Problem 77 - Intermediate Algebra

Question

Let \(h(t)=\sqrt{t+3}\) and \(k(t)=t-5 .\) Find each of the following. $$ (h \circ k)(18) $$

Step-by-Step Solution

Answer

The value of \((h \circ k)(18)\) is 4.

Step 1: Understand the Composition Function

The notation \((h \circ k)(t)\) represents the composition of the functions \(h\) and \(k\), which means \(h(k(t))\). In this step, we first need to evaluate the inner function \(k(t)\) for the given value of \(t\).

Step 2: Evaluate the Inner Function k(t)

Substitute \(t = 18\) into \(k(t) = t - 5\). This gives:\[ k(18) = 18 - 5 = 13 \].

Step 3: Substitute into Outer Function h(t)

Now substitute the result from Step 2 into the function \(h(t) = \sqrt{t+3}\). So, substitute \(t = 13\) in \(h(t)\):\[ h(13) = \sqrt{13 + 3} = \sqrt{16} = 4 \].

Step 4: Conclusion

After evaluating \(h(k(18))\), we find that the value of the composition \((h \circ k)(18)\) is 4.

Key Concepts

Square Root Function

A square root function is a type of function that involves the square root of an expression. It is written in the form \( h(t) = \sqrt{t} \). The square root function typically produces a non-negative result since we usually consider only the principal (non-negative) square root. For instance, if \( h(t) = \sqrt{t + 3} \), the expression inside the square root is \( t + 3 \). To ensure the function is defined, \( t + 3 \) must be zero or positive. This ensures we are not attempting to find the square root of a negative number, which isn't defined in the set of real numbers. The square root function creates a curve that starts from a specific point on the horizontal axis, moving to the right and steadily rising outward in a gentle curve. This gradual slope occurs because as \( t \) increases, \( \sqrt{t} \) grows at a decreasing rate.

Linear Function

Linear functions are among the simplest types of functions, characterized by a constant rate of change, which means they graph to a straight line. A linear function is generally expressed in the form \( k(t) = mt + b \), where \( m \) represents the slope and \( b \) is the y-intercept. In our exercise, the function \( k(t) = t - 5 \) is linear because \( t \) is multiplied by 1 (hence, the slope is 1) and shifted downwards by 5 due to the subtracted constant. Linear functions are straightforward: for every unit increase in \( t \), \( k(t) \) increases by the slope value. Thus, at \( t = 18 \), \( k(t) \) can be quickly calculated as \( 18 - 5 = 13 \). This simplicity makes linear functions an essential tool for modeling real-world situations where a constant rate of change is observed.

Evaluating Functions

Evaluating a function involves finding the output value for a particular input value. This process is the essence of using functions, as it allows us to determine specific values based on a formula or rule given by the function. In the context of our problem, evaluating \( k(t) = t - 5 \) at \( t = 18 \) involves substituting \( t = 18 \) into the function, which results in \( k(18) = 18 - 5 = 13 \). Similarly, with \( h(t) = \sqrt{t + 3} \), evaluating it at \( t = 13 \) means you calculate \( \sqrt{13 + 3} = \sqrt{16} = 4 \). Often, evaluating functions requires careful substitution and arithmetic to avoid errors. Remember, each step in the evaluation process should be completed with full attention to detail, particularly when working with nested or composed functions.

Composition of Functions

The composition of functions involves plugging one function into another, essentially combining them to create a new function. It's often denoted as \( (h \circ k)(t) = h(k(t)) \). In mathematics, function composition resembles assembling chain operations where the output of one function becomes the input of another. Our exercise is a classic example: first, evaluate the inner function \( k(t) = t - 5 \) with a given value of \( t = 18 \), yielding \( k(18) = 13 \). This result, \( 13 \), is then used as the input for the outer function \( h(t) = \sqrt{t + 3} \), so \( h(13) = \sqrt{16} = 4 \). Composition of functions is a powerful concept as it allows complex operations to be broken down into manageable steps, often revealing insights into the relationship between the involved functions.