Chapter 3: Q. 70 (page 276)

Sketch careful, labeled graphs of each function f in Exercises 63–82 by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f, f', a n d f'',$ and examine any relevant limits so that you can describe all key points and behaviors of f.
$f (x) = \frac{x}{x^{2} + 1}$

Short Answer

Expert verified

The sign chart is

The sketch of the graph is

Step by step solution

Step 1. Given Information.

The given function is $f (x) = \frac{x}{x^{2} + 1} .$

Step 2. Finding the roots and examining the relevant limit.

To find the roots we will put the given function equal to zero.

So,

$f (x) = \frac{x}{x^{2} + 1} 0 = \frac{x}{x^{2} + 1} x = 0$

Let's examine the limits of $f (x) = \frac{x}{x^{2} + 1} as x \to \pm \infty .$

$\lim_{x \to \infty} f (x) = 0 \lim_{x \to - \infty} f (x) = 0$

Step 3. Testing the signs.

To sketch the sign chart, let's differentiate the equation to find $f' .$

So,

$f' (x) = - \frac{x^{2} - 1}{{(x^{2} + 1)}^{2}} 0 = - \frac{x^{2} - 1}{{(x^{2} + 1)}^{2}} 0 = - x^{2} + 1 x^{2} = 1 x = \pm 1$

Testing the signs on both sides,

$f' (- 2) = - \frac{{(- 2)}^{2} - 1}{{({(- 2)}^{2} + 1)}^{2}} f' (- 2) = - \frac{3}{25} N o w, f' (0) = - \frac{{(0)}^{2} - 1}{{(0^{2} + 1)}^{2}} f' (0) = 1 N o w, f' (2) = - \frac{{(2)}^{2} - 1}{{(2^{2} + 1)}^{2}} f' (2) = - \frac{3}{25}$

Thus, $f'$ is positive on the interval $(- 1, 1)$ and negative on the interval $(- \infty, - 1) a n d (1, \infty) .$ Hence the graph of fwill be increasing on the positive intervals and decreasing on the negative intervals.

Step 4. Testing the signs.

Now, let's test the sign for $f'' .$

Let's differentiate again.

So,

$f'' (x) = \frac{2 x (x^{2} - 3)}{{(x^{2} + 1)}^{3}} 0 = \frac{2 x (x^{2} - 3)}{{(x^{2} + 1)}^{3}} 0 = 2 x (x^{2} - 3) x = 0 a n d 0 = x^{2} - 3 x^{2} = 3 x = \pm \sqrt{3}$

Thus, $f''$ is positive on the intervals $(- \sqrt{3}, 0), (\sqrt{3}, \infty)$ and negative on the intervals $(- \infty, - \sqrt{3}), (0, \sqrt{3}) .$ Hence the graph of fwill be concave up on the positive intervals and concave down on the negative intervals.