Problem 16
Question
Find all critical numbers of the given function. $$ f(x)=1 /\left(e^{x}-1\right) $$
Step-by-Step Solution
Verified Answer
The critical number is \( x = 0 \).
1Step 1: Understand the Function
The function given is \( f(x) = \frac{1}{e^x - 1} \). Our goal is to find the critical numbers, where the derivative is either zero or undefined.
2Step 2: Compute the First Derivative
Apply the quotient rule to find the derivative. Let \( u = 1 \) and \( v = e^x - 1 \). The quotient rule states \( f'(x) = \frac{v'u - vu'}{v^2} \), where \( u' = 0 \) and \( v' = e^x \). Substituting these, we get \( f'(x) = \frac{0 - 1 imes e^x}{(e^x - 1)^2} = \frac{-e^x}{(e^x - 1)^2} \).
3Step 3: Set the Derivative Equal to Zero
To find critical points, set the derivative equal to zero: \( \frac{-e^x}{(e^x - 1)^2} = 0 \). Since \( -e^x eq 0 \), there are no values of \( x \) that make the derivative zero.
4Step 4: Determine Where the Derivative is Undefined
The derivative \( f'(x) = \frac{-e^x}{(e^x - 1)^2} \) is undefined where the denominator \((e^x - 1)^2\) is zero. Set \( e^x - 1 = 0 \) which gives \( e^x = 1 \) or \( x = 0 \). Thus, \( x = 0 \) is a critical number.
Key Concepts
Derivative CalculationsQuotient Rule in CalculusUndefined Derivative
Derivative Calculations
Calculating derivatives is a crucial skill in calculus, especially when you're trying to find critical numbers of a function. A derivative helps us understand the rate at which a function changes. This involves finding the derivative of a function with respect to its variable.
To calculate the derivative efficiently, we often rely on rules like the power rule, product rule, quotient rule, and chain rule. In the context of our specific function, \( f(x) = \frac{1}{e^x - 1} \), one must compute its first derivative to identify critical points where the function's behavior changes. This is where changes such as peaks or troughs (local maxima or minima) occur. The first derivative tells us where the slope of the tangent to the function is zero (horizontal) or where it's undefined. These locations often mark critical numbers.
Critical numbers are where the derivative is zero or undefined, as these indicate potential maximum, minimum, or points of inflection on the graph. Understanding and computing derivatives correctly is key to identifying these characteristics of functions.
To calculate the derivative efficiently, we often rely on rules like the power rule, product rule, quotient rule, and chain rule. In the context of our specific function, \( f(x) = \frac{1}{e^x - 1} \), one must compute its first derivative to identify critical points where the function's behavior changes. This is where changes such as peaks or troughs (local maxima or minima) occur. The first derivative tells us where the slope of the tangent to the function is zero (horizontal) or where it's undefined. These locations often mark critical numbers.
Critical numbers are where the derivative is zero or undefined, as these indicate potential maximum, minimum, or points of inflection on the graph. Understanding and computing derivatives correctly is key to identifying these characteristics of functions.
Quotient Rule in Calculus
The quotient rule in calculus is essential when you need to find the derivative of a ratio of two functions. It is especially useful when dealing with functions defined as a fraction, like \( f(x) = \frac{1}{e^x - 1} \).
The quotient rule states that if you have functions \( u(x) \) and \( v(x) \), then the derivative \( f'(x) \) is given by:\[ f'(x) = \frac{v'(x)u(x) - u'(x)v(x)}{(v(x))^2} \] This formula may seem complex, but it breaks down into simpler steps:
The quotient rule states that if you have functions \( u(x) \) and \( v(x) \), then the derivative \( f'(x) \) is given by:\[ f'(x) = \frac{v'(x)u(x) - u'(x)v(x)}{(v(x))^2} \] This formula may seem complex, but it breaks down into simpler steps:
- Identify the function in the numerator \( u(x) \) and its derivative \( u'(x) \).
- Identify the function in the denominator \( v(x) \) and its derivative \( v'(x) \).
- Apply the quotient rule formula to compute \( f'(x) \).
Undefined Derivative
An undefined derivative occurs when division by zero happens in the derivative expression. For critical point analysis, identifying where a derivative is undefined is as vital as knowing where it equals zero.
In our example function, the derivative is \( f'(x) = \frac{-e^x}{(e^x - 1)^2} \). We observe that the derivative is undefined whenever the denominator \((e^x - 1)^2\) equals zero. Solving for this gives \( e^x - 1 = 0 \), resulting in \( e^x = 1 \) or \( x = 0 \).
Thus, \( x = 0 \) is a critical point because it makes the derivative undefined. Recognizing such points tells us where the function might have a sharp corner, discontinuity, or vertical tangent. These points become critical numbers to examine thoroughly when understanding a function's graph and identifying its peculiar behaviors.
In our example function, the derivative is \( f'(x) = \frac{-e^x}{(e^x - 1)^2} \). We observe that the derivative is undefined whenever the denominator \((e^x - 1)^2\) equals zero. Solving for this gives \( e^x - 1 = 0 \), resulting in \( e^x = 1 \) or \( x = 0 \).
Thus, \( x = 0 \) is a critical point because it makes the derivative undefined. Recognizing such points tells us where the function might have a sharp corner, discontinuity, or vertical tangent. These points become critical numbers to examine thoroughly when understanding a function's graph and identifying its peculiar behaviors.
Other exercises in this chapter
Problem 16
At 3 P.M. an oil tanker traveling west in the ocean at 15 kilometers per hour passes the same point as a luxury liner that arrived at the same spot at 2 P.M. wh
View solution Problem 16
Determine the function \(\mathrm{f}\) satisfying the given conditions. $$ f^{\prime}(x)=-\frac{3}{2} x^{2}, f(-1)=-\frac{1}{2} $$
View solution Problem 17
Find the given limit. $$ \lim _{x \rightarrow \infty} e^{-x} $$
View solution Problem 17
Find the intervals on which the graph of the function is concave upward and those on which it is concave downward. Then sketch the graph of the function. $$ f(x
View solution