Given The Piecewise Function Above Evaluate The Following Statements

Evaluating statements involving piecewise functions requires a careful understanding of how the function is defined across different intervals of its domain. Piecewise functions are defined by multiple sub-functions, each applying to a specific interval. The key to correctly evaluating statements is to identify the relevant interval for a given input value and then apply the corresponding sub-function. This process involves examining the conditions that define each piece of the function and substituting the input value into the appropriate sub-function. This article provides a comprehensive guide to evaluating statements involving piecewise functions, covering various scenarios and offering detailed examples.

Understanding Piecewise Functions

A piecewise function is a function defined by multiple sub-functions, each applying to a specific interval of the domain. These functions are essential in mathematics for modeling situations where different rules apply under different conditions.

Definition of Piecewise Functions

A piecewise function is defined as follows:

f(x) = {
  f1(x), if x ∈ I1
  f2(x), if x ∈ I2
  ...
  fn(x), if x ∈ In
}

• ( f(x) ) is the piecewise function.
• ( f_1(x), f_2(x), ..., f_n(x) ) are the sub-functions.
• ( I_1, I_2, ..., I_n ) are the intervals on which the sub-functions are defined.

Key Components

1. Sub-functions: Each ( f_i(x) ) is a standard function (e.g., linear, quadratic, exponential) that applies only within its specified interval.
2. Intervals: Each ( I_i ) is a subset of the real numbers, defining the domain for its corresponding sub-function. The intervals must be non-overlapping to ensure that the function is well-defined.
3. Conditions: The conditions ( x \in I_i ) determine which sub-function to use for a given value of ( x ).

Example

Consider the following piecewise function:

f(x) = {
  x^2,     if x < 0
  x + 1,   if 0 ≤ x ≤ 2
  3,       if x > 2
}

Here, ( f(x) ) is defined by three sub-functions over three intervals:

• ( f_1(x) = x^2 ) for ( x < 0 )
• ( f_2(x) = x + 1 ) for ( 0 \leq x \leq 2 )
• ( f_3(x) = 3 ) for ( x > 2 )

Evaluating Statements with Piecewise Functions

To evaluate statements involving piecewise functions, you must determine which sub-function applies based on the given input value and its corresponding interval.

Steps for Evaluation

1. Identify the Input Value: Determine the value of ( x ) for which you need to evaluate ( f(x) ).
2. Determine the Correct Interval: Find the interval ( I_i ) that contains the input value ( x ). This involves checking the conditions ( x \in I_i ) for each interval.
3. Apply the Corresponding Sub-function: Once the correct interval is identified, use the corresponding sub-function ( f_i(x) ) to compute the value of ( f(x) ).
4. Evaluate the Statement: Use the computed value of ( f(x) ) to evaluate the given statement.

Example 1: Evaluating ( f(x) )

Using the piecewise function defined earlier:

f(x) = {
  x^2,     if x < 0
  x + 1,   if 0 ≤ x ≤ 2
  3,       if x > 2
}

Evaluate ( f(-1) ), ( f(1) ), and ( f(3) ).

1. Evaluate ( f(-1) ):

   • Input value: ( x = -1 )
   • Interval: Since ( -1 < 0 ), the relevant interval is ( x < 0 ).
   • Sub-function: ( f_1(x) = x^2 )
   • Evaluation: ( f(-1) = (-1)^2 = 1 )

2. Evaluate ( f(1) ):

   • Input value: ( x = 1 )
   • Interval: Since ( 0 \leq 1 \leq 2 ), the relevant interval is ( 0 \leq x \leq 2 ).
   • Sub-function: ( f_2(x) = x + 1 )
   • Evaluation: ( f(1) = 1 + 1 = 2 )

3. Evaluate ( f(3) ):

   • Input value: ( x = 3 )
   • Interval: Since ( 3 > 2 ), the relevant interval is ( x > 2 ).
   • Sub-function: ( f_3(x) = 3 )
   • Evaluation: ( f(3) = 3 )

Example 2: Evaluating a Conditional Statement

Given the same piecewise function:

f(x) = {
  x^2,     if x < 0
  x + 1,   if 0 ≤ x ≤ 2
  3,       if x > 2
}

Evaluate the statement: "If ( x = -2 ), then ( f(x) = 4 )."

