Problem 70 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Substitute Into the Difference Quotient

We need to evaluate and simplify the expression \( \frac{f(x+h)-f(x)}{h} \). Substituting \( f(x) = x^4 \), we get:\[ \frac{(x+h)^4 - x^4}{h} \]

Step 2: Expand \((x+h)^4\) Using Binomial Expansion

Using the hint, write the expansion of \((x+h)^4\) as:\[ x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \]

Step 3: Calculate the Difference \(f(x+h) - f(x)\)

Subtract \(x^4\) from the expanded \((x+h)^4\) expression:\[ (x+h)^4 - x^4 = 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \]

Step 4: Simplify the Difference Quotient

Substitute the difference back into the quotient:\[ \frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h} \]Factor \(h\) out of each term to simplify:\[ \frac{h(4x^3 + 6x^2h + 4xh^2 + h^3)}{h} \]

Step 5: Cancel Out \(h\)

Cancel out \(h\) from the numerator and denominator:\[ 4x^3 + 6x^2h + 4xh^2 + h^3 \]

Step 6: Final Simplification

The expression simplifies to:\[ 4x^3 + 6x^2h + 4xh^2 + h^3 \]As \(h\) approaches 0, you can observe the primary term is \(4x^3\).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus that helps us understand how a function changes as its input changes. It forms the basis of the derivative calculation, which describes the rate of change of a function. In simple terms, the difference quotient is a fraction that measures the average rate of change of a function over a small interval. For a function \( f(x) \), it's given by:

• \( \frac{f(x+h)-f(x)}{h} \)

Here, \( h \) is a small change in the input \( x \). The purpose of this expression is to evaluate how the function \( f \) changes when \( x \) changes by a small amount \( h \), assuming \( h eq 0 \). The smaller \( h \) gets, the closer the difference quotient gets to the derivative of \( f \) at \( x \).
By simplifying this expression, as we did in the steps, we observe the behavior of the function in a precise manner. This concept becomes crucial when differentiating functions, especially when approaching the notion of instantaneous rate of change.

Binomial Expansion

Binomial expansion is a powerful algebraic tool used to expand expressions that are raised to a power, specifically involving polynomials. It's a way to expand \((x + y)^n\) into a sum involving terms of the form \(x^a y^b\). Each term in the expansion corresponds to a specific combination of these two variables raised to various powers.For the purpose of the exercise, we use the expansion for \((x+h)^4\). The binomial expansion yields:

• \( x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \)

Each coefficient in the expansion, like 4 in \( 4x^3h \), corresponds to a binomial coefficient, capturing the number of ways certain combinations occur. This expansion is essential, as it allows us to see precisely how each power of \( h \) contributes to the rate of change in the expression, preparing for simplifying the difference quotient.

Differentiation

Differentiation is the process in calculus of finding the derivative of a function. The derivative gives us a precise measure of how a function changes at any point. Essentially, it tells us the slope of the tangent line at any given point on the function's graph. The core idea is to take a function \( f(x) \) and figure out an expression that describes its instantaneous rate of change.After simplifying the difference quotient for our function, the result was \(4x^3 + 6x^2h + 4xh^2 + h^3\). When \( h \) approaches zero, the term \( 4x^3 \) remains, signifying the derivative \( f'(x) = 4x^3 \). The other terms fade as \( h \) becomes negligible. Thus, differentiation takes the notion of the difference quotient to the new level of providing exact rate measures, critical in real-world applications like physics, engineering, and economics.