Problem 16 - College Algebra

Question

Find \(f+g, f-g, f g,\) and \(\frac{f}{g}\). Determine the domain for each function. $$f(x)=\sqrt{x+6}, g(x)=\sqrt{x-3}$$

Step-by-Step Solution

Answer

The functions are: \(f+g = \sqrt{x+6} + \sqrt{x-3}\), \(f-g = \sqrt{x+6} - \sqrt{x-3}\), \(f*g = \sqrt{(x+6)*(x-3)}\), \(f/g = \sqrt{(x+6)/(x-3)}\). The domains are respectively: \([3, \infty)\), \([3, \infty)\), \([3, \infty)\), \((3, \infty)\).

Step 1: Addition of Functions

To find \(f+g\), simply add the functions together: \(f(x) + g(x) = \sqrt{x+6} + \sqrt{x-3}\). The domain of \(f+g\) is \([3, \infty)\) since both square-roots should be defined.

Step 2: Subtraction of Functions

To find \(f-g\), subtract \(g\) from \(f\): \(f(x) - g(x) = \sqrt{x+6} - \sqrt{x-3}\). The domain of \(f-g\) is \([3, \infty)\) since both square-roots need to be defined.

Step 3: Multiplication of Functions

To find \(f*g\), multiply \(f\) and \(g\) together: \(f(x)*g(x) = \sqrt{x+6} * \sqrt{x-3}\). This simplifies to \(f(x)*g(x) = \sqrt{(x+6)(x-3)}\). The domain of \(f*g\) is \([3, \infty)\), because both square-roots need to be defined (and thus, non-negative).

Step 4: Division of Functions

To find \(f/g\), divide \(f\) by \(g\): \(f(x)/g(x) = \sqrt{x+6}/\sqrt{x-3}\). This simplifies to \(f(x)/g(x) = \sqrt{(x+6)/(x-3)}\). The domain of \(f/g\) is \((3, \infty)\) because both square-roots need to be defined, and the denominator - the square-root in \(g(x)\) - needs to be non-zero to avoid division by zero.

Key Concepts

Domain of a Function

Understanding the domain of a function is key to solving many mathematical problems. The domain refers to all the possible input values (usually 'x' values) that a function can accept without causing any issues, like division by zero or taking the square root of a negative number.
For our functions \(f(x) = \sqrt{x+6}\) and \(g(x) = \sqrt{x-3}\), we must ensure the expressions under the square roots are non-negative.

• For \(f(x)\), \(x+6 \geq 0\), meaning \(x \geq -6\).
• For \(g(x)\), \(x-3 \geq 0\), which means \(x \geq 3\).

When combining functions, the domain is the intersection of these individual domains. Thus, the domain for the combined operations is typically \([3, \infty)\), ensuring both square roots remain valid.

Addition of Functions

Adding two functions is a straightforward process. You simply add the corresponding outputs of each function together. This is expressed as \((f+g)(x) = f(x) + g(x)\).
With our functions, this becomes \(\sqrt{x+6} + \sqrt{x-3}\).
The domain of \(f+g\) is important because both \(f\) and \(g\) need to be defined.
As calculated, the domain is \([3, \infty)\). This ensures that there are no imaginary numbers in the result.

Subtraction of Functions

Subtraction works similarly to addition, except you subtract one function's output from another. It is represented as \((f-g)(x) = f(x) - g(x)\).
For \(f(x) = \sqrt{x+6}\) and \(g(x) = \sqrt{x-3}\), we get \(\sqrt{x+6} - \sqrt{x-3}\).
The domain remains \([3, \infty)\) as both square roots need to be real numbers.
This means the values inside the square roots are non-negative, ensuring valid operations.

Multiplication of Functions

Multiplying functions involves multiplying their outputs for the same input \(x\). It is represented by \((f \, * \, g)(x) = f(x) \cdot g(x)\).
For our functions, this translates to \(\sqrt{x+6} \cdot \sqrt{x-3} = \sqrt{(x+6)(x-3)}\).
Like addition and subtraction, the domain is \([3, \infty)\).

• The reason is both square roots must be defined.
• We also ensure the product under the square root is not negative, maintaining real values.

Division of Functions

Dividing functions is a bit more complex compared to addition, subtraction, or multiplication. It involves taking the output of one function and dividing it by the output of another. This is expressed as \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\).
For \(f(x) = \sqrt{x+6}\) and \(g(x) = \sqrt{x-3}\), this results in \(\frac{\sqrt{x+6}}{\sqrt{x-3}}\).
The domain of \(f/g\) is \((3, \infty)\) instead of \([3, \infty)\) because the denominator \(g(x)\) must not be zero to avoid undefined behavior (division by zero).
Thus, \(x eq 3\), ensuring a valid real number for every input.