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Magazine Chapter 1: Problem 90
Find \(\frac{f(x+h)-f(x)}{h}\) for the indicated functions. $$ f(x)=x^{2}-1 $$
Short Answer
Expert verified
\(2x + h\)
Step by step solution
01
Identify the function
We are given the function \(f(x) = x^2 - 1\). Our task is to find \(\frac{f(x+h)-f(x)}{h}\).
02
Express \(f(x+h)\)
Substitute \(x + h\) into the function: \(f(x+h) = (x+h)^2 - 1\).
03
Expand \((x+h)^2\)
Calculate \((x+h)^2\) as \(x^2 + 2xh + h^2\). Thus, \(f(x+h) = x^2 + 2xh + h^2 - 1\).
04
Find \(f(x+h) - f(x)\)
Subtract \(f(x)\) from \(f(x+h)\): \(f(x+h) - f(x) = (x^2 + 2xh + h^2 - 1) - (x^2 - 1)\).
05
Simplify the expression
Simplify the expression: \(f(x+h) - f(x) = x^2 + 2xh + h^2 - 1 - x^2 + 1 = 2xh + h^2\).
06
Divide by \(h\)
Divide the simplified expression by \(h\): \(\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2}{h}\).
07
Factor and simplify
Factor \(h\) out of the numerator: \(\frac{h(2x + h)}{h}\). Simplify this to get \(2x + h\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Evaluation
Evaluating a function is a fundamental concept that involves replacing the variable in a function's formula with a given number or expression. Think of it like substituting ingredients in a recipe to get the final dish.
For example, given the function \(f(x) = x^2 - 1\), if we need to find \(f(3)\), we simply replace every \(x\) with 3:
For example, given the function \(f(x) = x^2 - 1\), if we need to find \(f(3)\), we simply replace every \(x\) with 3:
- \(f(3) = 3^2 - 1\)
- Calculate: \(9 - 1 = 8\)
- So, \(f(3) = 8\)
- Replace \(x\) with \(x+h\): \(f(x+h) = (x+h)^2 - 1\)
- Now expand: \((x+h)^2 = x^2 + 2xh + h^2\)
- So, \(f(x+h) = x^2 + 2xh + h^2 - 1\)
Polynomial Functions
Polynomial functions are expressions involving variables and constants combined using addition, subtraction, multiplication, and non-negative integer exponents. They form the backbone of many algebraic expressions and can be as simple as \(f(x) = x - 3\) or as complex as \(f(x) = 4x^3 - 2x^2 + x - 7\).
The function given in our exercise, \(f(x) = x^2 - 1\), is a polynomial function of degree 2, also known as a quadratic function. Here鈥檚 why polynomials are important:
The function given in our exercise, \(f(x) = x^2 - 1\), is a polynomial function of degree 2, also known as a quadratic function. Here鈥檚 why polynomials are important:
- **Smooth and Continuous:** Polynomial graphs are smooth curves without breaks.
- **Predictable Patterns:** They show consistent behavior, making them easy to study.
- **Versatile:** Used to model a wide range of real-world problems.
Limit Process
The limit process is a critical part of calculus that examines what happens to a function's output as the input gets closer to a specific value. This concept is fundamental when determining derivatives, which represent rates of change.
In the given exercise, the expression \(\frac{f(x+h) - f(x)}{h}\) is a difference quotient. It helps us approximate the derivative of the function at point \(x\) as \(h\) approaches zero.
Here's a closer look at the process:
In the given exercise, the expression \(\frac{f(x+h) - f(x)}{h}\) is a difference quotient. It helps us approximate the derivative of the function at point \(x\) as \(h\) approaches zero.
Here's a closer look at the process:
- **Difference Quotient:** Measures the average rate of change of the function between \(x\) and \(x+h\).
- **Approaching the Limit:** We simplify the expression to \(2x + h\). As \(h\) tends to zero, \(2x + h\) straightforwardly approaches \(2x\).
- **Derivative Insight:** This tells us that the instantaneous rate of change, or derivative of \(f(x) = x^2 - 1\), is \(2x\).
