Problem 11
Question
Use the definition of the derivative to find the derivative of the function. What is its domain? \(f(x)=\frac{1}{x+2}\)
Step-by-Step Solution
Verified Answer
The derivative of the function \(f(x) = \frac{1}{x+2}\) is \(f'(x) = \frac{1}{(x+2)^2}\). The domain of the function is \((-\infty, -2) \cup (-2, \infty)\).
1Step 1: Recalling the definition of the derivative
The definition of the derivative of a function \(f(x)\) at position \(x\) is given by:
\[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]
We have to use this definition to find the derivative of the given function \(f(x) = \frac{1}{x+2}\).
2Step 2: Evaluating the difference quotient
First, we evaluate the difference quotient:
\[\frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{(x+h)+2} - \frac{1}{x+2}}{h}\]
3Step 3: Simplifying the difference quotient
Next, we simplify the numerator by finding a common denominator:
\[\frac{(x+2) - [(x+h)+2]}{(x+2)[(x+h)+2]h} = \frac{x + 2 - x - h - 2}{(x+2)[(x+h)+2]h} = \frac{-h}{(x+2)(x+h+2)h}\]
Now, we can simplify the difference quotient by canceling out the -h term in the numerator and denominator:
\[\frac{1}{(x+2)(x+h+2)}\]
4Step 4: Applying the limit
Now, we apply the limit of the difference quotient as h approaches 0:
\[f'(x) = \lim_{h\to 0} \frac{1}{(x+2)(x+h+2)}\]
As h approaches 0, the second term in the denominator becomes (x + 2):
\[f'(x) = \frac{1}{(x+2)(x+2)} = \frac{1}{(x+2)^2}\]
5Step 5: Finding the domain
The function is \(f(x) = \frac{1}{x+2}\). The domain of a rational function is all real numbers except for the values that make the denominator equal to zero. To find the domain, we set the denominator to zero and solve for x:
\(x + 2 = 0\)
\(x = -2\)
The domain of the function is all real numbers except for -2. In interval notation, the domain is:
\((-\infty, -2) \cup (-2, \infty)\)
So, the derivative of the function is \(f'(x) = \frac{1}{(x+2)^2}\), and the domain of the function is \((-\infty, -2) \cup (-2, \infty)\).
Key Concepts
Difference QuotientRational FunctionsDomain of a FunctionLimit Process
Difference Quotient
The difference quotient is a foundational concept for understanding derivatives. This quotient is a formula used to approximate the slope of the tangent line to a curve at a specific point, giving insight into the function's behavior at that point. The formula is \[\frac{f(x+h) - f(x)}{h}\]where \( h \) is a small increment in \( x \), and \( f(x) \) is the function value at \( x \).
- This formula calculates the average rate of change of the function over the interval \([x, x+h]\).
- It forms the backbone of the derivative as it gradually becomes the instantaneous rate of change when \( h \) approaches zero.
Rational Functions
Rational functions are expressions formed by the ratio of two polynomials. The function \( f(x) = \frac{1}{x+2} \) is a simple example of a rational function, where the numerator is a constant polynomial (which is 1), and the denominator is the linear polynomial \( x+2 \).
- These functions are defined at all points where the denominator is not zero.
- The behavior of rational functions often includes asymptotes and discontinuities at points where their denominator equals zero.
Domain of a Function
The domain of a function defines the set of all possible input values (\( x \)-values) for which the function is defined. For rational functions, such as \( f(x) = \frac{1}{x+2} \), the domain includes all real numbers except those that make the denominator zero, because division by zero is undefined.
- In this exercise, to find the domain, we solve \( x+2=0 \) to get \( x=-2 \).
- Thus, the domain of \( f(x) \) is all real numbers except \( x = -2 \).
Limit Process
The limit process is a key tool in calculus used to find the derivative of a function. It involves resolving what happens to a function as the variable approaches a particular value. Specifically, for derivatives, this involves evaluating\[\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]
- This process enables us to determine the instantaneous rate of change or the slope of the tangent at a specific point.
- By applying the limit as \( h \) approaches zero, the difference quotient essentially "zooms in" on a very small interval around \( x \), giving an accurate slope of the curve at that point.
Other exercises in this chapter
Problem 11
Use the Quotient Rule to find the derivative of each function. \(F(x)=\frac{x}{1+x e^{x}}\)
View solution Problem 11
Find the derivative of the function. \(f(x)=x^{2}-2 x+8\)
View solution Problem 12
Find the derivative of the function. $$ f(x)=\left(\frac{x^{2}+3}{x}\right)^{-2} $$
View solution Problem 12
Find the differential of the function at the indicated number. $$ f(x)=x \tan x, \quad x=\frac{\pi}{4} $$
View solution