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Chapter 1: Problem 104

Use the derivative to help show whether each function is always increasing, always decreasing, or neither. $$f(x)=\frac{1}{x}, \quad x \neq 0$$

Short Answer

Expert verified
The function \( f(x) = \frac{1}{x} \) is always decreasing for \( x eq 0 \).

Step by step solution

01

Find the derivative of the function

The given function is \( f(x) = \frac{1}{x} \). To determine if the function is increasing or decreasing, find the first derivative using the quotient rule. The derivative \( f'(x) \) of the function \( f(x) = x^{-1} \) is calculated as follows:\[ f'(x) = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2} \]
02

Analyze the sign of the derivative

The sign of the derivative \( f'(x) \) indicates whether the function is increasing or decreasing:- If \( f'(x) > 0 \), the function is increasing.- If \( f'(x) < 0 \), the function is decreasing.Since \( f'(x) = -\frac{1}{x^2} \), and \( x^2 \) is always positive for all \( x eq 0 \), \( -\frac{1}{x^2} \) is always negative for all \( x eq 0 \).
03

Conclude whether the function is increasing or decreasing

Given that \( f'(x) = -\frac{1}{x^2} \) is always negative for all \( x eq 0 \), the function \( f(x) = \frac{1}{x} \) is always decreasing for \( x eq 0 \).

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Key Concepts

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

increasing and decreasing functions
When analyzing functions, knowing if a function is increasing or decreasing helps us understand how values change. A function is **increasing** if its values get larger as the input increases. Conversely, a function is **decreasing** if its values get smaller as the input increases.
To determine if a function is increasing or decreasing, we look at its **derivative**. The derivative tells us the rate at which the function's value changes as the input changes.
  • If the derivative is positive over an interval, the function is increasing in that interval.
  • If the derivative is negative over an interval, the function is decreasing in that interval.
In this exercise, we checked the behavior of the function by analyzing its derivative.
quotient rule
The **quotient rule** is used when differentiating functions that are fractions, where you have one function divided by another. It states that if you have a function of the form \( \frac{u(x)}{v(x)} \) where both u(x) and v(x) are differentiable, then the derivative f(x) is:
\[ f'(x) = \frac{v(x) * u'(x) - u(x) * v'(x)}{[v(x)]^2} \]
Let's simplify this with an example:
  • If u(x) = 1 and v(x) = x, then \[ f(x) = \frac{1}{x} \]
    • Use specific components:
    • u(x) = 1 and u'(x) = 0 (as it is a constant)
    • v(x) = x and v'(x) = 1
    Using our quotient rule formula, we get:

    \[ f'(x) = \frac{x * (0) - 1 * (1)}{x^2} = -\frac{1}{x^2} \]
    As we see, differentiating \( \frac{1}{x} \) gives us a result of \(-\frac{1}{x^2} \).
sign of the derivative
Understanding the **sign of the derivative** is crucial in determining the function's behavior. In this step-by-step solution, once we found the derivative, we analyzed its sign to see how the function behaves. For the derivative \( f'(x) = -\frac{1}{x^2} \), the sign is always negative since the square of any nonzero real number is always positive, leading to:

  • If \( x = 1 \), then \( f'(1) = -1 < 0 \)
  • If \( x = -1 \), then \( f'(-1) = -1 < 0 \)

    • It demonstrates that \( f'(x) \) remains negative for all non-zero values of x. Meaning, the function is always decreasing for all x 鈮� 0.
function behavior analysis
Understanding a function鈥檚 behavior is essential in calculus. Here, we determined whether the function \( f(x) = \frac{1}{x} \) is increasing or decreasing by analyzing its derivative and its sign.
Our analysis showed that the derivative \( f'(x) \) of the function is always negative as computed by \( - \frac{1}{x^2} \). This information reveals that:

  • The function decreases on its entire domain, which is all real numbers except zero.
  • More specifically, as x becomes larger or smaller, the value of \( f(x) \) approaches zero, but it never actually reaches zero.

    • This means the function \( f(x) = \frac{1}{x} \) is always **decreasing** for \( x eq 0 \), providing a clear understanding of its general behavior on the graph.

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