Chapter 4.5, Problem 1E

To determine

To find: the extrema’s of given function and sketch it’s graph.

Answer to Problem 1E

The function has no extrema on graph.

Explanation of Solution

Given:

The function is given as:

  $f(x)=\frac{2x-5}{x+3}$

Step 1:the domain is $\infty$ except (-3) and the function is discontinuous at (-3).

Step 2: the x-intercept is $\left(\frac{5}{2}\right)$ and y-intercept $f(0)=\left(\frac{-5}{3}\right)$ .

Step 3: The function is neither even nor odd.

Step 4:

Differentiating $f(x)=\frac{2x-5}{x+3}$ twice gives us

  $f\text{'}(x)=\frac{(x+3)2-(2x-5)1}{{(x+3)}^2}$

  $=\frac{11}{{(x+3)}^2}$

  $f\text{'}\text{'}(x)=\frac{-22}{{(x+3)}^3}$

Since $f\text{'}(x)>0$ for every $x$ . the function $f$ is increasing on $(-\infty ,\infty )$ .

Also $f\text{'}\text{'}(x)>0$ for $(-\infty ,3)$ .the function is concave upwards and $f\text{'}\text{'}(x)<0$ for $(-3,\infty )$ . the function will concave downwards.

Step 5:to find horizontal asymptotes, follow

  $\underset{x\to \infty }{\lim }\left(\frac{2x-5}{x+3}\right)=2$

Thus line $y=2$ is the horizontal asymptote. Vertical asymptotes corresponding to the zeroes of the denominator is $x=-3$ .

Graph of given function is shown below: