Problem 8
Question
For the following exercises, state the domain and range of the function. $$g(x)=\log _{5}(2 x+9)-2$$
Step-by-Step Solution
Verified Answer
Domain: \((-\frac{9}{2}, \infty)\); Range: \(( -\infty, \infty)\).
1Step 1: Understand the Function
The given function is a logarithmic function of the form \( g(x) = \log_{5}(2x + 9) - 2 \). Logarithmic functions are only defined when their argument is greater than zero.
2Step 2: Find the Domain
To determine the domain, we need \( 2x + 9 > 0 \). Solving this inequality gives:\[ 2x + 9 > 0 \]\[ 2x > -9 \]\[ x > -\frac{9}{2} \]Thus, the domain of the function is \( x > -\frac{9}{2} \) which can be expressed in interval notation as \( (-\frac{9}{2}, \infty) \).
3Step 3: Understand the Range of the Logarithmic Part
The range of \( \log_{5}(2x + 9) \) is all real numbers \(( -\infty, \infty)\) because the output of a logarithmic function can be any real number when the argument is positive.
4Step 4: Find the Range of the Entire Function
Since \( g(x) = \log_{5}(2x + 9) - 2 \), subtracting 2 from every output of the logarithmic function does not affect the range in terms of type of numbers. Hence, the range of \( g(x) \) remains \(( -\infty, \infty)\).
Key Concepts
Domain and RangeInequalitiesInterval Notation
Domain and Range
When working with logarithmic functions such as \(g(x) = \log_{5}(2x + 9) - 2\), it is crucial to establish both the domain and range. These are the sets of input values (domain) and output values (range) that make the function "work".
The domain of a logarithmic function is determined by the condition that the argument of the logarithm must be positive. For our function, we have \(2x + 9 > 0\). By solving this inequality, you simplify to \(x > -\frac{9}{2}\). Thus, the domain in interval notation is \((-\frac{9}{2}, \infty)\). This means that you can input any value for \(x\) greater than \(-\frac{9}{2}\) and the function will output a valid number.
The range of \(\log_{5}(2x + 9)\) is all real numbers \(( -\infty, \infty)\). This is because logarithmic functions, when their argument is positive, can output any real number. The subtraction of 2 in the function \(g(x) = \log_{5}(2x + 9) - 2\) simply shifts the function vertically but does not change the type of numbers the function can output. Hence, the range of the entire function remains \(( -\infty, \infty)\).
The domain of a logarithmic function is determined by the condition that the argument of the logarithm must be positive. For our function, we have \(2x + 9 > 0\). By solving this inequality, you simplify to \(x > -\frac{9}{2}\). Thus, the domain in interval notation is \((-\frac{9}{2}, \infty)\). This means that you can input any value for \(x\) greater than \(-\frac{9}{2}\) and the function will output a valid number.
The range of \(\log_{5}(2x + 9)\) is all real numbers \(( -\infty, \infty)\). This is because logarithmic functions, when their argument is positive, can output any real number. The subtraction of 2 in the function \(g(x) = \log_{5}(2x + 9) - 2\) simply shifts the function vertically but does not change the type of numbers the function can output. Hence, the range of the entire function remains \(( -\infty, \infty)\).
Inequalities
Understanding inequalities is critical when determining the domain of functions like logarithms. Logarithmic functions demand that their argument must be greater than zero. This requirement stems from the nature of logarithms, which cannot take negative numbers or zero as input.
Let's look at the inequality \(2x + 9 > 0\) from our function \(g(x) = \log_{5}(2x + 9) - 2\). Start by isolating \(x\) which yields, \(2x > -9\). Divide every term by 2 to solve for \(x\), that gives you \(x > -\frac{9}{2}\). This process of solving inequalities ensures that we only choose values for \(x\) that keep the logarithm defined.
Remember, regularly practicing inequalities will sharpen your ability to quickly identify valid domain constraints for different types of functions.
Let's look at the inequality \(2x + 9 > 0\) from our function \(g(x) = \log_{5}(2x + 9) - 2\). Start by isolating \(x\) which yields, \(2x > -9\). Divide every term by 2 to solve for \(x\), that gives you \(x > -\frac{9}{2}\). This process of solving inequalities ensures that we only choose values for \(x\) that keep the logarithm defined.
Remember, regularly practicing inequalities will sharpen your ability to quickly identify valid domain constraints for different types of functions.
Interval Notation
In mathematics, expressing solutions in interval notation is an efficient way to convey the domain and range of a function. This notation helps clearly state which numbers are included or excluded from these sets.
The domain of the function \(g(x) = \log_{5}(2x + 9) - 2\), calculated as \(x > -\frac{9}{2}\), is expressed in interval notation as \((-\frac{9}{2}, \infty)\). This tells us that the domain includes all numbers greater than but not equal to \(-\frac{9}{2}\) and ranges to infinity.
The domain of the function \(g(x) = \log_{5}(2x + 9) - 2\), calculated as \(x > -\frac{9}{2}\), is expressed in interval notation as \((-\frac{9}{2}, \infty)\). This tells us that the domain includes all numbers greater than but not equal to \(-\frac{9}{2}\) and ranges to infinity.
- The parenthesis "(" around \(-\frac{9}{2}\) indicates that \(-\frac{9}{2}\) is not included in the domain (non-inclusive).
- The infinity symbol \(\infty\) is always accompanied by a parenthesis because infinity is a concept, not an actual number that can be captured or reached.
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