Problem 5

Question

Exer. 3-8: 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 $f / g$ $$ f(x)=\sqrt{x+5}, \quad g(x)=\sqrt{x+5} $$

Step-by-Step Solution

Verified

Answer

(f+g)(x) = 2\sqrt{x+5}, (f-g)(x) = 0, (fg)(x) = x+5, (f/g)(x) = 1; domains: x ≥ -5 for f+g, f-g, fg; x > -5 for f/g.

1Step 1: Understanding the Functions

We are given two functions, $f(x) = \sqrt{x+5}$ and $g(x) = \sqrt{x+5}$. Both functions are actually the same, which implies that $f(x) = g(x)$. This symmetry will simplify our operations.

2Step 2: Addition of Functions

To find $(f+g)(x)$, we add the two functions: $(f+g)(x) = f(x) + g(x) = \sqrt{x+5} + \sqrt{x+5} = 2\sqrt{x+5}$.

3Step 3: Subtraction of Functions

To find $(f-g)(x)$, we subtract $g(x)$ from $f(x)$: $(f-g)(x) = f(x) - g(x) = \sqrt{x+5} - \sqrt{x+5} = 0$.

4Step 4: Multiplication of Functions

To find $(fg)(x)$, we multiply the functions: $(fg)(x) = f(x) \cdot g(x) = (\sqrt{x+5}) \cdot (\sqrt{x+5}) = x + 5$.

5Step 5: Division of Functions

To find $(f/g)(x)$, we divide $f(x)$ by $g(x)$: $(f/g)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+5}}{\sqrt{x+5}} = 1$, provided that $x+5 eq 0$ which means $x eq -5$.

6Step 6: Domain of Sum, Difference, and Product

The domain for $f+g$, $f-g$, and $fg$ is determined by the expression under the square root, $x+5\geq 0$, giving us $x \geq -5$.

7Step 7: Domain of Division

The domain for $f/g$ must also ensure that the denominator ($g(x)$) is not zero, which implies $x eq -5$. So, $x > -5$.

Key Concepts

Function AdditionFunction SubtractionFunction MultiplicationFunction DivisionDomain of a Function

Function Addition

Function addition is quite straightforward. When adding two functions, we essentially combine their output values for each point in their domain.
This operation is often represented as

$(f + g)(x) = f(x) + g(x)$
In the given exercise, both functions are identical, with $f(x) = \sqrt{x+5}$ and $g(x) = \sqrt{x+5}$.
Thus, the addition is $(f+g)(x) = \sqrt{x+5} + \sqrt{x+5} = 2\sqrt{x+5}$.

It is important to note that this operation is valid only where both $f(x)$ and $g(x)$ are defined. The restrictions on their domain affect the summation.

Function Subtraction

Subtraction of functions involves taking away the output of one function from another. For two functions, this is expressed as:

$(f - g)(x) = f(x) - g(x)$

In this exercise:

Since $f(x)$ and $g(x)$ are the same, their subtraction yields zero: $(f-g)(x) = \sqrt{x+5} - \sqrt{x+5} = 0$.

When subtracting functions, just like with addition, the result will also require that $x$ falls within the domain of both functions.

Function Multiplication

Multiplying functions involves taking their values at a certain point and multiplying them together. This operation is denoted as:

$(fg)(x) = f(x) \times g(x)$

For the exercise provided:

$(fg)(x) = \sqrt{x+5} \cdot \sqrt{x+5} = (x+5)$

Notice that multiplying two identical square roots results in removing the square root, yielding $(x + 5)$. The domain for this operation results from the restrictions on both functions.

Function Division

Function division is essentially finding the ratio of the two functions at given values of $x$. This is expressed as:

$(f/g)(x) = \frac{f(x)}{g(x)}$

It's important to emphasize in division:

The denominator should never be zero, which in this case means $g(x) eq 0$.

For our function $g(x) = \sqrt{x+5}$, the division works as:

$(f/g)(x) = \frac{\sqrt{x+5}}{\sqrt{x+5}} = 1$, provided that $x eq -5$.

The domain of a division operation is more restrictive than just the domain of the individual functions due to the potential for division by zero.

Domain of a Function

The domain of a function is the complete set of real numbers for which the function is defined.
To determine the domain, we consider all possible $x$ values that can be plugged into the function without causing any undefined operations, like division by zero or square roots of negative numbers.
For $f(x) = \sqrt{x+5}$ and $g(x) = \sqrt{x+5}$:

For function addition, subtraction, and multiplication, the expressions are defined when under the square root is not negative: $x+5 \geq 0$, resulting in $x \geq -5$.
For division, the domain excludes $x = -5$ because the denominator becomes zero. Hence, $x > -5$.

Domains can often establish if operations between functions are even possible.

Problem 5

Other exercises in this chapter

Problem 4

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Problem 5

Exer. 3-12: Determine whether $f$ is even, odd, or neither even nor odd. $$ f(x)=3 x^{4}+2 x^{2}-5 $$

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Problem 5

Exer. 5-12: Express $f(x)$ in the form $a(x-h)^{2}+k$. $$ f(x)=-x^{2}-4 x-8 $$

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Problem 5

Exer. 5-10: If $a$ and $h$ are real numbers, find (a) $f(a)$ (b) $f(-a)$ (c) $-f(a)$ (d) $f(a+h)$ (e) $f(a)+f(h)$ (f) $\frac{f(a+h)-f(a)}{h}$, i

View solution