Chapter 1: Problem 5
Evaluate each function. Given \(f(x)=\frac{1}{|x|},\) find a. \(f(2)\) b. \(f(-2)\) c. \(f\left(\frac{-3}{5}\right)\) d. \(f(2)+f(-2)\) e. \(f\left(c^{2}+4\right)\) f. \(f(2+h)\)
Short Answer
Expert verified
a. \(f(2) = \frac{1}{2}\) \nb. \(f(-2) = \frac{1}{2}\) \nc. \(f\left(-\frac{3}{5}\right) = \frac{5}{3}\) \nd. \(f(2) + f(-2) = 1\) \ne. \(f(c^2 + 4) = \frac{1}{c^2+4}\) \nf. \(f(2 + h) = \frac{1}{|2 + h|}\)
Step by step solution
01
Evaluating f(2)
Substitute \(x = 2\) into the function \(f(x)\): \n\(f(2)=\frac{1}{|2|} = \frac{1}{2}\)
02
Evaluating f(-2)
Substitute \(x = -2\) into the function \(f(x)\):\n \(f(-2)=\frac{1}{|-2|} = \frac{1}{2}\)
03
Evaluating f(-3/5)
Substitute \(x = -3/5\) into the function \(f(x)\):\n \(f(-3/5)=\frac{1}{|-3/5|} = \frac{1}{3/5} = \frac{5}{3}\)
04
Evaluating f(2)+f(-2)
Calculate \(f(2) + f(-2)\) using previously found values: \n\(f(2) + f(-2) = \frac{1}{2} + \frac{1}{2}= 1\)
05
Evaluating f(c^2+4)
Substitute \(x = c^2+4\) into the function \(f(x)\):\n \(f(c^2 + 4) = \frac{1}{|c^2+4|}\). Since \(c^2+4\) is always positive (as any real number squared is always positive or zero, and adding 4 will always keep it positive), this is equivalent to \(\frac{1}{c^2+4}\)
06
Evaluating f(2+h)
Substitute \(x = 2 + h\) into the function \(f(x)\):\n \(f(2+h)=\frac{1}{|2+h|}\). This is the final answer for this case since \(h\) can take different values and we cannot simplify the expression anymore
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Function
The absolute value function is a mathematical concept that gives the magnitude of a number irrespective of its sign. It is denoted by two vertical bars around the number or variable, like this:
- The absolute value of a positive number is the number itself.
- For example, \(|3| = 3\).
- The absolute value of a negative number is its positive counterpart.
- For example, \(|-3| = 3\).
Rational Functions
Rational functions are formed as the ratio of two polynomial functions. They are represented in the form \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)eq 0\). In this exercise, the function is a special kind of rational function:
- \(f(x) = \frac{1}{|x|}\), where \(P(x) = 1\) and \(Q(x) = |x|\).
Substitution Method
The substitution method is a useful technique in evaluating functions, solving equations, or simplifying mathematical expressions. It involves replacing a variable with a specific value or expression to find a particular result. This is how to effectively apply the substitution method:
- Identify the variable within the function that needs evaluation.
- Carefully substitute the specific value or algebraic expression into the variable’s place in the function.
- Proceed with any further simplification if necessary.
Piecewise Functions
Piecewise functions are defined by different expressions depending on the input value regions; that means, they "piece" together different formulas to handle distinct parts of a domain. Although the exercise might not directly entail a piecewise function, understanding them complements function evaluation.
- Piecewise functions are expressed as multiple sub-functions, each applying to a slice of the domain.
- They can be written in the format \(f(x) = \begin{cases} expression_1, & \text{if } condition_1 \ expression_2, & \text{if } condition_2 \end{cases} \).