Problem 68 - Algebra and Trigonometry

Question

Find the domain of the function. $$ h(x)=\sqrt{x-2}-\log _{5}(10-x) $$

Step-by-Step Solution

Answer

The domain of the function is \([2, 10)\).

Step 1: Analyze the Square Root

For the square root term \( \sqrt{x-2} \), the expression inside the square root cannot be negative. Therefore, \( x-2 \geq 0 \), which simplifies to \( x \geq 2 \). This means that \( x \) must be greater than or equal to 2 to keep the expression defined.

Step 2: Analyze the Logarithm

For the logarithmic term \( \log_{5}(10-x) \), the argument of the logarithm must be greater than zero. Therefore, \( 10-x > 0 \). This simplifies to \( x < 10 \). Hence, \( x \) must be less than 10.

Step 3: Combine the Conditions

To find the domain of the function \( h(x) \), we need to satisfy both conditions derived from the previous steps: \( x \geq 2 \) and \( x < 10 \). These can be combined into a single condition: \( 2 \leq x < 10 \). This interval represents all the values that \( x \) can take to ensure both the square root and logarithm are defined and real.

Key Concepts

Understanding the Square Root

When dealing with functions that include a square root, it's important to make sure that the expression inside the square root is not negative. This is because the square root of a negative number is not defined in the set of real numbers.In the function given, we have the square root component as \(\sqrt{x-2}\). To ensure this part of the function is valid, the expression \(x-2\) must be non-negative. This means we solve the inequality \(x-2 \geq 0\). Solving for \(x\), we find \(x \geq 2\), which tells us that \(x\) can be any number from 2 onwards, including 2 itself.This requirement ensures that the square root term stays defined and results in a real number.

Exploring the Logarithm

Logarithms are another key concept that influences the domain of many functions. The argument of a logarithm, which is the part inside the log function, must always be positive. This is because you cannot take the logarithm of zero or a negative number.In our function \(h(x)\), the logarithmic term is \(\log_{5}(10-x)\). For this term to be defined, we must have \(10-x > 0\). Solving this inequality gives us \(x < 10\). Thus, \(x\) must be less than 10 for the logarithm to be valid.This condition ensures that the argument of the log function remains positive, so the value of the logarithm stays defined and gives a real number.

Combining Inequalities for Domain

To find the domain of the function, we need to combine the conditions derived from both the square root and the logarithm parts of the function. The conditions are:

• For \(\sqrt{x-2}\), we have \(x \geq 2\)
• For \(\log_{5}(10-x)\), we have \(x < 10\)

By combining these inequalities, we create a domain that meets both requirements simultaneously. We form the interval \(2 \leq x < 10\), which includes all values of \(x\) that ensure both components of the function remain defined. This interval is the permitted range of \(x\) values that keep the entire function real and meaningful. Therefore, the domain of the function \(h(x)\) is \(x\) values between 2 and 10, where 2 is included, and 10 is not.