Problem 86 - Algebra and Trigonometry Real Mathematics, Real People

Question

Expand the expression in the difference quotient and simplify. \(\frac{f(x+h)-f(x)}{h}, h \neq 0\) \(f(x)=x^{4}\)

Step-by-Step Solution

Answer

The expanded form and simplified difference quotient for the function \( f(x) = x^{4} \) is \( 4x^{3} + 6x^{2}h + 4xh^{2} + h^{3} \).

Step 1: Identify the Function

Here, the function \( f(x) = x^{4} \), apply the function to both \( x+h \) and \( x \). This means \( f(x+h) = (x+h)^{4} \) and \( f(x) = x^{4} \).

Step 2: Substitute into Difference Quotient

Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient. This results in \(\frac{(x+h)^{4} - x^{4}}{h}\).

Step 3: Expand and Simplify

Expand the expression \( (x+h)^{4} \) using the binomial theorem to get \( x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4} \) and simplify the difference quotient. This gives \(\frac{x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4} - x^{4}}{h}\). Note that \( x^{4} \) terms cancel out. This leaves us with \(\frac{4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4}}{h}\).

Step 4: Simplify Further by Cancelling Out h

Factor out h in the numerator and cancel with the denominator: \(4x^{3} + 6x^{2}h + 4xh^{2} + h^{3}\).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in algebra that allows us to expand expressions raised to a power, such as \((x+h)^{n}\), where n is a nonnegative integer. The theorem tells us that we can write this expression as the sum of terms of the form \({n\choose k}x^{n-k}h^{k}\), where \({n\choose k}\) denotes the binomial coefficient. These coefficients can be found in Pascal's triangle or calculated using the formula \({n\choose k} = \frac{n!}{k!(n-k)!}\).

For the exercise \((x+h)^{4}\), we use the binomial theorem to expand the expression into a series of terms: \((x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4}\)). Each term represents a possible combination of \((x\)) and \((h\)) where the total number of \((x\)) and \((h\)) adds up to 4. It’s through this theorem that simplifying complex polynomial expressions becomes manageable.

Application to Difference Quotient

In our specific problem, after expanding \((x+h)^4\) using the binomial theorem, we effectively applied the formula to find the terms needed to simplify the difference quotient expression. Simplification of such expressions is critical when determining limits in calculus, particularly when there is a variable in the denominator that could approach zero.

Polynomial Functions

Polynomial functions are expressions that involve sums of powers of variables, with coefficients that are usually real numbers. The general form of a polynomial function in one variable \((x\)) is \((a_nx^n + a_{n-1}x^{n-1} + ... + a_2x^2 + a_1x + a_0\)), where \((a_0, a_1, ..., a_n\)) are constants and \((n\)) is a nonnegative integer called the degree of the polynomial.

Polynomial functions are especially important because they can model a vast number of real-life situations and can be easily differentiated or integrated when solving calculus problems. For example, the polynomial function \((f(x) = x^{4}\)) in our exercise is a simple monomial function with a single term and represents a polynomial of degree 4.

Understanding the Role of h

In calculus, the variable \((h\)) often represents an infinitesimally small change in \((x\)), which is why it's in the denominator of the difference quotient. The idea is to explore the behavior of the polynomial function as \((h\)) approaches zero, helping us understand the rate at which the function changes - in other words, its derivative.

Algebraic Simplification

Algebraic simplification is the process of reducing expressions into their simplest form using algebraic rules and properties. This process is crucial in solving many mathematical problems efficiently. By factoring, combining like terms, and canceling out common factors, we can often transform a complex expression into one that is much more manageable.

For instance, in the step-by-step solution provided in the exercise, after expanding the expression using the binomial theorem, we then simplified the difference quotient by canceling out common terms. Specifically, we noticed that the \((x^{4}\)) terms in \((x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4} - x^{4}\)) cancel each other out. Afterwards, by factoring out the common factor \((h\)) from the remaining terms, we were able to further simplify by canceling it with the \((h\)) in the denominator.

Final Simplified Form

Through algebraic simplification, the seemingly complex difference quotient turned into \((4x^{3} + 6x^{2}h + 4xh^{2} + h^{3}\)), which is much simpler to work with and analyze further. Mastering this process not only saves time but also reduces the likelihood of errors in computation and provides clearer insight into the nature of the algebraic expression under scrutiny.