1. Input value: ( x = -2 )
2. Interval: Since ( -2 < 0 ), the relevant interval is ( x < 0 ).
3. Sub-function: ( f_1(x) = x^2 )
4. Evaluation: ( f(-2) = (-2)^2 = 4 )

The statement "If ( x = -2 ), then ( f(x) = 4 )" is true.

Advanced Scenarios

Scenario 1: Discontinuities

Piecewise functions can be discontinuous at the boundaries of their intervals. Discontinuities occur when the sub-functions do not "meet" at the interval boundaries.

Consider the piecewise function:

f(x) = {
  x,       if x < 1
  x + 2,   if x ≥ 1
}

At ( x = 1 ), the function is discontinuous because:

• For ( x < 1 ), ( f(x) = x ), so ( \lim_{x \to 1^-} f(x) = 1 )
• For ( x \geq 1 ), ( f(x) = x + 2 ), so ( \lim_{x \to 1^+} f(x) = 3 )

Since the left-hand limit and the right-hand limit are not equal, the function is discontinuous at ( x = 1 ).

Scenario 2: Composite Functions

Evaluating composite functions involving piecewise functions requires careful application of the function definitions.

Consider ( f(x) ) defined as:

f(x) = {
  x^2,     if x < 0
  x + 1,   if 0 ≤ x ≤ 2
  3,       if x > 2
}

And ( g(x) = x - 1 ). Evaluate ( f(g(x)) ) at ( x = 2 ).

1. Evaluate ( g(x) ):
   • ( g(2) = 2 - 1 = 1 )
2. Evaluate ( f(g(2)) = f(1) ):
   • Since ( 0 \leq 1 \leq 2 ), use ( f(x) = x + 1 )
   • ( f(1) = 1 + 1 = 2 )

Thus, ( f(g(2)) = 2 ).

Scenario 3: Inverse Functions

Finding the inverse of a piecewise function involves finding the inverse of each sub-function and adjusting the intervals accordingly.

Consider the piecewise function:

f(x) = {
  x^2,     if x ≥ 0
  x,       if x < 0
}

1. For ( x \geq 0 ):

   • ( y = x^2 )
   • ( x = \sqrt{y} )
   • The inverse is ( f^{-1}(x) = \sqrt{x} ) for ( x \geq 0 )

2. For ( x < 0 ):

   • ( y = x )
   • ( x = y )
   • The inverse is ( f^{-1}(x) = x ) for ( x < 0 )

The inverse function is:

f^{-1}(x) = {
  √x,     if x ≥ 0
  x,       if x < 0
}

Common Mistakes

1. Incorrect Interval Selection: The most common mistake is selecting the wrong interval for a given input value. Always double-check the conditions to ensure the correct sub-function is used.
2. Ignoring Discontinuities: When dealing with discontinuous piecewise functions, be careful when evaluating limits or derivatives at the points of discontinuity.
3. Misapplication of Composite Functions: Ensure that the output of the inner function is correctly used as the input for the outer function, paying attention to the intervals.
4. Algebraic Errors: Simple algebraic errors can lead to incorrect evaluations. Always double-check calculations.

Practical Applications

Piecewise functions are used in various fields to model real-world situations:

1. Economics: Modeling tax brackets, where different income levels are taxed at different rates.
2. Physics: Describing the motion of an object with varying forces or conditions.
3. Computer Science: Defining conditional logic in programming, such as if-else statements.
4. Engineering: Modeling systems with different operational modes or thresholds.

Examples with Detailed Explanations

Example 3: Evaluating a Piecewise Function with Absolute Values

Consider the piecewise function:

f(x) = {
  |x|,     if x < -1
  x^2,     if -1 ≤ x ≤ 1
  2x - 1,  if x > 1
}

Evaluate ( f(-2) ), ( f(0) ), and ( f(2) ).

