Problem 19 - The Calculus with Analytic Geometry

Question

\(f(x)= \begin{cases}(x-2)^{2}-3 & \text { if } x \leq 5 \\ \frac{1}{2}(x+7) & \text { if } 5

Step-by-Step Solution

Answer

For . For . For : , substitute into :

Step 1: Identify the piecewise function

The function given is a piecewise function, which means it's defined by different expressions depending on the value of x. The function is defined as:

Step 2: Expression for x ≤ 5

For values of x that are less than or equal to 5, the expression for the function is This means for any value of x that is 5 or less, we use the expression .

Step 3: Expression for x > 5

For values of x that are greater than 5, the expression for the function is This means for any value of x that is greater than 5, we use the expression .

Step 4: Evaluate at a specific point (Example: x = 4)

Substitute : .

Step 5: Evaluate at another point (Example: x = 6)

When , use the second expression since : . Substitute : .

Step 6: Evaluate at the boundary (x = 5)

At , use the first expression since it includes an equality condition: . Substitute : .

Key Concepts

Continuity in Piecewise Functions

A piecewise function is defined differently over various intervals in its domain.
It is common to check whether such functions are continuous at the points where the expressions change.
This is important because continuity ensures there are no breaks or jumps in the function.
For example, with the function provided: \( f(x)= \begin{cases}(x-2)^{2}-3 & \text { if } x \ \frac{1}{2}(x+7) & \text { if } x > 5 \end{cases} \),
we need to examine the point where the expressions switch, at \( x = 5 \).To determine if \( f(x) \) is continuous at \( x = 5 \), we must check three conditions:

• The limit of the expression as \( x \) approaches 5 from the left and the right must exist.


• The function's value at \( x = 5 \) must be defined.


• Both the limit from the left and right must equal the function's value at \( x = 5 \).

In this case:

• From the left, using \( (x-2)^{2}-3 \): \( \text {lim}_{{x \rightarrow 5-}}( (x-2)^{2}-3 ) = 6 \).


• From the right, using \( \frac{1}{2}(x+7) \): \( \text {lim}_{{x \rightarrow 5+}}( \frac{1}{2}(x+7) ) = 6 \).


• The value at \( x = 5 \) using \( (x-2)^{2}-3 \) is 6.

Since these three values are equal, the function is continuous at \( x = 5 \).

Evaluating Functions

Evaluating piecewise functions involves substituting the input value into the appropriate expression.
The selection of which expression to use depends on the given value.Let's see how it works with our function. For \( f(x) \) defined as:\( f(x)= \begin{cases}(x-2)^{2}-3 & \text { if } x \ \frac{1}{2}(x+7) & \text { if } x > 5 \end{cases} \),
we follow these steps:

• Identify your \( x \) value.


• Determine the interval in which \( x \) falls.


• Use the corresponding expression to compute the function's value.

For example:

• When \( x = 4 \): \( 4 \leq 5 \), so use \( (x-2)^{2}-3 \)
  Substitute and calculate: \( (4-2)^{2}-3 = 1 \).


• When \( x = 6 \): \( 6 > 5 \), so use \( \frac{1}{2}(x+7) \)
  Substitute: \( \frac{1}{2}(6+7) = 6.5 \).


• At the boundary \( x = 5 \),use \( (x-2)^{2}-3 \)
  Substitute: \( (5-2)^{2}-3 = 6 \).

By carefully following these steps, evaluating piecewise functions becomes straightforward and manageable.

Function Expressions

Function expressions play a crucial role in defining piecewise functions.
They describe the behavior of the function over different intervals of the domain.
Typically, a piecewise function will have more than one expression connected by conditions that specify which input values to use.Understanding these expressions is fundamental in working with such functions.Let's break down the expressions in our example:

• For \( x \): The expression is \( (x-2)^{2}-3 \).
  It represents a quadratic function shifted to the right by 2 units and down by 3 units.


• For \( x > 5 \): The expression is \( \frac{1}{2}(x+7) \).
  It represents a linear function with a slope of \( \frac{1}{2} \) and a y-intercept of \( \frac{7}{2} \).

By examining each expression separately, we gain a clear understanding of how they contribute to the entire piecewise function.
Each part of the function serves a specific segment of the domain, ensuring the entire piecewise function accurately represents the relationship between inputs and outputs over its range.