Problem 64 - Brief Applied Calculus

Question

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=3 x^{2} $$

Step-by-Step Solution

Answer

The simplified expression is \(6x + 3h\).

Step 1: Write the Expression

Start by writing the difference quotient expression for the function \[ f(x) = 3x^2 \] The expression is: \[ \frac{f(x+h)-f(x)}{h} \] This represents the average rate of change of the function over the interval \(h\).

Step 2: Substitute with Function

Substitute \(f(x+h)\) and \(f(x)\) in the expression. Given \[ f(x) = 3x^2 \]\(f(x+h)\) is calculated by substituting \(x+h\) into the function: \[ f(x+h) = 3(x+h)^2 \] Now substitute in the difference quotient: \[ \frac{3(x+h)^2 - 3x^2}{h} \]

Step 3: Expand the Squared Term

Expand the squared term in the numerator: \[ (x+h)^2 = x^2 + 2xh + h^2 \] So, \[ 3(x+h)^2 = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2 \]

Step 4: Simplify the Numerator

Substitute the expanded term back into the expression and simplify the numerator: \[ \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} \] Combine like terms: the \(3x^2\) terms cancel out, leaving: \[ \frac{6xh + 3h^2}{h} \]

Step 5: Factor and Cancel Terms

Factor out \(h\) from the numerator: \[ \frac{h(6x + 3h)}{h} \] As long as \(heq 0\), cancel \(h\) from the numerator and the denominator: \[ 6x + 3h \] This is the simplified form of the difference quotient for the given function.

Key Concepts

Average Rate of Change

The average rate of change is a concept that's quite similar to finding the slope between two points on a graph. It tells us how much the function is changing on average over a specific interval. In essence, it's how much the function's value differs as we transition between these two points. For a function defined as \( f(x) \), the average rate of change from \( x \) to \( x+h \) is given by the difference quotient:

• \( \frac{f(x+h) - f(x)}{h} \)

This formula essentially computes the difference in function values at two points and divides it by the change in the \( x \)-coordinate, which is \( h \). It's important because it gives us a tangible measure of how a function behaves over an interval, and it is the foundation for the derivative in calculus. When \( h \) approaches zero, the average rate of change becomes the instantaneous rate of change, representing the slope at a single point.

Polynomial Functions

Polynomial functions are mathematical expressions involving variables raised to integer powers and coefficients. They look like this:

• \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \)

Where \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are constants, and \( n \) is a non-negative integer. In our exercise, the polynomial function given is \( f(x) = 3x^2 \), which is a simple quadratic polynomial.
Polynomial functions are versatile and appear frequently in various calculations, modeling phenomena ranging from simple shapes to complex curves. They can be added, subtracted, multiplied, and divided, much like numbers, which makes them highly useful in algebraic manipulations. Understanding how to work with polynomials is crucial as it allows you to explore their properties like factorization, roots, and behavior at different points.

Simplifying Expressions

Simplifying expressions is a fundamental skill in algebra that involves rewriting expressions in their simplest form. This means eliminating complex fractions, reducing terms, and factoring where possible to make the expression easier to work with. In the context of a difference quotient, the goal is to write the expression in a form that is not only simpler but also more understandable.
The exercise problem required expanding and simplifying the expression:

• Expand: \( (x+h)^2 = x^2 + 2xh + h^2 \)
• Substitute and simplify: \( 3(x+h)^2 = 3x^2 + 6xh + 3h^2 \)
• Cancel out terms: Combine like terms and factor where applicable

The final step involves canceling terms, especially where \( h \) appears both in the numerator and the denominator. In the end, you get a streamlined expression \( 6x + 3h \), which is much easier to work with. The ability to simplify expressions not only makes solving problems more straightforward but also uncovers the underlying structure of the algebraic operation.