Problem 34 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the function \(f\) at the indicated values \(f(-3), f(2), f(-a),-f(a), f(a+h)\). Given the function \(k(t)=2 t-1:\) (a) Evaluate \(k(2)\). (b) Solve \(k(t)=7\).

Step-by-Step Solution

Answer

(a) 3, (b) t = 4

Step 1: Substitute 2 into the function

We are given the function \(k(t) = 2t - 1\). To evaluate \(k(2)\), substitute 2 for \(t\) in the function. This gives us:\[k(2) = 2(2) - 1.\]

Step 2: Simplify the expression

Simplify the expression from Step 1:\[k(2) = 4 - 1 = 3.\] Thus, \(k(2) = 3\).

Step 3: Set the function equal to 7

To solve the equation \(k(t) = 7\), we substitute the expression for \(k(t)\) into the equation to get:\[2t - 1 = 7.\]

Step 4: Isolate the variable t

Add 1 to both sides of the equation to isolate the term with \(t\):\[2t - 1 + 1 = 7 + 1,\] which simplifies to \[2t = 8.\]

Step 5: Solve for t

Divide both sides of the equation by 2 to solve for \(t\):\[t = \frac{8}{2},\] which simplifies to \(t = 4\).

Key Concepts

Linear Function

A linear function is a type of mathematical function which, when graphed, gives a straight line. It can be expressed in the form \( f(x) = mx + b \), where:

• \( m \) is the slope of the line, which indicates how steep the line is.
• \( b \) is the y-intercept, where the line crosses the y-axis.

For example, given the function \( k(t) = 2t - 1 \), the slope \( m \) is 2, and the y-intercept \( b \) is -1. This means for every increase of 1 in \( t \), the function value increases by 2. Linear functions are straightforward because their rate of change is constant, they make the evaluation, graphing, and solving of equations a simpler process.

Substitution Method

The substitution method is a useful technique for evaluating functions or solving equations. It involves replacing a variable with a given value or another expression. This method is straightforward when working with linear functions.
To evaluate \( k(2) \) in the function \( k(t) = 2t - 1 \), replace \( t \) with 2. Thus, you substitute 2 into the function and calculate:

• \( k(2) = 2(2) - 1 \)
• Simplify: \( k(2) = 4 - 1 = 3 \)

By substituting a specific value into the function, you directly evaluate the function at that point. This method is also helpful when you need to find unknown values because it allows you to simplify and manipulate equations easily.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. For linear functions, this is usually a simple process, as linear equations typically have one solution.
To solve \( k(t) = 7 \) using the function \( k(t) = 2t - 1 \), follow these steps:

• Write the equation: \( 2t - 1 = 7 \)
• Use algebraic methods, especially focusing on isolating the variable.

In linear equations, the operations required are generally straightforward, comprised mainly of addition, subtraction, multiplication, and division. This method efficiently finds the value of the variable that satisfies the equation.

Variable Isolation

Variable isolation is the process of rearranging an equation to express one variable explicitly. This involves using basic arithmetic operations to simplify the equation step by step.
Let's isolate \( t \) in the equation \( 2t - 1 = 7 \):

• Add 1 to both sides to remove the constant term on the side of the variable: \( 2t = 8 \)
• Divide both sides by 2 to solve for \( t \): \( t = \frac{8}{2} = 4 \)

By isolating \( t \), you pivot it as the subject of the equation, and this operation allows for a clear and systematic solution approach. This is crucial for solving linear equations as it simplifies the equation to its simplest terms.