Question

For the given functions fand g. find the following. For parts \((a)-(d),\) also find the domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f \cdot g)(x)\) (d) \(\left(\frac{f}{g}\right)(x)\) (e) \((f+g)\) (3) (f) \((f-g)\) (4) (g) \((f \cdot g)\) ( 2 ) \((h)\left(\frac{f}{g}\right)(1)\) \(f(x)=\sqrt{x+1} ; \quad g(x)=\frac{2}{x}\)

Short answer

(a) \( \sqrt{x+1} + \frac{2}{x} \) Domain: \( x \in [-1, 0) \cup (0, \infty) \) (b) \( \sqrt{x+1} - \frac{2}{x} \) Domain: \( x \in [-1, 0) \cup (0, \infty) \) (c) \( \frac{2 \sqrt{x+1}}{x} \) Domain: \( x \in [-1, 0) \cup (0, \infty) \) (d) \( \frac{x \sqrt{x+1}}{2} \) Domain: \( x \in [-1, 0) \cup (0, \infty) \) (e) \( \frac{8}{3} \) (f) \( \sqrt{5} - \frac{1}{2} \) (g) \( \sqrt{3} \) (h) \( \frac{\sqrt{2}}{2} \).

Step-by-step solution

Define the functions

Given functions are: - \( f(x) = \sqrt{x+1} \) - \( g(x) = \frac{2}{x} \)

Calculate \((f+g)(x)\) and find the domain

To find \((f+g)(x)\): \[ (f+g)(x) = f(x) + g(x) = \sqrt{x+1} + \frac{2}{x} \] The domain is where both functions are defined. \( f(x) \) is defined when \( x+1 \geq 0 \), i.e., \( x \geq -1 \), and \( g(x) \) is defined when \( x eq 0 \). Hence, the domain is \( x \in [-1, 0) \cup (0, \infty) \).

Calculate \((f-g)(x)\) and find the domain

To find \((f-g)(x)\): \[ (f-g)(x) = f(x) - g(x) = \sqrt{x+1} - \frac{2}{x} \] The domain remains the same as above. Therefore, the domain is \( x \in [-1, 0) \cup (0, \infty) \).

Calculate \((f \cdot g)(x)\) and find the domain

To find \((f \cdot g)(x)\): \[ (f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x+1} \cdot \frac{2}{x} = \frac{2 \sqrt{x+1}}{x} \] Again, the domain is where both functions are defined, thus it is \( x \in [-1, 0) \cup (0, \infty) \).

Calculate \( \left( \frac{f}{g} \right)(x) \) and find the domain

To find \( \left( \frac{f}{g} \right)(x) \): \[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+1}}{\frac{2}{x}} = \frac{x \sqrt{x+1}}{2} \] The domain remains the same as before: \( x \in [-1, 0) \cup (0, \infty) \).

Calculate \((f+g)(3)\)

Substitute \( x = 3 \) into \( (f+g)(x) \): \[ (f+g)(3) = \sqrt{3+1} + \frac{2}{3} = \sqrt{4} + \frac{2}{3} = 2 + \frac{2}{3} = \frac{8}{3} \]

Calculate \((f-g)(4)\)

Substitute \( x = 4 \) into \( (f-g)(x) \): \[ (f-g)(4) = \sqrt{4+1} - \frac{2}{4} = \sqrt{5} - \frac{1}{2} \]

Calculate \((f \cdot g)(2)\)

Substitute \( x = 2 \) into \( (f \cdot g)(x) \): \[ (f \cdot g)(2) = \frac{2 \sqrt{2+1}}{2} = \sqrt{3} \]

Calculate \( \left( \frac{f}{g} \right)(1) \)

Substitute \( x = 1 \) into \( \left( \frac{f}{g} \right)(x) \): \[ \left( \frac{f}{g} \right)(1) = \frac{1 \sqrt{1+1}}{2} = \frac{1 \cdot \sqrt{2}}{2} = \frac{\sqrt{2}}{2} \]

Key concepts

domain of functions

The domain of a function is the set of all possible input values (x) that make the function well-defined. These values must satisfy any constraints the functions may have.
For example, consider the function \( f(x) = \sqrt{x+1} \). Here, the square root requires that \( x+1 \geq 0 \), so the domain is all values \( x \geq -1 \).
Similarly, the function \( g(x) = \frac{2}{x} \) has a restriction because division by zero is undefined, so \( x eq 0 \).
When dealing with combination functions like \( f(x) + g(x) \), \( f(x) - g(x) \), \( f(x) \cdot g(x) \), or \( \frac{f(x)}{g(x)} \), their domains are the intersection of the individual domains. For the given functions, the combined domain would be \( x \in [-1, 0) \cup (0, \infty) \). This ensures the values work for both \( f(x) \) and \( g(x) \).

function addition

Adding functions simply means combining their outputs for each input value.
Given \( f(x) = \sqrt{x+1} \) and \( g(x) = \frac{2}{x} \), the addition is:
\[ (f+g)(x) = f(x) + g(x) = \sqrt{x+1} + \frac{2}{x} \]
To add these functions, each must be defined for the same input values.
Consequently, as mentioned earlier, the domain for \( (f+g)(x) \) is \( x \in [-1, 0) \cup (0, \infty) \).
When asked to evaluate at a particular point, like \( (f+g)(3) \), substitute \( x \) into the expression and solve:
\[ (f+g)(3) = \sqrt{3+1} + \frac{2}{3} = 2 + \frac{2}{3} = \frac{8}{3} \]

function subtraction

Subtracting functions works very similarly to adding them; it involves subtracting the output values for a given input.
For the functions given, \( f(x) = \sqrt{x+1} \) and \( g(x) = \frac{2}{x} \), function subtraction is:
\[ (f-g)(x) = f(x) - g(x) = \sqrt{x+1} - \frac{2}{x} \]
Just like addition, the domain for subtraction is the same, \( x \in [-1, 0) \cup (0, \infty) \).
To find a specific value, such as \( (f-g)(4) \):
\[ (f-g)(4) = \sqrt{4+1} - \frac{2}{4} = \sqrt{5} - \frac{1}{2} \]

function multiplication

Multiplying functions involves taking the product of their outputs for each input value. For the given functions, multiplication looks like:
\[ (f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x+1} \cdot \frac{2}{x} = \frac{2 \sqrt{x+1}}{x} \]
The combined domain must ensure both functions are defined. Thus, the domain remains \( x \in [-1, 0) \cup (0, \infty) \).
To evaluate the function at a certain point, say \( x = 2 \):
\[ (f \cdot g)(2) = \frac{2 \sqrt{2+1}}{2} = \sqrt{3} \]

function division

Dividing functions means taking the quotient of their outputs for each input value. For functions \( f \) and \( g \), it is important that the denominator (\( g(x) \)) is not zero to avoid undefined results.
The division is:
\[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+1}}{\frac{2}{x}} = \frac{x \sqrt{x+1}}{2} \]
The domain must exclude points where \( g(x) = 0 \), so it is the same as before, \( x \in [-1, 0) \cup (0, \infty) \).
For example, to find the value at \( x=1 \):
\[ \left( \frac{f}{g} \right)(1) = \frac{\sqrt{1+1}}{\frac{2}{1}} = \frac{1 \cdot \sqrt{2}}{2} = \frac{\sqrt{2}}{2} \]