Problem 26 - Calculus Single Variable

Question

Let \(y=f(x)\) where \(f(x)=m x+b\) for constants \(m \neq 0\) and b. Show that a change in the value of \(x\) from \(x_{0}\) to \(x_{0}+\Delta x\) results in a change \(\Delta y\) in the value of \(f(x)\) that does not depend on the initial value \(x_{0}\). In other words, the increment \(\Delta y=f\left(x_{0}+\Delta x\right)-f\left(x_{0}\right)\) depends on the increment \(\Delta x\) but not on the value of \(x_{0}\).

Step-by-Step Solution

Answer

The change \(\Delta y\) is \(m \Delta x\), depending only on \(\Delta x\), not on \(x_0\).

Step 1: Write the Expression for Function Increment

We need to find the change in the function value when \(x\) changes from \(x_0\) to \(x_0 + \Delta x\). So, calculate \(f(x_0 + \Delta x)\) using the linear function formula \(f(x) = mx + b\).

Step 2: Evaluate f(x_0 + Δx)

Substitute \(x_0 + \Delta x\) into the function: \(f(x_0 + \Delta x) = m(x_0 + \Delta x) + b\). This simplifies to \(mx_0 + m \Delta x + b\).

Step 3: Calculate f(x_0)

Now, calculate the function value at \(x_0\): \(f(x_0) = mx_0 + b\).

Step 4: Compute the Change Δy

The change in \(y\), denoted \(\Delta y\), is given by \(\Delta y = f(x_0 + \Delta x) - f(x_0)\). Substitute the expressions from Step 2 and Step 3: \(\Delta y = (mx_0 + m \Delta x + b) - (mx_0 + b)\).

Step 5: Simplify the Expression for Δy

Simplify the expression: \(\Delta y = mx_0 + m \Delta x + b - mx_0 - b\). The terms \(mx_0\) and \(b\) cancel out, leaving \(\Delta y = m \Delta x\).

Step 6: Conclusion: Dependence of Δy on Δx

Notice that \(\Delta y = m \Delta x\) depends only on \(\Delta x\) and the constant \(m\), not on \(x_0\). Therefore, the change in \(y\) does not depend on the initial value \(x_0\).

Key Concepts

Differential Calculus

Differential Calculus is all about understanding how functions change. It's like asking, "What happens when I tweak a variable a little bit?" Imagine you have a car, and you want to know what happens to the speed if you press the gas pedal a little more. In mathematical terms, that tweak or change is called a "differential." For linear functions, this means examining how changes in the input, say from \(x_0\) to \(x_0 + \Delta x\), affect the output \(y\).

In our exercise, we explored how \(f(x) = mx + b\) changes. Even though Differential Calculus is often used with curves, it applies to linear equations too. It tells us how \(y\) changes with \(x\) in a very straightforward way: a change in \(x\) directly impacts \(y\) by the slope \(m\). This is the heart of what differential calculus deals with, at least for linear equations.

Slope-Intercept Form

The Slope-Intercept Form is a way of writing linear equations, making them easy to read and understand. The form looks like this: \(y = mx + b\). Here, \(m\) represents the slope of the line, while \(b\) is the y-intercept, where the line crosses the y-axis.

Let's break it down a bit more:

• **Slope** \(m\): This shows how steep the line is. If \(m\) is positive, the line goes upwards as \(x\) increases. If negative, it slopes downwards.
• **Y-intercept** \(b\): The point where the line hits the y-axis. It's the value of \(y\) when \(x = 0\).

In the context of our problem, this form helps us understand that any change in \(x\) is simply multiplied by \(m\), the slope. This knowledge is vital in finding out how \(y\) changes as we move along the line.

Incremental Change

Incremental Change is a key concept when dealing with functions. It's like taking baby steps: How does something change when we alter a small part of it? In our exercise, we are shifting \(x\) from \(x_0\) to \(x_0 + \Delta x\), where \(\Delta x\) represents the small increment.

For linear functions, this incremental change is straightforward and easy to calculate because the function doesn’t curve. The change in \(y\) is simply \(\Delta y = m \Delta x\), which we figured out by subtracting the initial function value from the new one. Thus, the change in \(y\) hinges entirely on \(\Delta x\) and the slope \(m\), not on \(x_0\).

In simple terms, for linear functions, each step we take in \(x\) results in a consistent step size in \(y\), defined by the slope \(m\). This is what makes linear equations beautifully predictable.

Algebraic Manipulation

Algebraic Manipulation is like solving puzzles. It involves using the rules of algebra to modify and simplify expressions, helping us to find solutions or understand relationships. In our exercise, we used algebraic manipulation to uncover how \(\Delta y\) changes.

We started by writing down the expressions for \(f(x_0 + \Delta x)\) and \(f(x_0)\). Through simple addition and subtraction:

• Finding \(f(x_0 + \Delta x) = mx_0 + m \Delta x + b\)
• Finding \(f(x_0) = mx_0 + b\)
• Subtracting the two to obtain \(\Delta y = (mx_0 + m \Delta x + b) - (mx_0 + b)\)

The algebraic manipulation allowed us to cancel out common terms like \(mx_0\) and \(b\), leading us to the simplified expression \(\Delta y = m \Delta x\).

This shows us how the math works underneath, revealing that \(\Delta y\) relies solely on \(\Delta x\) and \(m\) and not on the particular initial value of \(x\). Algebraic manipulation is a powerful tool to see these relationships clearly.