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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 8: Q. 14 (page 557)

Suppose that $f (x) = 4 x + 5 and g (x) = x^{2} + 5 x - 24$
(a) Solve $f (x) = 0$
(b) Solve $f (x) = 13$
(c) Solve $f (x) = g (x)$
(d) Solve $f (x) > 0$

Short Answer

Expert verified

The solution is

$(a) x = - \frac{5}{4} (b) x = 2 (c) x = \frac{- 1 卤 3 \sqrt{13}}{2} (d) x 鈭� (- \frac{5}{4}, 鈭�)$

Step by step solution

01

Part (a) Step 1. Given Information

We are given two functions,

$f (x) = 4 x + 5 and g (x) = x^{2} + 5 x - 24$

We need to find the solution of $f (x) = 0$

02

Part (a) Step 2, Solution of the equation

The given function is,

$f (x) = 4 x + 5$

For $f (x) = 0,$

$f (x) = 4 x + 5 = 0 4 x + 5 = 0 4 x = - 5 x = - \frac{5}{4}$

03

Part (b) Step 1. Solution of the equation

The given function is,

$f (x) = 4 x + 5 f o r f (x) = 13 4 x + 5 = 13 4 x = 13 - 5 4 x = 8 x = 2$

04

Part (c) Step 1. Solution of the equation

The given functions are, $f (x) = 4 x + 5 and g (x) = x^{2} + 5 x - 24$

For $f (x) = g (x)$

$4 x + 5 = x^{2} + 5 x - 24 x^{2} + x - 29 = 0 U \sin g x = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a}, = \frac{- 1 卤 \sqrt{1^{2} - 4 (1) (- 29)}}{2 (1)} = \frac{- 1 卤 \sqrt{1} + 116}{2} = \frac{- 1 卤 \sqrt{117}}{2} = \frac{- 1 卤 3 \sqrt{13}}{2}$

05

Part (d) Step 1. Solution of the inequality

The given function is $f (x) = 4 x + 5$

For $f (x) > 0$

$f (x) = 4 x + 5 > 0 4 x + 5 > 0 4 x > - 5 x > - \frac{5}{4} x 鈭� (- \frac{5}{4}, 鈭�)$

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