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Chapter 1: Problem 32

Let \(f(x)=x^{2}-4\) \(g(x)=x^{3},\) and \(F(x)=1 /(x-3) .\) Simplify or evaluate the following expressions. $$\frac{f(2+h)-f(2)}{h}$$

Short Answer

Expert verified
Answer: The simplified expression is \(4 + h\).

Step by step solution

01

- Evaluate f(2+h) and f(2)

Given the function \(f(x) = x^2 - 4\), we can calculate \(f(2+h)\) and \(f(2)\) as follows: $$ f(2+h) = (2+h)^2 - 4\\ f(2) = 2^2 - 4 $$
02

- Calculate f(2+h) - f(2)

Next, we find the difference between \(f(2+h)\) and \(f(2)\): $$ f(2+h) - f(2) = [(2+h)^2 - 4] - [2^2 - 4] $$
03

- Simplify the expression

Simplify the expression we found in step 2: $$ [(2+h)^2 - 4] - [2^2 - 4] = (2^2 + 4h + h^2 - 4) - (4 - 4) = 4 + 4h + h^2 - 4 = 4h + h^2 $$
04

- Calculate the whole fraction expression

Now that we have found the simplified expression for \(f(2+h) - f(2)\), we can divide it by \(h\) as required: $$ \frac{f(2+h) - f(2)}{h} =\frac{4h + h^2}{h} $$
05

- Simplify the final expression

Finally, we can simplify the expression further by factoring out the common term \(h\): $$ \frac{4h + h^2}{h} = h(4 + h) \div h = 4 + h $$ So the expression is simplified to \(4 + h\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
A derivative represents the rate at which a function is changing at any given point. It's a cornerstone concept in calculus that helps us understand how one variable changes in relation to another. Let's unpack what it actually means and how it uses the Difference Quotient.
  • The derivative of a function at a specific point is defined as the limit of the difference quotient as the change in the input approaches zero.
  • In calculus notation, the derivative of a function \( f \) at a point \( x \) is \( f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} \).
  • This essentially captures the slope of the tangent line to the function \( f \) at the point \( x \).
Understanding the derivative through the lens of the difference quotient helps in appreciating how exact and specific the calculations become as \( h \) approaches zero, giving us precise information about the function's behavior.
Limit Definition
The limit is a fundamental concept that forms the backbone of calculus. It deals with analyzing the behavior of a function as its input approaches a specific value.
  • The limit of a function \( f(x) \) as \( x \) approaches some value \( a \) is the value that \( f(x) \) gets closer to as \( x \) gets closer to \( a \).
  • In the limit definition of the derivative, we use this concept to understand how small changes in the input affect the output or slope.
  • For example, in the expression \( \frac{{f(2+h) - f(2)}}{h} \), the limit is applied as \( h \to 0 \) to find the derivative of \( f(x) \) at \( x=2 \).
Limits help us grasp the notion of continuity and instantaneous rates of change, making them crucial for defining derivatives and other dynamic quantities in calculus.
Function Evaluation
Function evaluation is the process of finding the output of a function for a specific input. This is an essential skill in mathematics as it lets us understand how functions behave and allows for practical application.
  • For any given function, say \( f(x) = x^2 - 4 \), evaluating \( f \) at \( x = 2 \) means substituting 2 into the function to find \( f(2) \).
  • In the context of the difference quotient, evaluation is done twice: once with \( x = 2 \) and once with \( x = 2+h \).
  • For example, \( f(2+h) = (2+h)^2 - 4 \) gives us an expression in terms of \( h \) to work with further.
This process of substitution and simplification is a necessary step in calculus to manipulate expressions and gain insights into how functions change and interact under different scenarios.

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Most popular questions from this chapter

Consider the general quadratic function \(f(x)=a x^{2}+b x+c,\) with \(a \neq 0\) a. Find the coordinates of the vertex in terms of \(a, b,\) and \(c\) b. Find the conditions on \(a, b,\) and \(c\) that guarantee that the graph of \(f\) crosses the \(x\) -axis twice.

The surface area of a sphere of radius \(r\) is \(S=4 \pi r^{2} .\) Solve for \(r\) in terms of \(S\) and graph the radius function for \(S \geq 0\)

Sketch a graph of the given pair of functions to conjecture a relationship between the two functions. Then verify the conjecture. $$\tan ^{-1} x ; \frac{\pi}{2}-\cot ^{-1} x$$

Without using a calculator, evaluate or simplify the following expressions. $$\sec ^{-1} 2$$

Without using a calculator, evaluate or simplify the following expressions. $$\cot ^{-1}(-1 / \sqrt{3})$$
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