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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 14: Q. 50 (page 883)

find the limit as x approaches c of the average rate of change of each function from c to x.
$c = 0; f (x) = 4 x^{3} - 5 x + 8$

Short Answer

Expert verified

The limit is $- 5$

Step by step solution

01

Step 1. The given information is

$x = 0; f (x) = 4 x^{3} - 5 x + 8$

02

Step 2. An average rate change of function f from point c to x

$\frac{鈭� y}{鈭� x} = \frac{f (x) - f (c)}{x - c} \frac{鈭� y}{鈭� x} = \frac{4 x^{3} - 5 x + 8 - (4 c^{3} - 5 c + 8)}{x - 0} \frac{鈭� y}{鈭� x} = \frac{4 x^{3} - 5 x + 8 - 8)}{x} \frac{鈭� y}{鈭� x} = \frac{4 x^{3} - 5 x}{x} \frac{鈭� y}{鈭� x} = \frac{x (4 x^{2} - 5)}{x} \frac{鈭� y}{鈭� x} = 4 x^{2} - 5$

03

Step 3. Solving the limit we get,

$\lim_{x 鈫� c} \frac{鈭� y}{鈭� x} = \lim_{x 鈫� c} \frac{f (x) - f (c)}{x - c} \lim_{x 鈫� c} \frac{鈭� y}{鈭� x} = \lim_{x 鈫� c} (4 x^{2} - 5) \lim_{x 鈫� c} \frac{鈭� y}{鈭� x} = 4 {(0)}^{2} - 5 \lim_{x 鈫� c} \frac{鈭� y}{鈭� x} = - 5$

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Most popular questions from this chapter

In Problems 7鈥� 42, find each limit algebraically

$\lim_{x 鈫� - 2} (5 x)$

In Problems 53-56, use the properties of limits and the facts that

$\lim_{x 鈫� 0} \frac{\sin x}{x} = 1 \lim_{x 鈫� 0} \frac{\cos x - 1}{x} = 0 \lim_{x 鈫� 0} \sin x = 0 \lim_{x 鈫� 0} \cos x = 1$

$\lim_{x 鈫� 0} \frac{\sin (2 x)}{x}$

In Problems 7鈥� 42, find each limit algebraically

$\lim_{x 鈫� 1} (- 3)$

True or False The limit of a polynomial function as $x$ approaches $5$ equals the value of the polynomial at $5$

Use a graphing utility to find the indicated limit rounded to two decimal places.

$\lim_{x 鈫� - 1} \frac{x^{3} + 2 x^{2} + x}{x^{4} + x^{3} + 2 x + 2}$

See all solutions

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