Problem 36 - Algebra and Trigonometry

Question

Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0\). $$ f(x)=x^{3} $$

Step-by-Step Solution

Answer

1. \(f(a) = a^3\); 2. \(f(a+h) = a^3 + 3a^2h + 3ah^2 + h^3\); 3. \(\frac{f(a+h)-f(a)}{h} = 3a^2 + 3ah + h^2\).

Step 1: Find f(a)

To find \(f(a)\), we substitute \(a\) into the function \(f(x) = x^3\). This means we compute \(a^3\). Thus, \(f(a) = a^3\).

Step 2: Find f(a+h)

Next, we substitute \(a+h\) into the function \(f(x) = x^3\). This gives us \((a+h)^3\). Using the binomial theorem, we expand this to \(a^3 + 3a^2h + 3ah^2 + h^3\). So, \(f(a+h) = (a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3\).

Step 3: Compute the Difference Quotient

Now, we calculate \(\frac{f(a+h) - f(a)}{h}\). Substitute \(f(a+h)\) and \(f(a)\) found from the previous steps: \(\frac{a^3 + 3a^2h + 3ah^2 + h^3 - a^3}{h}\). Simplifying, \(a^3\) cancels out, leaving \(\frac{3a^2h + 3ah^2 + h^3}{h}\). Divide each term by \(h\) to get \(3a^2 + 3ah + h^2\).

Key Concepts

Polynomial Function

A polynomial function is an algebraic expression consisting of variables raised to whole number powers, coefficients, and constants, combined using addition, subtraction, and multiplication. In mathematics, polynomial functions are essential because they are used to approximate more complex functions and model various real-world situations. The most basic form of a polynomial function is a monomial (a single term), while more complex forms can have multiple terms (e.g., binomials or trinomials).

For example, the function given in the exercise, \( f(x) = x^3 \), is a simple polynomial of degree 3, known as a cubic polynomial. It is comprised only of one term: \( x^3 \), with no additional coefficients or constants. Understanding polynomial functions allows you to work with expressions and equations in algebra, calculus, and beyond.

Binomial Theorem

The binomial theorem is a powerful tool in algebra that provides a formula for expanding expressions raised to a power, specifically binomials (expressions with two terms). It can be particularly useful when dealing with polynomial functions where expanding terms is necessary. The theorem tells us how to expand \((a+b)^n\) in terms of individual powers of \(a\) and \(b\).

The binomial theorem formula is given as:

• \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)

Where \( \binom{n}{k} \) represents a binomial coefficient, equivalent to choosing \(k\) elements from \(n\) without regard to order. In this exercise, to find \( f(a+h) = (a+h)^3 \), we apply the theorem to expand it into terms: \( a^3 + 3a^2h + 3ah^2 + h^3 \). This expansion helps in calculating the difference quotient by expressing the polynomial in simpler terms.

Derivatives

Derivatives are a fundamental concept in calculus, representing how a function changes when its input changes. The derivative of a function at a certain point measures the function's rate of change, often interpreted as the slope of the tangent line at that point. The difference quotient is a central idea in finding derivatives, especially when calculating them from first principles.

In this exercise, we work with the difference quotient: \( \frac{f(a+h) - f(a)}{h} \). This expression estimates the derivative of the function \( f(x) \) at the point \( a \) as \( h \) approaches zero. The difference quotient simplifies to \( 3a^2 + 3ah + h^2 \), which approaches the derivative of \( f(x) = x^3 \), ultimately leading to \( f'(x) = 3x^2 \) as \( h \to 0 \). Understanding derivatives helps analyze rates of change, optimize functions, and solve problems in physics and engineering.

Algebraic Expression

An algebraic expression is a combination of numbers, variables, operators, and sometimes constants related through various mathematical operations. These expressions form the basis of algebraic equations and are used extensively in mathematics to represent relationships or solve problems.

In our exercise, expressions such as \( f(a) = a^3 \) and \( f(a+h) = a^3 + 3a^2h + 3ah^2 + h^3 \) are algebraic expressions. They demonstrate how different operations on symbols can transform polynomial expressions. By substituting a value, algebraic expressions become numeric expressions and can be evaluated. Understanding how to manipulate and simplify algebraic expressions is crucial for solving equations, optimizing functions, and conducting algebraic proofs.