Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Problem 23
If \(f(x)=\frac{1}{x}, x \neq 0,\) evaluate \(\frac{f(x+h)-f(x)}{h}\) and express the answer in simplest form.
Short Answer
Expert verified
Answer: The simplified expression is \(\frac{-1}{x(x + h)}\).
Step by step solution
01
Find \(f(x+h)\) and \(f(x)\)
Substitute \(x + h\) into the function \(f(x)\) to find \(f(x + h)\), and simply use \(f(x)\) itself:
$$
f(x + h) = \frac{1}{x + h}, \quad f(x) = \frac{1}{x}
$$
02
Substitute \(f(x+h)\) and \(f(x)\) into the expression
Now we substitute the values for \(f(x + h)\) and \(f(x)\) into the expression \(\frac{f(x+h)-f(x)}{h}\):
$$
\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x + h} - \frac{1}{x}}{h}
$$
03
Simplify the numerator
To simplify the expression further, we need to find a common denominator for the fractions in the numerator:
$$
\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}
$$
04
Simplify the entire expression
Now, substitute the simplified numerator back into the expression and simplify further:
$$
\frac{f(x+h)-f(x)}{h} = \frac{\frac{-h}{x(x + h)}}{h}
$$
Divide by \(h\):
$$
\frac{f(x+h)-f(x)}{h} = \frac{-h}{x(x + h)} \cdot \frac{1}{h} = \frac{-1}{x(x + h)}
$$
So, after simplifying, we get:
$$
\frac{f(x+h)-f(x)}{h} = \frac{-1}{x(x + h)}
$$
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference Quotient
The difference quotient is a fundamental concept in calculus, often used to find the derivative of a function at a certain point. It's essentially a formula that represents the slope of the secant line between two points on a function. The formula typically appears as \(\frac{f(x+h) - f(x)}{h}\). The difference quotient gives us a way to measure how a function changes as its input changes by a small amount \(h\).
In simple terms, it's trying to capture what happens to the output of a function as you nudge the input slightly. This is a stepping stone to finding the derivative when \(h\) approaches zero, which gives the exact slope of the tangent line at a point.
In simple terms, it's trying to capture what happens to the output of a function as you nudge the input slightly. This is a stepping stone to finding the derivative when \(h\) approaches zero, which gives the exact slope of the tangent line at a point.
Function Evaluation
Function evaluation is a process of determining the output of a function for a specific input. When a function is expressed as \(f(x)\), evaluating \(f(x)\) at a certain \(x\) means substituting \(x\) into the function and computing the corresponding result.
For example, if you have \(f(x) = \frac{1}{x}\) and you want to evaluate it at \(x = 2\), you replace \(x\) with 2, which will yield \(f(2) = \frac{1}{2}\).
In the context of the example we are dealing with, you substitute \(x + h\) into the function instead of \(x\) for \(f(x+h)\). So, \(f(x+h)\) would be \(\frac{1}{x+h}\). It helps figure out how the function behaves for inputs near a specific point.
For example, if you have \(f(x) = \frac{1}{x}\) and you want to evaluate it at \(x = 2\), you replace \(x\) with 2, which will yield \(f(2) = \frac{1}{2}\).
In the context of the example we are dealing with, you substitute \(x + h\) into the function instead of \(x\) for \(f(x+h)\). So, \(f(x+h)\) would be \(\frac{1}{x+h}\). It helps figure out how the function behaves for inputs near a specific point.
Rational Functions
Rational functions are functions of the form \(\frac{p(x)}{q(x)}\), where both \(p(x)\) and \(q(x)\) are polynomials. They are prevalent in calculus because they can model a wide range of phenomena.
The function \(f(x)=\frac{1}{x}\) is a simple rational function. Its characteristics include
The function \(f(x)=\frac{1}{x}\) is a simple rational function. Its characteristics include
- Identifying a vertical asymptote where the denominator is zero (here, \(xeq0\) ensures no division by zero).
- Horizontal and vertical asymptotic behavior, which describes how the function approaches certain lines as \(x\) becomes very large or very small.
- Finding common denominators is often necessary when working with rational functions, particularly during the simplification process, to easily manage and simplify the expression.
Simplifying Expressions
Simplifying expressions involves rewriting a complex expression in a simpler, equivalent form. This process often involves combining like terms, eliminating unnecessary components, and factoring.
In our example, simplifying the expression \(\frac{\frac{1}{x+h} - \frac{1}{x}}{h}\) starts by finding a common denominator for the fractions in the numerator, resulting in \(\frac{x - (x + h)}{x(x + h)}\), which further simplifies to \(\frac{-h}{x(x+h)}\).
We then divide by \(h\), leading to \(\frac{-1}{x(x+h)}\).
In our example, simplifying the expression \(\frac{\frac{1}{x+h} - \frac{1}{x}}{h}\) starts by finding a common denominator for the fractions in the numerator, resulting in \(\frac{x - (x + h)}{x(x + h)}\), which further simplifies to \(\frac{-h}{x(x+h)}\).
We then divide by \(h\), leading to \(\frac{-1}{x(x+h)}\).
- Finding common denominators helps in rewriting sums or differences of fractions as a single fraction, easing the simplification process.
- Division by \(h\) ultimately cancels out terms in the numerator and denominator, streamlining the expression.
- Simplifying expressions can make it easier to understand and solve problems, especially in calculus, where interpretations are often required at a glance.
