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Chapter 0: Problem 34

In Exercises 29-36, evaluate the function for the indicated values. $$ k(x)=\left\lfloor\frac{1}{2} x+6 \rrbracket\right. $$ (a) \(k(5)\) (b) \(k(-6.1)\) (c) \(k(0.1)\) (d) \(k(15)\)

Short Answer

Expert verified
The evaluated solutions are: \(k(5) = 8\), \(k(-6.1) = 3\), \(k(0.1) = 6\), and \(k(15) = 13\)

Step by step solution

01

Evaluate k(5)

Plug '5' for 'x' in the function, \(k(x) = \left\lfloor\frac{1}{2}x + 6\right\rfloor\). So, \(k(5) = \left\lfloor\frac{1}{2}(5) + 6\right\rfloor\). Simplify the expression to find the answer.
02

Evaluate k(-6.1)

Similarly, plug '-6.1' for 'x'. So, \(k(-6.1) = \left\lfloor\frac{1}{2}(-6.1) + 6\right\rfloor\). Simplify to find the answer.
03

Evaluate k(0.1)

Now plug '0.1' for 'x'. So, \(k(0.1) = \left\lfloor\frac{1}{2}(0.1) + 6\right\rfloor\). Simplify to find the answer.
04

Evaluate k(15)

Finally, plug '15' for 'x'. So, \(k(15) = \left\lfloor\frac{1}{2}(15) + 6\right\rfloor\). Simplify to find the answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation involves determining the output value of a mathematical function, given an input value. In this specific exercise, you deal with the floor function, denoted by \( \lfloor \cdot \rfloor \). The floor function takes a real number and rounds it down to the nearest integer. It is sometimes called the "greatest integer function."
  • First, identify the function you are working with. In this case, it is \( k(x) = \left\lfloor \frac{1}{2}x + 6 \right\rfloor \).
  • To evaluate the function, substitute the indicated value of \( x \) into the expression.
  • After substitution, simplify the expression inside the floor operator.
  • Finally, apply the floor function to get the integer part of the result.
For example, when evaluating \( k(5) \), you would substitute 5 for \( x \) resulting in \( \frac{1}{2}(5) + 6 \), which gives you \( 2.5 + 6 = 8.5 \). Applying the floor function, \( \lfloor 8.5 \rfloor \), gives you \( 8 \).
Trigonometric Functions
Trigonometric functions, such as sine, cosine, and tangent, are based on the properties of triangles and circles. Though the exercise does not directly involve trig functions, it's essential to understand their common usage in math problems.
  • These functions often require evaluating aspects of angles and triangles.
  • They can be periodic, which means they repeat values at regular intervals.
  • While trig functions themselves are not evaluated here, the logic of substituting values and simplifying expressions is similar across different function types.
  • Understanding different functions, like floor and trigonometric, can deepen comprehension of mathematical evaluation.
In math exercises like these, familiarity with a variety of functions enables students to apply similar evaluation techniques effectively across diverse problems.
Algebraic Expressions
Algebraic expressions combine numbers and variables using arithmetic operations such as addition, subtraction, multiplication, and division. The key focus here is simplifying these expressions to evaluate them correctly.
  • These expressions can contain multiple terms, making simplification crucial for accurate function evaluation.
  • Start by identifying parts of the expression: constants, coefficients, and variables.
  • Simplify by performing operations in the correct order; a valuable guideline is "PEMDAS" (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).
  • For the given exercise, the expression \( \frac{1}{2}x + 6 \) is simplified before applying the floor function.
Mastering these simplification techniques allows students to comfortably approach various algebraic and mathematical challenges. In this exercise, noticing that adding or subtracting a number to or from \( \frac{1}{2}x \) prepares students for simplifying and evaluating more complex expressions effectively.

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