Problem 11 - Contemporary Precalculus

Question

Fill in the entries in the following table. If it is impossible to fill in an entry, put an X in it. $$\begin{array}{|c|c|c|c|c|}\hline t & f(t) & g(t)=f(t)-3 & h(t)=4 f(-t) & i(t)=f(t-1)-2 \\\\\hline-2 & 3 & & & \\\\\hline-1 & 6 & & & \\\\\hline 0 & 8 & & & \\\\\hline 1 & 0 & & & \\\\\hline 2 & 5 & & & \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Question: For each function, find the value for t = 2: g(2), h(2), and i(2). Answer: For t = 2, the function values are g(2) = 2, h(2) = 12, and i(2) = -2.

Step 1: Complete g(t) column

To compute g(t), we need to subtract 3 from the given f(t) values. Apply the formula g(t) = f(t) - 3 to each of the f(t) values in the table: - For t = -2, g(-2) = 3 - 3 = 0 - For t = -1, g(-1) = 6 - 3 = 3 - For t = 0, g(0) = 8 - 3 = 5 - For t = 1, g(1) = 0 - 3 = -3 - For t = 2, g(2) = 5 - 3 = 2

Step 2: Complete h(t) column

To compute h(t), we need to substitute -t into f(t) and then multiply by 4. Apply the formula h(t) = 4 * f(-t) using the given f(t) values in the table: - For t = -2, h(-2) = 4 * f(2) = 4 * 5 = 20 - For t = -1, h(-1) = 4 * f(1) = 4 * 0 = 0 - For t = 0, h(0) = 4 * f(0) = 4 * 8 = 32 - For t = 1, h(1) = 4 * f(-1) = 4 * 6 = 24 - For t = 2, h(2) = 4 * f(-2) = 4 * 3 = 12

Step 3: Complete i(t) column

To compute i(t), we need to substitute t-1 into f(t) and then subtract 2. Apply the formula i(t) = f(t-1) - 2 to each of the t values in the table. If f(t-1) value is not given, fill the cell with "X": - For t = -2, we cannot find f(-3), so i(-2) = X - For t = -1, i(-1) = f(-2) - 2 = 3 - 2 = 1 - For t = 0, i(0) = f(-1) - 2 = 6 - 2 = 4 - For t = 1, i(1) = f(0) - 2 = 8 - 2 = 6 - For t = 2, i(2) = f(1) - 2 = 0 - 2 = -2 With all computations complete, we now have the filled table: $$\begin{array}{|c|c|c|c|c|}\hline t & f(t) & g(t)=f(t)-3 & h(t)=4 f(-t) & i(t)=f(t-1)-2 \\\\\hline-2 & 3 & 0 & 20 & X \\\\\hline-1 & 6 & 3 & 0 & 1 \\\\\hline 0 & 8 & 5 & 32 & 4 \\\\\hline 1 & 0 & -3 & 24 & 6 \\\\\hline 2 & 5 & 2 & 12 & -2 \\\\\hline\end{array}$$

Key Concepts

Function Transformations

Understanding function transformations is essential for analyzing the behavior of functions and predicting their graphical representation. In the given exercise, we transform the base function, f(t), to create new functions g(t) and h(t).

When we subtract a constant from a function, we are moving its graph vertically. For instance, g(t) = f(t) - 3 represents a vertical shift downward by 3 units of the original function f(t). This can be visualized as shifting the y-coordinates of the graph of f(t) down by 3.

Multiplying a function by a constant, such as in h(t) = 4f(-t), results in two transformations: a horizontal reflection and a vertical stretch. The f(-t) part reflects the function about the y-axis, changing the direction of the function as t becomes negative t. The multiplication by 4 then stretches the function vertically, making it 'taller' as every point's y-coordinate is multiplied by 4.

Piecewise Functions

A piecewise function is defined by multiple sub-functions, where each sub-function applies to a certain interval in the domain of the whole function. Essentially, piecewise functions can be seen as building a single function out of several 'pieces.'

In our exercise, although not explicitly a piecewise function, the function i(t) = f(t-1) - 2 introduces the concept of handling different intervals uniquely. For a piecewise function, we would define different expressions for different intervals of t. For example, if f(t) were defined differently for positive and negative t, we'd have a piecewise function. When we cannot find a value to compute, such as f(-3) in i(-2), in a piecewise situation, we would consider if there's another 'piece' of the function that could provide the required value.

Understanding piecewise functions is crucial for more complex mathematical analysis, such as integrating different function behaviors over a range of values.

Function Composition

The composition of functions is a process where the output of one function becomes the input of another. The notation h(f(t)), read as 'h of f of t,' represents the function h composed with f.

In the exercise, the function i(t) = f(t-1) - 2 is not an explicit composition of functions but hints at the idea. We shifted the input of f by subtracting 1 from t before applying the original function f. If we had another function to apply to the output of f(t-1), before subtracting 2, this would form a composition. Composing functions adds layers of transformation, allowing for more sophisticated functions that model complex phenomena. As compositions increase, it's crucial to remember order matters profoundly—the output can change significantly based on the function's order in the composition.