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Chapter 9: Chapter 9

Problem 168

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If \(\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}+1\), find the given element \(\mathrm{m}\) the range, a) \(g(-2)\) b) \(\mathrm{g}(0)\) c) \(g(a+1)\) d) \(g(a-1)\)

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The values of g(x) for the given input values are: a) \(g(-2) = 9\) b) \(g(0) = 1\) c) \(g(a+1) = a^2\) d) \(g(a-1) = a^2 - 4a + 4\)
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find g(-2)

First, let's find the value of g(-2) by substituting -2 in place of x in the given formula: g(-2) = (-2)^2 - 2(-2) + 1 g(-2) = 4 + 4 + 1 g(-2) = 9 So, g(-2) = 9.

Step 2: Find g(0)

Next, let's find the value of g(0) by substituting 0 in place of x: g(0) = (0)^2 - 2(0) + 1 g(0) = 0 + 0 + 1 g(0) = 1 So, g(0) = 1.

Step 3: Find g(a+1)

Now, let's find the value of g(a+1) by substituting a+1 in place of x: g(a+1) = (a+1)^2 - 2(a+1) + 1 g(a+1) = (a^2 + 2a + 1) - 2(a + 1) + 1 g(a+1) = a^2 + 2a + 1 - 2a - 2 + 1 g(a+1) = a^2 So, g(a+1) = a^2.

Step 4: Find g(a-1)

Lastly, let's find the value of g(a-1) by substituting a-1 in place of x: g(a-1) = (a-1)^2 - 2(a-1) + 1 g(a-1) = (a^2 - 2a + 1) - 2(a - 1) + 1 g(a-1) = a^2 - 2a + 1 - 2a + 2 + 1 g(a-1) = a^2 - 4a + 4 So, g(a-1) = a^2 - 4a + 4. Now, we have found the values of g(x) for all given input values: a) g(-2) = 9 b) g(0) = 1 c) g(a+1) = a^2 d) g(a-1) = a^2 - 4a + 4

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