Chapter 13: Problem 28
Find the following for the given function \(f:\) (a) \(f^{\prime}(a),\) where \(a\) is in the domain of \(f,\) and (b) \(f^{\prime}(3)\) and \(f^{\prime}(4)\) $$f(x)=-\frac{1}{x^{2}}$$
Short Answer
Expert verified
(a) \( f'(a) = \frac{2}{a^3} \); (b) \( f'(3) = \frac{2}{27} \) and \( f'(4) = \frac{1}{32} \).
Step by step solution
01
Understand the Function
The given function is \( f(x) = -\frac{1}{x^2} \), which is a rational function where the denominator is \( x^2 \). This tells us that \( x \) cannot be 0 because it would make the denominator zero, so the domain of \( f \) is \( x eq 0 \).
02
Differentiate the Function
To find \( f'(x) \), use the power rule for differentiation. Rewrite \( f(x) \) as \( f(x) = -x^{-2} \). The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Apply this to \( f(x) = -x^{-2} \):\[ f'(x) = (-1) \cdot (-2)x^{-3} = 2x^{-3} \]This can be rewritten as \( f'(x) = \frac{2}{x^3} \).
03
Evaluate \( f'(a) \)
For part (a), where \( a \) is any point in the domain of \( f \), the derivative \( f'(a) \) can be expressed by replacing \( x \) with \( a \) in \( f'(x) \):\[ f'(a) = \frac{2}{a^3} \].
04
Calculate \( f'(3) \)
Substitute \( x = 3 \) into the derivative \( f'(x) = \frac{2}{x^3} \) to find \( f'(3) \):\[ f'(3) = \frac{2}{3^3} = \frac{2}{27} \].
05
Calculate \( f'(4) \)
Substitute \( x = 4 \) into the derivative \( f'(x) = \frac{2}{x^3} \) to find \( f'(4) \):\[ f'(4) = \frac{2}{4^3} = \frac{2}{64} = \frac{1}{32} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Functions
A rational function is a type of function that is expressed as the ratio of two polynomials. In mathematical terms, it's written as \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). This condition on \( Q(x) \) ensures the function is defined for all values where the denominator is not zero.
For example, if we consider the function \( f(x) = -\frac{1}{x^2} \), it is a rational function because it can be expressed as \( \frac{-1}{x^2} \). The numerator is a constant polynomial (\(-1\)), and the denominator is \( x^2 \), a polynomial with a degree of 2.
Key aspects of rational functions to remember include:
For example, if we consider the function \( f(x) = -\frac{1}{x^2} \), it is a rational function because it can be expressed as \( \frac{-1}{x^2} \). The numerator is a constant polynomial (\(-1\)), and the denominator is \( x^2 \), a polynomial with a degree of 2.
Key aspects of rational functions to remember include:
- They can have vertical asymptotes, which occur where the denominator is zero.
- The domain of a rational function consists of all real numbers except where the denominator is zero.
- They can also have horizontal or oblique asymptotes, giving insight into the function's behavior as \( x \) approaches positive or negative infinity.
Power Rule
The power rule is a basic principle in differentiation used to find the derivative of functions of the form \( x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \). This rule allows for quick differentiation without the need for more complex techniques.
To apply the power rule to the function \( f(x) = -x^{-2} \), we follow these steps:
To apply the power rule to the function \( f(x) = -x^{-2} \), we follow these steps:
- Identify the exponent: here, \( n = -2 \).
- Differentiate using the power rule: the derivative is \( (-2)x^{-2-1} = -2x^{-3} \).
- Simplify the derivative: \( -2x^{-3} \) can be rewritten as \( \frac{-2}{x^3} \).
Domain of a Function
The domain of a function is a set of all input values (typically \( x \)-values) for which the function is defined. In simpler terms, it's all the numbers you can plug into a function without causing mathematical errors, such as division by zero or taking the square root of a negative number.
For the function \( f(x) = -\frac{1}{x^2} \), determining the domain involves finding all the values of \( x \) that do not make the denominator zero. Because the denominator \( x^2 \) equals zero when \( x = 0 \), we have to exclude this value.
For the function \( f(x) = -\frac{1}{x^2} \), determining the domain involves finding all the values of \( x \) that do not make the denominator zero. Because the denominator \( x^2 \) equals zero when \( x = 0 \), we have to exclude this value.
- The domain of \( f(x) \) is \( x eq 0 \), which can also be expressed as all real numbers except zero.