Chapter 2, Problem 38RE

(a)

To determine

To find: The asymptotes of the graph of $f\left(x\right)=\frac{4-x}{3+x}$.

Answer to Problem 38RE

The vertical asymptote is $x=-3$ and horizontal asymptote is $y=-1$.

Explanation of Solution

Calculation:

Recall that the line $y=a$ is horizontal asymptote if $\underset{x\to \pm \infty }{\lim }f\left(x\right)=a$.

$f\left(x\right)=\frac{4-x}{3+x}$

Take limit on the above.

$\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{4-x}{3+x}\\ =\underset{x\to \infty }{\lim }\frac{\frac{4}{x}-1}{\frac{3}{x}+1}\\ =\frac{\frac{4}{\infty }-1}{\frac{3}{\infty }+1}\\ =-1\end{array}$

Thus, the horizontal asymptote is $y=-1$.

Recall that the line $x=b$ is vertical asymptote if $\underset{x\to b}{\lim }f\left(x\right)=\infty$.

$f\left(x\right)=\frac{4-x}{3+x}$ approaches infinity if $3+x$ approaches which implies x approaches $-3$.

$\begin{array}{c}\underset{x\to -3}{\lim }f\left(x\right)=\frac{\underset{x\to -3}{\lim }\left(4-x\right)}{\underset{x\to -3}{\lim }\left(3+x\right)}\\ =\frac{7}{\underset{x\to -3}{\lim }\left(3+x\right)}\\ =\infty \end{array}$

Thus, the vertical asymptote is $x=-3$.

Graph:

Use the above information and trace the graph of $f\left(x\right)=\frac{4-x}{3+x}$ as shown in Figure 1

Observation:

$x=-3$ is a discontinuous point of f.

(b)

To determine

To sketch: The graph of $f^{\prime }$ using the graph of part a.

Explanation of Solution

Calculation:

From the above graph in part a,

it is cleared that $x=-3$ is a discontinuous point of f.

Both the two disjoint curves of f have negative slope.

Thus, the graph of $f^{\prime }$ will have negative value.

For the curve present below the y-axis,

the value of the slope decreases as the curve moves from left to right

approaching the line $x=-3$.

Thus, the graph of $f^{\prime }$ in this part will be decreasing from left to right.

For the curve present above the y-axis,

the value of the slope increases as the curve moves from left to right

approaching the line $y=-1$.

Thus, the graph of $f^{\prime }$ in this part will be increasing from left to right.

Graph:

Use the information above and trace the graph of $f^{\prime }$ as shown if Figure 2

(c)

To determine

To find: The derivative of $f$ using the definition.

Answer to Problem 38RE

The derivative is $f^{\prime }\left(x\right)=-\frac{7}{{\left(x+3\right)}^2}$.

Explanation of Solution

Formula used:

The derivative of a function f, denoted by $f^{\prime }\left(x\right)$, is

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (1)

Calculation:

Obtain the derivative of the function $f\left(x\right)$.

Compute $f^{\prime }\left(x\right)$ by using the equation (1),

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{4-\left(x+h\right)}{3+\left(x+h\right)}-\frac{4-x}{3+x}}{h}\end{array}$

Simplify the numerator in the above expression

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{-h\left(3+x\right)-\left(4-x\right)h}{h\left(3+x+h\right)\left(3+x\right)}$

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{-\left(3+x\right)-\left(4-x\right)}{\left(3+x+h\right)\left(3+x\right)}\\ =\underset{h\to 0}{\lim }\frac{-7}{\left(3+x+h\right)\left(3+x\right)}\\ =\frac{-7}{\left(3+x+0\right)\left(3+x\right)}\\ =-\frac{7}{{\left(3+x\right)}^2}\end{array}$

Thus, the value of the derivative is $f^{\prime }\left(x\right)=-\frac{7}{{\left(x+3\right)}^2}$.

(d)

To determine

To sketch: The graph of $f^{\prime }$ and compare with the sketch in part b.

Explanation of Solution

Graph:

Use an online graphing calculator to sketch the graph of $f^{\prime }\left(x\right)=-\frac{7}{{\left(x+3\right)}^2}$ as shown in Figure 3

Observation:

From Figure 3, it has been observed that $y=0$ is a horizontal asymptote of $f^{\prime }$

Comparison:

The function in Figure 3 seems to be the same graph which was sketched in part b.