Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the function \(f\) at the indicated values \(f(-3), f(2), f(-a),-f(a), f(a+h)\). \(f(x)=2 x-5 \quad\)

Step-by-Step Solution

Answer

\(f(-3) = -11\), \(f(2) = -1\), \(f(-a) = -2a - 5\), \(-f(a) = -2a + 5\), \(f(a+h) = 2a + 2h - 5\).

Step 1: Evaluate f(-3)

To find \(f(-3)\), substitute \(-3\) for \(x\) in the function \(f(x) = 2x - 5\). Calculate as follows:\[ f(-3) = 2(-3) - 5 = -6 - 5 = -11 \] So, \(f(-3) = -11\).

Step 2: Evaluate f(2)

To find \(f(2)\), substitute \(2\) for \(x\) in the function \(f(x) = 2x - 5\). Calculate as follows:\[ f(2) = 2(2) - 5 = 4 - 5 = -1 \] So, \(f(2) = -1\).

Step 3: Evaluate f(-a)

To find \(f(-a)\), substitute \(-a\) for \(x\) in the function \(f(x) = 2x - 5\). Calculate as follows:\[ f(-a) = 2(-a) - 5 = -2a - 5 \] So, \(f(-a) = -2a - 5\).

Step 4: Evaluate -f(a)

To find \(-f(a)\), first calculate \(f(a)\) by substituting \(a\) for \(x\):\[ f(a) = 2a - 5 \]Then, multiply by \(-1\):\[-f(a) = -(2a - 5) = -2a + 5 \] So, \(-f(a) = -2a + 5\).

Step 5: Evaluate f(a+h)

To find \(f(a+h)\), substitute \(a + h\) for \(x\) in the function \(f(x) = 2x - 5\). Calculate as follows:\[ f(a+h) = 2(a+h) - 5 = 2a + 2h - 5 \] So, \(f(a+h) = 2a + 2h - 5\).

Key Concepts

Substitution

Substitution is like solving a puzzle, where you replace each variable in an expression with given values. Imagine you have a function, like a magic box, that takes inputs and gives outputs. For function evaluation, you need to substitute the variables inside the box with specific numbers or expressions. This helps you find out what the box will give as an output for those particular inputs.
For example, in our exercise, you have the function \(f(x) = 2x - 5\). When you're asked to find \(f(-3)\), you're substituting \(-3\) in place of \(x\). Essentially, you replace every \(x\) in the function with \(-3\):

• Start with \(2(-3) - 5\).
• Multiply 2 by -3 to get -6.
• Subtract 5, resulting in -11.

So, substituting \(-3\) into the function gives \(-11\) as the output. Substitution is a powerful tool in algebra that allows you to see how a function behaves for different inputs.

Algebraic Expressions

Algebraic expressions are like a special language for talking about mathematics. They're made up of numbers, variables, and operations like addition and multiplication. In the given exercise, the function \(f(x) = 2x - 5\) is an algebraic expression. It's an expression because it combines different elements:

• "2x" - this part involves multiplication of 2 and the variable \(x\).
• "-5" - this is a constant term, simply an integer added to the expression.

Together, they tell you the rule for getting from an input (\(x\)) to an output (\(f(x)\)). When you substitute values for \(x\), you're actually doing arithmetic within the boundaries set by these algebraic symbols. This exercise also has variations like \(f(a + h)\), which show how algebraic expressions can become more complex when you start adding more variables or terms into them, but the idea remains the same.
Working with these expressions helps you see patterns and relationships between numbers, turning complex problems into simpler ones.

Linear Functions

Linear functions are a simple, yet powerful concept in mathematics. They are called "linear" because their graph is a straight line. The most common form of a linear function is \(f(x) = mx + b\), where \(m\) represents the slope and \(b\) is the y-intercept. In our exercise, \(f(x) = 2x - 5\) is a linear function.
Let's break it down:

• "2x" indicates the slope of the line, which tells how steep the line is. Here, the slope \(m\) is 2, meaning for every unit increase in \(x\), \(f(x)\) increases by 2 units.
• "-5" is the y-intercept. This means the line crosses the y-axis at -5. It's the value of \(f(x)\) when \(x\) is 0.

Linear functions are crucial because they simplify the analysis of relationships between variables. In problems like our exercise, evaluating the function at different points shows how the linear relationship behaves for different \(x\)-values, demonstrating both the simplicity and predictability of linear functions.