Problem 70 - Brief Applied Calculus

Question

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ \begin{array}{l} f(x)=x^{4} \\ \text { [Hint: Use }(x+h)^{4}=x^{4}+4 x^{3} h+6 x^{2} h^{2}+ \\ \left.4 x h^{3}+h^{4} .\right] \end{array} $$

Step-by-Step Solution

Answer

The simplified expression is \( 4x^3 + 6x^2h + 4xh^2 + h^3 \).

Step 1: Substitute and Expand

Substitute \( x + h \) for \( x \) in \( f(x) \). Therefore, \[ f(x+h) = (x+h)^4. \]Using the hint, expand this to get: \[ f(x+h) = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4. \]

Step 2: Compute Difference

Calculate the difference \( f(x+h) - f(x) \): \[ f(x+h) - f(x) = (x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - x^4. \]Simplifying, the \( x^4 \) terms cancel out, leaving: \[ 4x^3h + 6x^2h^2 + 4xh^3 + h^4. \]

Step 3: Simplify the Expression

Divide the expression from Step 2 by \( h \): \[ \frac{f(x+h) - f(x)}{h} = \frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h}. \]Cancel out \( h \) from the numerator and denominator: \[ = 4x^3 + 6x^2h + 4xh^2 + h^3. \]

Step 4: Final Simplified Expression

The simplified expression for \( \frac{f(x+h)-f(x)}{h} \) is: \[ 4x^3 + 6x^2h + 4xh^2 + h^3. \]

Key Concepts

Function Evaluation

Evaluating a function involves substituting a specific value or expression in place of the variable in the function. Let's take a closer look at our exercise: the function is given by \( f(x) = x^4 \). To evaluate \( f(x+h) \), substitute \( x+h \) for \( x \) across the function.

Why substitute? Substitution helps us see how the function behaves for inputs slightly different from \( x \). This is crucial when working with operations like the difference quotient, which helps to understand the concept of a derivative in calculus.

• Start with the original function: \( f(x) = x^4 \).
• Substitute \( x+h \) into the function: \( f(x+h) = (x+h)^4 \).
• Expand the substituted function: \( x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \).

This process, while simple, lays the foundation for understanding changes in functions.

Binomial Expansion

Binomial expansion is a powerful algebraic tool that allows us to expand expressions that are raised to a power, such as \((x+h)^4\). Using the binomial theorem, an expression of the form \((a+b)^n\) can be expanded.

In our exercise, we deal with the expansion of \((x+h)^4\). Using binomial expansion:

• The formula becomes: \( x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \).
• The expansion uses coefficients from Pascal's triangle: \( 1, 4, 6, 4, 1 \).

These coefficients multiply each term incrementally involving powers of \( x \) and \( h \). This results in terms that reflect combined effects of these variables on the function.

Understanding binomial expansion simplifies the process of handling powers of binomial expressions, making calculations for the difference quotient manageable and precise.

Simplification of Expressions

Simplifying expressions is a fundamental skill in algebra that involves reducing an expression to its simplest form. In our problem, simplification plays a key role.

Start with the difference we obtained: \( f(x+h) - f(x) = 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \). Division by \( h \) (provided \( h eq 0 \)) cancels out the common \( h \) in all terms.

• The expression \( \frac{f(x+h) - f(x)}{h} \) becomes \( 4x^3 + 6x^2h + 4xh^2 + h^3 \).
• Every term has been divided by \( h \), simplifying the overall expression.

This step highlights the process of extracting the core structure of the changing relationship in the original function.

Through simplification, understanding the behavior of functions near a point becomes more intuitive, paving the way to grasp concepts like derivatives more effectively.