Problem 80 - Elementary and Intermediate Algebra

Question

Let \(h(t)=\sqrt{t+3}\) and \(k(t)=t-5 .\) Find each of the following. $$ (k \circ h)(-2) $$

Step-by-Step Solution

Answer

The value of \((k \circ h)(-2)\) is \(-4\).

Step 1: Understand the Composition of Functions

The composition \((k \circ h)(t)\) means we need to apply the function \(h\) first and then use the result as the input for the function \(k\). Thus, it is expressed as \(k(h(t))\).

Step 2: Execute the Inner Function \(h(t)\)

Substitute \(t = -2\) into the function \(h(t) = \sqrt{t+3}\). Calculate \(h(-2) = \sqrt{-2 + 3} = \sqrt{1} = 1\).

Step 3: Evaluate the Outer Function \(k(t)\) with the Result from Step 2

Now substitute the result from Step 2 into the function \(k(t) = t - 5\). Since \(h(-2) = 1\), we find \(k(1) = 1 - 5 = -4\).

Step 4: Conclude the Calculation

The result of the composition \((k \circ h)(-2)\) is \(-4\).

Key Concepts

Square Root Function

A square root function is a type of function that includes a square root of a variable expression. This function is typically expressed as \( f(x) = \sqrt{x} \). In our exercise, the square root function is given by \( h(t) = \sqrt{t + 3} \). Square root functions are crucial in understanding various mathematical models and appear often in problems involving geometry and physics.

Key characteristics of the square root function include:

• Non-negative inputs: You can only use non-negative numbers under the square root. Hence, the expression inside the root must be \( \geq 0 \).
• Domain: For \( h(t) = \sqrt{t + 3} \), the domain is all real numbers \(t \geq -3 \).
• Range: The output or range is non-negative, as square roots do not produce negative results.

When working with the composition of functions involving a square root, it is first necessary to ensure that any substituted input meets the function's domain conditions.

Linear Function

Linear functions create straight lines when graphed and are one of the simplest types of functions. In this exercise, the linear function is defined by \( k(t) = t - 5 \). This implies that for any input \( t \), the output will be found by subtracting 5 from \( t \).

Characteristics of linear functions include:

• Constant Rate of Change: The graph has a constant slope. For \( k(t) = t - 5 \), the slope is 1, indicating a steady increase or decrease.
• Domain and Range: Both are all real numbers since there are no restrictions like in the square root function.
• Graph: A straight line that extends infinitely in both directions.

Understanding linear functions is essential for performing compositions like \((k \circ h)(t)\). After evaluating \( h(t) \), the output serves as the input for \( k(t) \), allowing us to see the sequential impact of the two distinct operations.

Algebraic Evaluation

Algebraic evaluation involves plugging in numbers and simplifying expressions. It is a core skill in solving math problems like our function composition exercise. This specific problem requires algebraic evaluation of \((k \circ h)(-2)\).

The critical steps for algebraic evaluation are:

• Substitution: Insert \( t = -2 \) into the inner function \( h(t) \), then simplify \( h(-2) = \sqrt{-2 + 3} = \sqrt{1} = 1 \).
• Follow the Order: Using the result of \( h(-2) \), substitute \( t = 1 \) into the outer function \( k(t) = t - 5 \).
• Simplification: Calculate \( k(1) = 1 - 5 = -4 \).

Accurate algebraic evaluation, especially in composed functions, helps ensure each sequential step is based on correct input, arriving at the right solution \((k \circ h)(-2) = -4\). Consistent practice of this method allows you to solve more complex problems with confidence.