91Ó°ÊÓ

Function evaluation is the process of finding the value of a function for a specific input. When we solve mathematical problems involving functions, we often need to evaluate these functions to obtain concrete results. Here, the exercise involves two functions, \(h(t)\) and \(k(t)\).
For \(h(t) = \sqrt{t + 3}\), when we want to find \(h(22)\), we replace \(t\) with 22. This substitution gives us \(h(22) = \sqrt{22 + 3}\). Solving this, we find \(h(22) = \sqrt{25} = 5\).
Similarly, for \(k(t) = t - 5\), the value \(k(5)\) is found by replacing \(t\) with 5, resulting in \(k(5) = 5 - 5 = 0\). This approach is fundamental since it allows us to extract meaningful quantitative results from functions.

The square root function is a specific type of function where the output is the positive square root of the input. Mathematically, for a function \(h(t) = \sqrt{t + 3}\), it involves taking the square root of the expression \(t + 3\).
This function is defined only for values of \(t\) for which the expression under the square root, \(t + 3\), is non-negative. This means \(t\) must be such that \(t + 3 \geq 0\), or \(t \geq -3\). Therefore, we evaluate \(h(t)\) only within this domain to ensure the function is well-defined and returns real numbers.

• The square root function is continuous and smooth.
• It increases gradually as \(t\) increases.

Understanding how to manipulate and evaluate this function is crucial, especially when dealing with composite functions.

The substitution method is a helpful technique used to simplify complex problems by temporarily replacing parts of the expression with intermediate variables. It is particularly useful in evaluating composite functions.
In this exercise, we use substitution in two ways:

• To evaluate \(h(22)\), we first substitute 22 into the expression \(t + 3\) inside \(h(t)\), resulting in \(h(22) = \sqrt{25} = 5\).
• Then, to find \(k(5)\), the output of \(h(22)\) substitutes as the new input in \(k(t)\), leading to \(k(5) = 0\).

A systematic application of substitution makes it easier to evaluate composite functions step-by-step, ensuring clarity and accuracy in solving the problem.