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Q29.

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Precalculus Mathematics for Calculus

Found in: Page 156

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

Find the value of $f (- 2), f (0), f (\frac{1}{2}), f (2), f (x + 1), f (x^{2} + 2)$ for the function $f (x) = 2 |x - 1|$ .

The value of function are

$f (- 2) = 6, f (0) = 2, f (\frac{1}{2}) = 1, f (2) = 2, f (x + 1) = 2 |x|, f (x^{2} + 2) = 2 | x^{2} + 1 |$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Evaluate f−2,f0,f12 for the given function.

From the given function,

$f (x) = 2 | x - 1 | .... (1)$

Evaluate the value of function at $x = - 2, 0, \frac{1}{2}$ gives,

role="math" localid="1644516780178" $f (- 2) = 2 | (- 2) - 1 | .... (Substitute x = - 2 in (1)) = 2 | - 3 | = 6 f (0) = 2 | (0) - 1 | .... (Substitute x = 0 in (1)) = 2 .... (∵ | - 1 | = 1) f (\frac{1}{2}) = 2 |(\frac{1}{2}) - 1| .... (Substitute x = \frac{1}{2} in (1)) = 2 |- \frac{1}{2}| = 1$

Step 2. Evaluate f2,fx+1,fx2+2 for the given function.

Similarly, evaluate the value of function (1) at $x = 2, x + 1, x^{2} + 2$ gives,

role="math" localid="1644517286232" $f (2) = 2 | (2) - 1 | .... (Substitute x = 2 in (1)) = 2 f (x + 1) = 2 | (x + 1) - 1 | .... (Substitute x = x + 1 in (1)) = 2 |x| f (x^{2} + 2) = 2 |(x^{2} + 2) - 1| .... (Substitute x = x^{2} +2 in (1)) = 2 |x^{2} + 1|$

Step 3. Description of steps.

From the obtained values,

$f (- 2) = 6, f (0) = 2, f (\frac{1}{2}) = 1, f (2) = 2, f (x + 1) = 2 |x|, f (x^{2} + 2) = 2 | x^{2} + 1 |$ .

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