How do I work with the radicals? ### Transcription of Math Problem Solving1. **Expression for $ h(x) $:** \[ h(x) = \sqrt{2 - x} - 2 \]2. **Simplification Steps:** \[ 2 \cdot \frac{1}{2} - \frac{x}{2} \]3. **Derivative Approximation:** \[ \frac{f(x+h) - f(x)}{h} \]4. **Simplifying the Expression:** \[ \frac{\sqrt{2-(x+h)} - \sqrt{2-x}}{h} \]5. **Rewriting the Expression:** \[ \frac{\sqrt{2-(x+h)} - \sqrt{2-x}}{h} = \frac{\left( \sqrt{2-(x+h)} - \sqrt{2-x} \right) \left( \sqrt{2-(x+h)} + \sqrt{2-x} \right)}{\sqrt{2-(x+h)} + \sqrt{2-x}} \]6. **Solving the Difference of Squares:** \[ 2 - (x+h) - 2 + x \] \[ 2 \cdot (-x - h) \] \[ -2x - 2h - 2x \]7. **Further Simplification:** \[ \frac{-4 - 2h}{h} \] \[ \frac{-x - 2h - 2}{h} \]### ExplanationThe work involves simplifying a mathematical expression involving radicals and using the concept of difference quotients for derivatives. The student approaches the problem by manipulating the expression to remove radicals through rationalization. This common algebraic technique is used to facilitate the computation of limits, as is typical in calculus when determining derivatives. The image contains a mathematical problem involving the function $ h(x) = \sqrt{2-x} $, with $ a = -2 $.### Problem Setup:1. **Function**: $ h(x) = \sqrt{2-x} $2. **Point**: $ a = -2 $### Derivation Process:- **Objective**: Determine the derivative $ h'(x) $ and evaluate at specific points.### Steps:1. **Expression for limit definition of a derivative**: \[ \frac{f(x+h) - f(x)}{h} \] Specifically applied to: \[ \frac{\sqrt{2-(x+h)} - \sqrt{2-x}}{h} \]2. **Simplification using conjugates**: Multiply the numerator and the denominator by the conjugate: \[ \frac{(\sqrt{2-(x+h)} - \sqrt{2-x})(\sqrt{2-(x+h)} + \sqrt{2-x})}{h(\sqrt{2-(x+h)} + \sqrt{2-x})} \]3. **Resulting expression**: - The numerator simplifies to: $-x-h$ - The full expression simplifies: \[ \frac{-x-h}{h(\sqrt{2-(x+h)} + \sqrt{2-x})} \]4. **Derivative $ h'(x) $**: Evaluating the derivative: \[ h'(x) = -\frac{1}{2\sqrt{2-x}} \] (Based on substitution and simplification.)5. **Specific evaluations**: - $ h'(-2) = -\frac{1}{2} $ - Further evaluations are suggested, but the steps aren’t filled in.### Additional Notes:- This exercise involves both algebraic manipulation and understanding of the derivative concept using limits.- The steps may include checking the rationalization of expressions using conjugates to achieve the simplified form.This solution provides a comprehensive example of applying the derivative concept to square root functions and verifying points as part of derivative evaluation.

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3) A)-V2-a-2 3a.) (r)D 3b.) N(-2)-2 3c.) fxth)-f(x) 12-x+h) =D72-(x) (Jz-Cth)+

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Power Operation

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Exponents

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Question

How do I work with the radicals?

### Transcription of Math Problem Solving

1. **Expression for $ h(x) $:**
\[
h(x) = \sqrt{2 - x} - 2
\]

2. **Simplification Steps:**
\[
2 \cdot \frac{1}{2} - \frac{x}{2}
\]

3. **Derivative Approximation:**
\[
\frac{f(x+h) - f(x)}{h}
\]

4. **Simplifying the Expression:**
\[
\frac{\sqrt{2-(x+h)} - \sqrt{2-x}}{h}
\]

5. **Rewriting the Expression:**
\[
\frac{\sqrt{2-(x+h)} - \sqrt{2-x}}{h} = \frac{\left( \sqrt{2-(x+h)} - \sqrt{2-x} \right) \left( \sqrt{2-(x+h)} + \sqrt{2-x} \right)}{\sqrt{2-(x+h)} + \sqrt{2-x}}
\]

6. **Solving the Difference of Squares:**
\[
2 - (x+h) - 2 + x
\]
\[
2 \cdot (-x - h)
\]
\[
-2x - 2h - 2x
\]

7. **Further Simplification:**
\[
\frac{-4 - 2h}{h}
\]
\[
\frac{-x - 2h - 2}{h}
\]

### Explanation

The work involves simplifying a mathematical expression involving radicals and using the concept of difference quotients for derivatives. The student approaches the problem by manipulating the expression to remove radicals through rationalization. This common algebraic technique is used to facilitate the computation of limits, as is typical in calculus when determining derivatives.

The image contains a mathematical problem involving the function $ h(x) = \sqrt{2-x} $, with $ a = -2 $.

### Problem Setup:
1. **Function**: $ h(x) = \sqrt{2-x} $
2. **Point**: $ a = -2 $

### Derivation Process:
- **Objective**: Determine the derivative $ h'(x) $ and evaluate at specific points.

### Steps:

1. **Expression for limit definition of a derivative**:
\[
\frac{f(x+h) - f(x)}{h}
\]
Specifically applied to:
\[
\frac{\sqrt{2-(x+h)} - \sqrt{2-x}}{h}
\]

2. **Simplification using conjugates**:
Multiply the numerator and the denominator by the conjugate:
\[
\frac{(\sqrt{2-(x+h)} - \sqrt{2-x})(\sqrt{2-(x+h)} + \sqrt{2-x})}{h(\sqrt{2-(x+h)} + \sqrt{2-x})}
\]

3. **Resulting expression**:
- The numerator simplifies to: $-x-h$
- The full expression simplifies:
\[
\frac{-x-h}{h(\sqrt{2-(x+h)} + \sqrt{2-x})}
\]

4. **Derivative $ h'(x) $**:
Evaluating the derivative:
\[
h'(x) = -\frac{1}{2\sqrt{2-x}}
\]
(Based on substitution and simplification.)

5. **Specific evaluations**:
- $ h'(-2) = -\frac{1}{2} $
- Further evaluations are suggested, but the steps aren’t filled in.

### Additional Notes:
- This exercise involves both algebraic manipulation and understanding of the derivative concept using limits.
- The steps may include checking the rationalization of expressions using conjugates to achieve the simplified form.

This solution provides a comprehensive example of applying the derivative concept to square root functions and verifying points as part of derivative evaluation.

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