Problem 6 - Algebra and Trigonometry

Question

\(1-6=\) An expression is given. (a) Evaluate it at the given value. (b) Find its domain. $$ \frac{1}{\sqrt{x-1}}, \quad t=5 $$

Step-by-Step Solution

Answer

(a) Evaluation at t=5 is 1/2. (b) Domain is x > 1.

Step 1: Substitute the value into the expression

To evaluate the expression \( \frac{1}{\sqrt{x-1}} \) at \( t = 5 \), substitute \( x = 5 \) into the expression. This gives us: \( \frac{1}{\sqrt{5-1}} \).

Step 2: Simplify the expression

Simplify \( \sqrt{5-1} \) to \( \sqrt{4} \), which equals \( 2 \). The expression then becomes \( \frac{1}{2} \).

Step 3: Evaluate the expression

Since the expression simplifies to \( \frac{1}{2} \), we evaluate it to determine that for \( t = 5 \), the result is \( \frac{1}{2} \).

Step 4: Determine the domain of the expression

The expression \( \frac{1}{\sqrt{x-1}} \) involves a square root in the denominator. For the expression to be defined, the radicand \( x - 1 \) must be positive (since square roots of non-positive numbers are undefined in the real number system). Thus, \( x - 1 > 0 \), which simplifies to \( x > 1 \). Therefore, the domain is all real numbers \( x \) such that \( x > 1 \).

Key Concepts

Domain of a Function

The domain of a function includes all the possible input values (typically "x" values) that allow the function to produce real, valid outputs. To find the domain of the expression \( \frac{1}{\sqrt{x-1}} \), we need to make sure that the operations within the expression are valid and do not produce undefined results.

For this particular expression, there are two crucial points to consider:

• Square roots require non-negative radicands. This means in our case, the expression under the square root, \(x - 1\), must be greater than zero. So we solve the inequality \(x - 1 > 0\), which gives \(x > 1\).
• In addition to this, since the square root is in the denominator, it also cannot equal zero (as division by zero is undefined).

Combining these conditions, we determine the domain of our expression to be all real numbers greater than 1. In interval notation, this is expressed as \((1, \infty)\).

Square Roots

Square roots are a type of radical expression. They involve finding a number which, when multiplied by itself, gives the original number inside the root. The operation itself is simple but requires careful consideration about the numbers we apply it to.

• Non-negative radicands: Square roots of negative numbers are undefined in the set of real numbers because there is no real number that squared will result in a negative number.
• Simplifying square roots: If a perfect square lies inside the root, it can be evaluated directly. For instance, \( \sqrt{4} = 2 \) because \( 2 \times 2 = 4 \).

In the expression \(\frac{1}{\sqrt{x-1}}\), when \(x = 5\), we calculate \(\sqrt{5-1}\), simplifying the expression under the root to \(\sqrt{4}\). This results in \(2\). Thus, square roots transform the expression into a simpler, more workable form.

Evaluating Expressions

Evaluating expressions involves determining the result of an algebraic formula by substituting variable values into it. For the expression \( \frac{1}{\sqrt{x-1}} \), let's examine how it's done step by step.

• Substitute the value of the variable: When asked to evaluate at \( t = 5 \), substitute \(x\) with \(5\) resulting in \( \frac{1}{\sqrt{5-1}} \).
• Simplify inside the expression: Compute inside the square root, which gives \( \frac{1}{\sqrt{4}} \).
• Simplify the entire expression: Since \( \sqrt{4} = 2 \), the expression simplifies to \( \frac{1}{2} \).
• Final result: The expression evaluates to \( \frac{1}{2} \) when \(x = 5\). This process reveals how expressions can be simplified and evaluated for a given input.

Each step unfolds the expression into manageable parts, ensuring accuracy and comprehension in evaluating any algebraic formula.