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Chapter 4: Problem 68

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

Short Answer

Expert verified
The domain is \([2, 10)\).

Step by step solution

01

Consider the Square Root Function

The function involves a square root, \(\sqrt{x-2}\). This expression is only defined for values where the expression inside the square root is non-negative. Therefore, solve \(x-2 \geq 0\), which simplifies to \(x \geq 2\). This means \(x\) must be greater than or equal to 2.
02

Consider the Logarithmic Function

The function also includes a logarithm, \(\log_5(10-x)\). A logarithmic expression \(\log_b(y)\) is defined only for positive \(y\). Thus, \(10-x > 0\), which simplifies to \(x < 10\). This means \(x\) must be less than 10.
03

Determine the Intersection of Both Conditions

To find the domain of the combined expression, we need the intersection of the two conditions from the previous steps: \(x \geq 2\) and \(x < 10\). Combining these conditions gives \(2 \leq x < 10\). Therefore, the domain of the function is all \(x\) values in the interval \([2, 10)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Root Function
The square root function is a common mathematical function defined as the non-negative value of a number that, when multiplied by itself, gives the original number. In the expression \( \sqrt{x-2} \), this function requires the term inside the square root, here \( x-2 \), to be non-negative.
This is because the square root of a negative number is not defined within the set of real numbers.
To determine when a square root is defined, set the expression inside the square root greater than or equal to zero:
  • Solve for \( x \) in \( x-2 \geq 0 \).
  • This simplifies to \( x \geq 2 \).
In simple terms, \( x \) should be 2 or more for the square root function to work in the real number system.
Logarithmic Function
A logarithmic function is based on the principle of quantifying how many times you can multiply one number (the base) to get another number (the argument). In this context, the function \( \log_5(10-x) \) represents the base-5 logarithm of the expression \( 10-x \). For a logarithmic function to be defined, the argument (inside the logarithm) must be positive.
Thus, \( 10-x \) needs to satisfy:
  • Solve \( 10-x > 0 \).
  • It simplifies to \( x < 10 \).
This means that \( x \) must be less than 10 to keep the logarithmic function valid.
Logarithms of non-positive numbers are undefined in the real numbers.
Interval Notation
Interval notation provides a concise method of describing a range of values, often representing the domain of a function. When analyzing functions, especially when involving square roots and logarithms, using interval notation helps in collectively visualizing all valid inputs.
In the function \( h(x) = \sqrt{x-2} - \log_5(10-x) \), we've identified the set of \( x \) satisfying both conditions:
  • \( x \geq 2 \) from the square root requirement.
  • \( x < 10 \) from the logarithmic requirement.
The overlap or intersection of these two conditions is \( 2 \leq x < 10 \).
This is efficiently expressed in interval notation as \([2, 10)\), meaning \( x \) can take any value from 2 to just shy of 10.

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Most popular questions from this chapter

Draw the graph of the function in a suitable viewing rectangle, and use it to find the domain, the asymptotes, and the local maximum and minimum values. $$y=\ln \left(x^{2}-x\right)$$

Interest Rate \(A\) sum of \(\$ 1000\) was invested for 4 years, and the interest was compounded semiannually. If this sum amounted to \(\$ 1435.77\) in the given time, what was the interest rate?

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