-1.\overline{001}+1}{-0.001}\\ =\frac{-0.\overline{001}}{-0.001}\\ \approx 1.001001\end{array}$

Thus, the slope of the secant line PQ when $x=1.999$ is 1.001001.

To determine

To find: The slope of the secant line PQ for $x=2.5$.

Answer to Problem 3E

The slope of the secant line PQ when $x=2.5$ is 0.666667.

Explanation of Solution

Obtain the slope of the secant line PQ when $x=2.5$.

Substitute 2.5 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-2.5\right)}\\ =\frac{1}{-1.5}\\ =-0.\overline{6}\end{array}$

Substitute 2.5 for x and $-0.\overline{6}$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-0.\overline{6}-\left(-1\right)}{2.5-2}\\ =\frac{-0.\overline{6}+1}{0.5}\\ =\frac{-0.\overline{3}}{0.5}\\ \approx -0.666667\end{array}$

Thus, the slope of the secant line PQ when $x=2.5$ is 0.666667.

To determine

To find: The slope of the secant line PQ for $x=2.1$.

Answer to Problem 3E

The slope of the secant line PQ when $x=2.1$ is 0.909091.

Explanation of Solution

Obtain the slope of the secant line PQ when $x=2.1$.

Substitute 2.1 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-2.1\right)}\\ =\frac{1}{-1.1}\\ =-0.\overline{90}\end{array}$

Substitute 2.1 for x and $-0.\overline{90}$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-0.\overline{90}-\left(-1\right)}{2.1-2}\\ =\frac{-0.\overline{90}+1}{0.1}\\ =\frac{0.\overline{09}}{0.1}\\ \approx 0.909091\end{array}$

Thus, the slope of the secant line PQ when $x=2.1$ is 0.909091.

To determine

To find: The slope of the secant line PQ for $x=2.01$.

Answer to Problem 3E

The slope of the secant line PQ when $x=2.01$ is 0.990099.

Explanation of Solution

Obtain the slope of the secant line PQ when $x=2.01$.

Substitute 2.01 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-2.01\right)}\\ =\frac{1}{-1.01}\\ =-0.\overline{9900}\end{array}$

Substitute 2.01 for x and $-0.\overline{9900}$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-0.\overline{9900}-\left(-1\right)}{2.01-2}\\ =\frac{-0.\overline{9900}+1}{0.01}\\ =\frac{0.\overline{0099}}{0.01}\\ \approx 0.990099\end{array}$

Thus, the slope of the secant line PQ when $x=2.01$ is 0.990099.

To determine

To find: The slope of the secant line PQ for $x=2.001$.

Answer to Problem 3E

The slope of the secant line PQ when $x=2.001$ is 0.999001.

Explanation of Solution

Obtain the slope of the secant line PQ when $x=2.001$.

Substitute 2.001 for x in $\frac{1}{\left(1-x\right)}$.

$\begin{array}{c}\frac{1}{\left(1-x\right)}=\frac{1}{\left(1-2.001\right)}\\ =\frac{1}{-1.001}\\ =-0.\overline{999000}\end{array}$

Substitute 2.001 for x and $-0.\overline{999000}$ for $\frac{1}{\left(1-x\right)}$ in equation (1).

$\begin{array}{c}m=\frac{-0.\overline{999000}-\left(-1\right)}{2.001-2}\\ =\frac{-0.\overline{999000}+1}{0.01}\\ =\frac{0.\overline{000999}}{0.01}\\ \approx 0.999001\end{array}$

Thus, the slope of the secant line PQ when $x=2.001$ is 0.999001.

To determine

To guess: The value of the slope of the tangent line to the curve at $P\left(2,-1\right)$.

Answer to Problem 3E

The estimated value of the slope of the tangent line to the curve at $P\left(2,-1\right)$ is 1.

Explanation of Solution

The secant line is closer to the tangent line at $P\left(2,-1\right)$ when $t=1.999$ and $t=2.001$.

From part (a), the slope of the secant line PQ when $x=1.999$ is 1.001001 and the slope of the secant line PQ when $x=2.001$ is 0.999001.

The value of the slope of the tangent line to the curve at $P\left(2,-1\right)$ is closer to the average value of the slope of the secant lines near to P.

Take the average of the two secant lines when $x=1.999$ and $x=2.001$.

$\begin{array}{c}m=\frac{1.001001+0.999001}{2}\\ =\frac{2.000002}{2}\\ =1.000001\\ \approx 1\end{array}$

Thus, the estimated value of the slope of the tangent line to the curve at $P\left(2,-1\right)$ is 1.

To determine

To find: The equation of the tangent line to the curve at $P\left(2,-1\right)$.

Answer to Problem 3E

The equation of the tangent line is $y=x-3$.

Explanation of Solution

Formula used:

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(x_1,y_1\right)$ is,

$y-y_1=m\left(x-x_1\right)$ (2)

Calculation:

From part (b), the slope of the tangent line to the curve is 1.

Substitute $m=1$ and $\left(x_1,y_1\right)=\left(2,-1\right)$ in equation (2).

$\begin{array}{c}y-\left(-1\right)=1\left(x-2\right)\\ y+1=\left(x-2\right)\end{array}$

Isolate y.

$\begin{array}{l}y=x-2-1\\ y=x-3\end{array}$

Thus, the equation of the tangent line to the curve at $P\left(2,-1\right)$ is $y=x-3$.