1. Evaluate ( f(-2) ):

   • Input value: ( x = -2 )
   • Interval: Since ( -2 < -1 ), the relevant interval is ( x < -1 ).
   • Sub-function: ( f_1(x) = |x| )
   • Evaluation: ( f(-2) = |-2| = 2 )

2. Evaluate ( f(0) ):

   • Input value: ( x = 0 )
   • Interval: Since ( -1 \leq 0 \leq 1 ), the relevant interval is ( -1 \leq x \leq 1 ).
   • Sub-function: ( f_2(x) = x^2 )
   • Evaluation: ( f(0) = (0)^2 = 0 )

3. Evaluate ( f(2) ):

   • Input value: ( x = 2 )
   • Interval: Since ( 2 > 1 ), the relevant interval is ( x > 1 ).
   • Sub-function: ( f_3(x) = 2x - 1 )
   • Evaluation: ( f(2) = 2(2) - 1 = 4 - 1 = 3 )

Example 4: Evaluating a Piecewise Function with Trigonometric Functions

Consider the piecewise function:

f(x) = {
  sin(x),   if x < π/2
  cos(x),   if π/2 ≤ x ≤ π
  tan(x),   if x > π
}

Evaluate ( f(0) ), ( f(π/2) ), and ( f(π) ).

1. Evaluate ( f(0) ):

   • Input value: ( x = 0 )
   • Interval: Since ( 0 < π/2 ), the relevant interval is ( x < π/2 ).
   • Sub-function: ( f_1(x) = sin(x) )
   • Evaluation: ( f(0) = sin(0) = 0 )

2. Evaluate ( f(π/2) ):

   • Input value: ( x = π/2 )
   • Interval: Since ( π/2 \leq π/2 \leq π ), the relevant interval is ( π/2 \leq x \leq π ).
   • Sub-function: ( f_2(x) = cos(x) )
   • Evaluation: ( f(π/2) = cos(π/2) = 0 )

3. Evaluate ( f(π) ):

   • Input value: ( x = π )
   • Interval: Since ( π/2 \leq π \leq π ), the relevant interval is ( π/2 \leq x \leq π ).
   • Sub-function: ( f_2(x) = cos(x) )
   • Evaluation: ( f(π) = cos(π) = -1 )

Example 5: Evaluating a Piecewise Function in a Real-World Scenario

Consider a parking fee structure defined as a piecewise function:

C(t) = {
  5,       if 0 < t ≤ 2
  5 + 2(t - 2),  if 2 < t ≤ 6
  13,      if t > 6
}

Where ( C(t) ) is the cost in dollars and ( t ) is the time in hours. Evaluate the cost for parking for 1 hour, 4 hours, and 8 hours.

1. Evaluate ( C(1) ):

   • Input value: ( t = 1 )
   • Interval: Since ( 0 < 1 \leq 2 ), the relevant interval is ( 0 < t \leq 2 ).
   • Sub-function: ( C(t) = 5 )
   • Evaluation: ( C(1) = 5 )

2. Evaluate ( C(4) ):

   • Input value: ( t = 4 )
   • Interval: Since ( 2 < 4 \leq 6 ), the relevant interval is ( 2 < t \leq 6 ).
   • Sub-function: ( C(t) = 5 + 2(t - 2) )
   • Evaluation: ( C(4) = 5 + 2(4 - 2) = 5 + 2(2) = 5 + 4 = 9 )

3. Evaluate ( C(8) ):

   • Input value: ( t = 8 )
   • Interval: Since ( 8 > 6 ), the relevant interval is ( t > 6 ).
   • Sub-function: ( C(t) = 13 )
   • Evaluation: ( C(8) = 13 )

Conclusion

Evaluating statements involving piecewise functions requires careful attention to detail and a thorough understanding of how the function is defined across different intervals. By following the systematic approach outlined in this article, you can accurately evaluate piecewise functions and avoid common mistakes. This skill is essential in various fields, from mathematics and physics to computer science and economics, where piecewise functions are used to model complex and dynamic systems.