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Chapter 2: Problem 6

Find (a) \((f+g)(x),(f-g)(x),(f g)(x),\) and \((f / g)(x)\) (b) the domain of \(f+g, f-g,\) and \(f g\) (c) the domain of \(f / g\) $$f(x)=\sqrt{3-2 x}, \quad g(x)=\sqrt{x+4}$$

Short Answer

Expert verified
(a) \((f+g)(x) = \sqrt{3-2x} + \sqrt{x+4},\) \((f-g)(x) = \sqrt{3-2x} - \sqrt{x+4},\) \((fg)(x) = \sqrt{3-2x} \cdot \sqrt{x+4},\) \(\left( \frac{f}{g} \right)(x) = \frac{\sqrt{3-2x}}{\sqrt{x+4}}\). (b) Domain of \(f+g, f-g, fg: [-4, \frac{3}{2}]\). (c) Domain of \(\frac{f}{g}: (-4, \frac{3}{2}]\).

Step by step solution

01

Find (f+g)(x)

To find \((f+g)(x)\), we need to add the functions \(f(x)\) and \(g(x)\). Therefore,\[(f+g)(x) = f(x) + g(x) = \sqrt{3-2x} + \sqrt{x+4}. \]
02

Find (f-g)(x)

To find \((f-g)(x)\), subtract the function \(g(x)\) from \(f(x)\). Therefore,\[(f-g)(x) = f(x) - g(x) = \sqrt{3-2x} - \sqrt{x+4}.\]
03

Find (f*g)(x)

To multiply the functions \(f(x)\) and \(g(x)\), we have:\[(fg)(x) = f(x) \cdot g(x) = \sqrt{3-2x} \cdot \sqrt{x+4}.\]
04

Find (f/g)(x)

For division of functions, divide \(f(x)\) by \(g(x)\):\[\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{3-2x}}{\sqrt{x+4}}, \quad g(x) eq 0.\]
05

Determine the domain of f(x) and g(x)

The domain of a function involving a square root is the set of values for which the expression inside the square root is non-negative.- For \(f(x) = \sqrt{3-2x}\), set up the inequality: \[3-2x \geq 0 \Rightarrow x \leq \frac{3}{2}.\]- For \(g(x) = \sqrt{x+4}\), set up the inequality: \[x+4 \geq 0 \Rightarrow x \geq -4.\]Therefore, the domain of \(f(x)\) is \(x \leq \frac{3}{2}\) and the domain of \(g(x)\) is \(x \geq -4\).
06

Find the domain of (f+g), (f-g), and (f*g)

The domain of \(f+g\), \(f-g\), and \(f*g\) is the intersection of the domains of \(f(x)\) and \(g(x)\).Since \(x \leq \frac{3}{2}\) for \(f(x)\) and \(x \geq -4\) for \(g(x)\), their intersection is: \[-4 \leq x \leq \frac{3}{2}.\]
07

Find the domain of (f/g)

For \((f/g)(x)\), ensure \(g(x) eq 0\). Since \(g(x) = \sqrt{x+4}\), set \(\sqrt{x+4} eq 0\), which implies:\[x+4 eq 0 \Rightarrow x eq -4.\]Thus, the domain of \(\frac{f}{g}\) is \[-4 < x \leq \frac{3}{2}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Operations
In mathematics, it's common to operate with functions to create new functions. These operations include addition, subtraction, multiplication, and division. When combining functions using these operations:
  • For \( (f+g)(x) \), you simply add the outputs of \( f(x) \) and \( g(x) \)
  • For \( (f-g)(x) \), you subtract \( g(x) \)'s output from \( f(x) \)
  • To multiply functions, as in \( (fg)(x) \), multiply the outputs of \( f(x) \) and \( g(x) \)
  • For division, \( \left( \frac{f}{g} \right)(x) \), divide \( f(x) \) by \( g(x) \), keeping in mind \( g(x) eq 0 \) to avoid division by zero.
Combining functions using these operations is straightforward but requires attention to the domain of the resulting function.
Square Root Functions
Square root functions often appear in mathematics and have important constraints. The key characteristic of square root functions, like \( f(x) = \sqrt{3-2x} \) and \( g(x) = \sqrt{x+4} \), is that they can only accept non-negative arguments inside the square root:
  • The expression inside a square root, \( \sqrt{a} \), must be \( a \geq 0 \)
  • For \( \sqrt{3-2x} \), we solve \( 3-2x \geq 0 \), narrowing \( x \) to \( x \leq \frac{3}{2} \)
  • For \( \sqrt{x+4} \), set \( x+4 \geq 0 \) to get \( x \geq -4 \)
Understanding these constraints ensures we know where the square root functions are defined (also known as the function's domain). Be cautious with the input values to maintain the validity of the function.
Intersection of Domains
The intersection of domains is critical when performing operations on multiple functions. This involves finding the values common to the domains of the individual functions:
  • Let's say \( f(x) \) defines its domain as \( x \leq \frac{3}{2} \) and \( g(x) \) as \( x \geq -4 \)
  • The intersection comprises values satisfying both conditions: \( -4 \leq x \leq \frac{3}{2} \)
Only within this intersection can operations such as \( f + g, f - g, \) and \( fg \) be validly computed. This ensures any function operation doesn't involve an undefined input for either function.
Domain Restrictions
Domain restrictions are a notable aspect of function operations to prevent undefined conditions. Factors leading to domain restrictions include:
  • Square roots, which must maintain non-negative inside values
  • Division operations, which can't have zero denominators
For \( \left( \frac{f}{g} \right)(x) \), the domain must exclude points where \( g(x) = 0 \). For \( g(x) = \sqrt{x+4} \), exclude \( x = -4 \) since this renders \( g(x) \) as zero:
  • The valid domain for \( \frac{f}{g} \) becomes \(-4 < x \leq \frac{3}{2}\)
Clarifying domain restrictions, especially in operations like division, ensures mathematical operations remain valid without errors.

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