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Q. 69

Expert-verified

Calculus

Found in: Page 276

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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.
$f (x) = x^{3} (x + 2)$

The sign chart is

The sketch of the graph is

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information.

The given function is $f (x) = x^{3} (x + 2) .$

Step 2. Finding the roots.

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

So,

$f (x) = x^{3} (x + 2) 0 = x^{3} (x + 2) x = 0 a n d x + 2 = 0 x = - 2$

Therefore, the given function have roots at $x = 0, - 2 .$

Step 3. Testing the signs of f.

To sketch the sign chart, let's test the signs on both sides.

For f

$f (- 3) = {(- 3)}^{3} (- 3 + 2) f (- 3) = 27 Now, f (- 1) = {(- 1)}^{3} (- 1 + 2) f (- 1) = - 1 Now, f (1) = {(1)}^{3} (1 + 2) f (1) = 3$

Step 4. Testing the signs.

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

Let's differentiate the equation to find $f' .$

So,

$f' (x) = 4 x^{3} + 6 x^{2} 0 = 4 x^{3} + 6 x^{2} 0 = 2 x^{2} (2 x + 3) x = 0 and 0 = 2 x + 3 x = - \frac{3}{2}$

Testing the signs on both sides,

$f' (- 2) = 4 {(- 2)}^{3} + 6 {(- 2)}^{2} f' (- 2) = - 8 N o w, f' (- 1) = 4 {(- 1)}^{3} + 6 {(- 1)}^{2} f' (- 1) = 2 N o w, f' (1) = 4 {(1)}^{3} + 6 {(1)}^{2} f' (1) = 10$

Thus, $f'$ is negative on the interval $(- \infty, - \frac{3}{2})$ and positive on the interval $(- \frac{3}{2}, \infty) .$ Hence the graph of f will be increasing on the positive intervals and decreasing on the negative intervals.

Let's differentiate again.

So,

$f'' (x) = 12 x^{2} + 12 x 0 = 12 x (x + 1) 0 = x and x + 1 = 0 x = - 1$

Testing the sign on both sides,

role="math" localid="1648474789089" $f'' (- 2) = 12 {(- 2)}^{2} + 12 (- 2) f'' (- 2) = 24 A n d f'' (- 0.5) = 12 {(- 0.5)}^{2} + 12 (- 0.5) f'' (- 0.5) = - 3 A n d f'' (1) = 12 {(1)}^{2} + 12 (1) f'' (1) = 24$

Thus, $f''$ is positive on the interval role="math" localid="1648474872810" $(- \infty, - 1) a n d (0, \infty)$ and negative on the interval $(- 1, 0) .$ Hence, the graph of f will be concave up on the positive interval and concave down on the negative interval. Inflection point at $x = - 1, 0 .$

Step 5. Sketch the sign chart.

The sign chart is

Step 6. Examine the relevant limit.

Let's examine the limits of $f (x) = x^{3} (x + 2) as x \to \pm \infty .$

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

Step 7. Sketch the graph of function f.

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