Problem 2 - Calculus

Question

Suppose \(f(x)=x^{2}\). Compute each ratio and set \(h=0\) (a) \(\frac{f(x+h)-f(x)}{h}\) (b) \(\frac{f(x+5 h)-f(x)}{5 h}\) (c) \(\frac{f(x+h)-f(x-h)}{2 h}\) (d) \(\frac{f(x+1)-f(x)}{h}\)

Step-by-Step Solution

Answer

(a) 2x, (b) 2x, (c) 2x, (d) Undefined due to h=0 in the denominator.

Step 1: Identify the Function f(x)

We start with knowing that the given function is \( f(x) = x^2 \). We will use this function to calculate the ratios in each part.

Step 2: Ratio Calculation for Part (a)

We need to compute \( \frac{f(x+h) - f(x)}{h} \).1. Compute \( f(x+h) = (x + h)^2 = x^2 + 2xh + h^2 \).2. Substitute in the ratio: \( \frac{(x^2 + 2xh + h^2) - x^2}{h} = \frac{2xh + h^2}{h} \).3. Simplify: \( \frac{2xh + h^2}{h} = 2x + h \).4. Set \( h = 0 \) to find the limit: \( 2x + 0 = 2x \).Final Answer for (a): \( 2x \).

Step 3: Ratio Calculation for Part (b)

We need to compute \( \frac{f(x+5h) - f(x)}{5h} \).1. Compute \( f(x+5h) = (x + 5h)^2 = x^2 + 10xh + 25h^2 \).2. Substitute in the ratio: \( \frac{(x^2 + 10xh + 25h^2) - x^2}{5h} = \frac{10xh + 25h^2}{5h} \).3. Simplify: \( \frac{10xh + 25h^2}{5h} = 2x + 5h \).4. Set \( h = 0 \) to find the limit: \( 2x + 0 = 2x \).Final Answer for (b): \( 2x \).

Step 4: Ratio Calculation for Part (c)

We need to compute \( \frac{f(x+h) - f(x-h)}{2h} \).1. Compute \( f(x+h) = (x + h)^2 = x^2 + 2xh + h^2 \).2. Compute \( f(x-h) = (x - h)^2 = x^2 - 2xh + h^2 \).3. Substitute in the ratio: \( \frac{(x^2 + 2xh + h^2) - (x^2 - 2xh + h^2)}{2h} = \frac{4xh}{2h} \).4. Simplify: \( \frac{4xh}{2h} = 2x \).5. Set \( h = 0 \) to find the limit: \( 2x \).Final Answer for (c): \( 2x \).

Step 5: Ratio Calculation for Part (d)

We need to compute \( \frac{f(x+1) - f(x)}{h} \).1. Compute \( f(x+1) = (x + 1)^2 = x^2 + 2x + 1 \).2. Substitute in the ratio: \( \frac{(x^2 + 2x + 1) - x^2}{h} = \frac{2x + 1}{h} \).3. Since we're setting \( h = 0 \) directly leads to an undefined ratio, consider the expression as a function of \( x \) apart from \( h \) instead.Final Answer for (d): The expression doesn't simplify with \( h = 0 \) as written, it remains \( \frac{2x + 1}{h} \). However, re-evaluation of context implied might visualize solving under error.

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus representing the average rate of change of a function over a small interval. Mathematically, it is expressed as \( \frac{f(x+h) - f(x)}{h} \), where \( h \) is a small increment. This expression essentially estimates the slope of the secant line between two points, namely \( x \) and \( x+h \), on the curve of the function \( f(x) \).

• The difference quotient is used to approximate the derivative of a function, a concept we'll dive into further.
• By using the difference quotient, we can determine how a polynomial function's output changes as its input changes slightly.

Understanding and calculating the difference quotient provides insights into how functions behave locally and sets the groundwork for finding exact slopes or rates of change.

Limits

In calculus, limits help us to understand the behavior of functions as inputs approach certain values. By examining the limit of the difference quotient as \( h \) approaches zero, we can find the derivative of a function.

• The limit process allows smoothing out the approximate rate of change calculated by the difference quotient into an exact rate of change, known as the derivative.
• It also provides a way to solve indeterminate forms, such as those that might occur when a substitution makes the denominator of a fraction zero.

In our specific exercise, taking the limit as \( h \to 0 \) is crucial. It refines the approximate slope between two points into the exact slope or derivative at a single point.

Derivative

A derivative of a function at a point provides the instantaneous rate of change or the slope of the tangent line at that point. It is derived from the limit of the difference quotient.

• The derivative of a polynomial function, like \( f(x) = x^2 \), can be systematically found using differentiation rules.
• For the function \( f(x) = x^2 \), the derivative \( f'(x) = 2x \) gives the slope of the tangent line at any point \( x \) on the parabola.

When solving problems in calculus, finding the derivative makes it possible to analyze and predict how changes in the input affect changes in the output quickly.

Polynomial Function

A polynomial function is a mathematical expression involving sums of powers of a variable, with each term consisting of a coefficient and the variable raised to a non-negative integer exponent. The general form is \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \).

• Polynomials are among the simplest types of functions and are widely used because of their smoothness and continuity.
• In this exercise, we have \( f(x) = x^2 \), which is a simple second-degree polynomial (quadratic function).

Polynomials are important in calculus because they enable straightforward differentiation and offer insights into the behavior of more complex functions.