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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, 14 (page 907)

For the function $f (x) = 4 x^{2} - 11 x - 3$
(a) Find the derivative of f at $x = 2$ .
(b) Find the equation of the tangent line to the graph of $f$ at the point $(2, - 9)$ .
(c) Graph $f$ and the tangent line.

Short Answer

Expert verified

The derivative of the $f (2) = 5$ and the equation at $(2, - 9) = y = 5 x - 19$

Step by step solution

01

Part(a) Step 1, Given Information

We are given a function $f (x) = 4 x^{2} - 11 x - 3$

We need to find the derivative of the function at $x = 2$ $x = 2 .$

02

Part(a) Step 2. Explanation

We know,

$f^{'} (c) = \lim_{x 鈫� c} \frac{f (x) - f (c)}{x - c}$

Also,

$f (2) = 4 (2)^{2} - 11 (2) - 3 = - 9$

Then, using the formula,

$f^{'} (2) = \lim_{x 鈫� 2} \frac{4 x^{2} - 11 x - 3 - (- 9)}{x - 2} = \lim_{x 鈫� 2} \frac{4 x^{2} - 11 x + 6}{x - 2} = \lim_{x 鈫� 2} \frac{(4 x - 3) (x - 2)}{x - 2} = \lim_{x 鈫� 2} (4 x - 3) = 4 (2) - 3 = 5$

03

Part(b) Step 1. Explanation

We know,

$m_{tan} = \lim_{x 鈫� c} \frac{f (x) - f (c)}{x - c} w h e r e m = s l o p e .$

On substituting the point $(2, - 9)$

$m_{tan} = \lim_{x 鈫� 2} \frac{4 x^{2} - 11 x - 3 - (- 9)}{x - 2} = \lim_{x 鈫� 2} \frac{4 x^{2} - 11 x + 6}{x - 2} = \lim_{x 鈫� 2} \frac{(4 x - 3) (x - 2)}{x - 2} = \lim_{x 鈫� 2} 4 x - \lim_{x 鈫� 2} 3 = 5$

The equation of a tangent line is,

$y - f (c) = m_{\tan} (x - c)$

On substituting,

$y - (- 9) = 5 (x - 2) y + 9 = 5 x - 10 y = 5 x - 19$

04

Part(c) Step 1. Explanation

The equation of the tangent line for the function

$f (x) = 4 x^{2} - 11 x - 3$ at $(2, - 9)$ is $y = 5 x - 19$ .

Plotting both the lines, we get the graph,

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