Problem 4 - Calculus and its applications

Question

(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=-4 x^{2} $$

Step-by-Step Solution

Answer

Simplified difference quotient: \(-8x - 4h\). Table values: -48, -44, -40.4, -40.04.

Step 1: Understand the Problem

We have a function \( f(x) = -4x^2 \). We need to find the simplified form of the difference quotient \( \frac{f(x+h)-f(x)}{h} \) and then complete the table.

Step 2: Write the Expression for \( f(x+h) \)

Substitute \( x+h \) into the function: \[ f(x+h) = -4(x+h)^2 \]. Expanding this gives: \[ f(x+h) = -4(x^2 + 2xh + h^2) = -4x^2 - 8xh - 4h^2 \].

Step 3: Calculate the Difference Quotient

The difference quotient is \[ \frac{f(x+h) - f(x)}{h} = \frac{-4x^2 - 8xh - 4h^2 + 4x^2}{h} \]. Simplify the numerator: \[ = \frac{-8xh - 4h^2}{h} \]. Factor out \( h \) to get: \[ \frac{h(-8x - 4h)}{h} = -8x - 4h \].

Step 4: Simplified Difference Quotient

The simplified form of the difference quotient is: \[ -8x - 4h \].

Step 5: Use the Difference Quotient to Fill the Table

For each row in the table, substitute \( x = 5 \) and the corresponding \( h \) value into the simplified quotient, \(-8x - 4h\).- For \(h = 2\): \(-8(5) - 4(2) = -40 - 8 = -48 \).- For \(h = 1\): \(-8(5) - 4(1) = -40 - 4 = -44 \).- For \(h = 0.1\): \(-8(5) - 4(0.1) = -40 - 0.4 = -40.4 \).- For \(h = 0.01\): \(-8(5) - 4(0.01) = -40 - 0.04 = -40.04 \).

Key Concepts

Simplified Form

The simplified form of the difference quotient can often seem intimidating at first. It's simply a way to measure how a function changes as its input changes slightly. For our function, \( f(x) = -4x^2 \), the difference quotient is initially expressed as \( \frac{f(x+h) - f(x)}{h} \). The idea is to evaluate the change in the function's value as \( x \) increases by a small amount \( h \).

To simplify, we first substitute \( x+h \) into \( f \) to get \( f(x+h) = -4(x+h)^2 \), which expands to a more extensive polynomial form \( -4x^2 - 8xh - 4h^2 \). This expression represents the new value when \( x+h \) is used instead of \( x \).

By substituting back into the difference quotient, we get:

• \( \frac{-4x^2 - 8xh - 4h^2 + 4x^2}{h} = \frac{-8xh - 4h^2}{h} \)
• Cancel out \( h \) from numerator and denominator to simplify.
• Resulting in \( -8x - 4h \) after factoring and canceling \( h \).

The simplified form \( -8x - 4h \) makes it easy to calculate how quickly \( f(x) \) is changing.

Function Evaluation

Function evaluation is a crucial step in finding the difference quotient. It ensures that we understand how to substitute into the function properly. For the given function, \( f(x) = -4x^2 \), we first need to find \( f(x+h) \).

Here's a step-by-step guide:

• Start by replacing \( x \) with \( x+h \) in the function, which gives: \( -4(x+h)^2 \).
• To expand, remember \( (x+h)^2 = x^2 + 2xh + h^2 \), leading to: \(-4x^2 - 8xh - 4h^2 \).
• This result represents \( f(x+h) \), the value of the function when evaluating at \( x+h \).

By knowing how to evaluate the function, you are ready to find the difference with \( f(x) \) and simplify the resulting expression.

Table Completion

Completing a table using the difference quotient helps visualize how a function's output changes as \( h \) changes. We use the simplified form \( -8x - 4h \) with \( x = 5 \) and various values of \( h \).

Let's see how this works in practice:

• For \( h = 2 \): Substitute into the simplified formula: \(-8(5) - 4(2) = -40 - 8 = -48\).
• For \( h = 1 \): Substitute into the formula: \(-8(5) - 4(1) = -40 - 4 = -44\).
• For \( h = 0.1 \): \(-8(5) - 4(0.1) = -40 - 0.4 = -40.4\).
• For \( h = 0.01 \): \(-8(5) - 4(0.01) = -40 - 0.04 = -40.04\).

Each calculation shows how the function's change becomes smaller as \( h \) approaches zero, revealing an insight into the function's slope at \( x = 5 \). This exercise in table completion is a visual representation of understanding the difference quotient.