Problem 84 - Calculus of a Single Variable

Question

Finding a Limit In Exercises \(83-88\) , find \(\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\) $$ f(x)=-6 x+3 $$

Step-by-Step Solution

Answer

The limit of the given expression as \(\Delta x\) approaches 0 is -6.

Step 1: Evaluate \(f(x+ \Delta x)\)

Substitute \(x+\Delta x\), for every instance of \(x\) in \(f(x) = -6x+3\). This gives \(f(x + \Delta x) = -6(x + \Delta x) + 3 = -6x - 6 \Delta x + 3\).

Step 2: Plug values into the limit expression

Substitute \(f(x + \Delta x) = -6x - 6\Delta x + 3\) and \(f(x) = -6x + 3\) into the expression for the limit. This gives \(\lim _{\Delta x \rightarrow 0} \frac{-6x - 6 \Delta x + 3 - (-6x + 3)}{\Delta x} = \lim _{\Delta x \rightarrow 0} \frac{-6 \Delta x}{\Delta x}\).

Step 3: Simplify the expression

The \(\Delta x\) in the numerator and denominator cancel out, so the expression simplifies to \(\lim _{\Delta x \rightarrow 0} -6.\)

Step 4: Evaluate the resulting limit

The resulting expression does not depend on \(\Delta x\), so the limit as \(\Delta x\) approaches 0 is -6.

Key Concepts

Limit

In calculus, a limit helps us understand what happens to a function as we get really close to a specific point. It's like trying to see what value a function approaches. Here, we're interested in what happens when \( \Delta x \) gets very close to zero.
We use notation like \( \lim_{\Delta x \to 0} \) to express this idea. It's about examining the behavior of a function and predicting its output without actually reaching the point. Limits are foundational in calculus because they help define more advanced concepts, such as derivatives.

Difference Quotient

The difference quotient is a way to measure how a function's output changes as its input changes. It's essentially a fraction that shows the average rate of change of the function over a small interval. When we compute the difference quotient, we use:

• \( f(x + \Delta x) \) which represents a small change in \( x \).
• Subtract \( f(x) \), the original function value.
• Divide by \( \Delta x \), the small change in \( x \).

The formula looks like \( \frac{f(x + \Delta x) - f(x)}{\Delta x} \), and as \( \Delta x \) approaches zero, this helps us find the derivative, a precise measure of the rate of change.

Algebraic Simplification

Algebraic simplification involves reducing expressions to their simplest form to make calculations easier. In this exercise, we substitute and simplify:

• First, plug in \( x + \Delta x \) into the function to find \( f(x + \Delta x) = -6(x + \Delta x) + 3 \).
• Simplify to get \( -6x - 6 \Delta x + 3 \).
• Subtract \( f(x) = -6x + 3 \) from this expression.

The result, \( \frac{-6\Delta x}{\Delta x} \), can be simplified by canceling \( \Delta x \), showing the power of algebra in solving calculus problems smoothly and accurately.

Rate of Change

The rate of change tells us how one variable changes in relation to another. It's similar to speed, describing how quickly or slowly something is changing. In the context of functions, it helps us understand how the output of a function changes as the input changes.
When we use the limit of the difference quotient, we're finding the instantaneous rate of change, also known as the derivative. For this exercise, simplifying and calculating the derivative gives us \(-6\). This value tells us the function is decreasing at a constant rate, meaning for every 1 unit increase in \( x \), the function's output decreases by 6.