Problem 65 - Brief Applied Calculus

Question

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=2 x^{2}-5 x+1 $$

Step-by-Step Solution

Answer

\( 4x + 2h - 5 \)

Step 1: Understanding the Problem

We need to find the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x) = 2x^2 - 5x + 1 \). This expression is used in calculus to find the derivative of a function.

Step 2: Substitute \( x + h \) into the Function

First, calculate \( f(x + h) \) by substituting \( x + h \) into the function \( f(x) = 2x^2 - 5x + 1 \):\[ f(x + h) = 2(x + h)^2 - 5(x + h) + 1 \]

Step 3: Expand \( f(x + h) \)

Expand the expression to simplify:- \( (x + h)^2 = x^2 + 2xh + h^2 \)- Therefore, \( 2(x + h)^2 = 2x^2 + 4xh + 2h^2 \)Now substitute back:\[ f(x + h) = 2x^2 + 4xh + 2h^2 - 5x - 5h + 1 \]

Step 4: Subtract \( f(x) \) from \( f(x+h) \)

Now, compute \( f(x+h) - f(x) \): \[ f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 5x - 5h + 1) - (2x^2 - 5x + 1) \]Simplify by canceling similar terms:\[ f(x+h) - f(x) = 4xh + 2h^2 - 5h \]

Step 5: Divide by \( h \) and Simplify

Divide the result by \( h \) to obtain the difference quotient:\[ \frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 - 5h}{h} \]Factor \( h \) from the numerator:\[ = \frac{h(4x + 2h - 5)}{h} \]Simplify by canceling \( h \) in the numerator and denominator:\[ = 4x + 2h - 5 \]

Step 6: Final Solution

The simplified expression for the difference quotient \( \frac{f(x+h)-f(x)}{h} \) is \( 4x + 2h - 5 \).

Key Concepts

Difference Quotient

The difference quotient is an essential concept in calculus, especially when it comes to understanding the derivative of a function. It serves as a formula for the derivative, providing a way to see how a function changes as you make small adjustments to its input value. This is critical in analyzing how fast something changes—in other words, its rate of change.

The general formula for the difference quotient is:

• \( \frac{f(x+h) - f(x)}{h} \)

Here, \( f(x) \) represents the function, and \( h \) is a small change in the input value \( x \). To make use of the difference quotient, you substitute \( x + h \) into the function, calculate \( f(x+h) \), subtract \( f(x) \), and divide the result by \( h \).

This approach is like taking an average rate of change between two points that are very close together. As \( h \) approaches zero, the difference quotient approaches the exact, instantaneous rate of change at \( x \), which is the function’s derivative.

Function Derivative

The derivative of a function is a fundamental concept in calculus that measures how a function changes as its input changes. It is often denoted by \( f'(x) \) or \( \frac{dy}{dx} \). The derivative provides the slope of the tangent line to the graph of the function at any given point.

One of the key uses of the difference quotient is to derive the function's derivative. By simplifying the difference quotient and then letting \( h \) approach zero, you get the exact value of the derivative. This process is called "taking the limit."

• \( f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

For polynomial functions like the one given, finding the derivative usually results in terms of decreasing powers of \( x \). Once you have the derivative, you can solve a wide range of real-world problems, such as finding maxima and minima of functions, which helps in optimization problems.

Polynomial Functions

Polynomial functions are algebraic expressions that include terms of variables powered to whole numbers. They are one of the most common types of functions in calculus and algebra.

The general form of a polynomial function is:

• \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \)

where \( a_n, a_{n-1}, \ldots, a_0 \) are constants, and \( n \) is a non-negative integer.

Polynomial functions are continuous and smooth, which makes them an excellent candidate for derivative calculations. In the given exercise, \( f(x) = 2x^2 - 5x + 1 \), we see a quadratic function, which is a polynomial of degree two. Calculating the difference quotient for such functions involves substitution, expansion, simplification, and factorization—all standard procedures that reflect the structured nature of polynomials. Furthermore, while finding the derivative of polynomial functions, you simplify the quadratic or cubic expressions, reducing the power by one for each derivative term.