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Use a derivative to show that \(f(x)=\ln \left(x^{3}-1\right)\) is one-to-one.

Short Answer

Expert verified
The derivative \( f'(x) = \frac{3x^2}{x^3 - 1} \) shows \( f(x) \) changes sign at \( x = 1 \), ensuring one-to-one behavior overall.

Step by step solution

01

Understand the One-to-One Function Concept

A function is one-to-one if, for every pair of distinct inputs, the outputs are also distinct. Essentially, different inputs produce different outputs.
02

Derive f(x)

To find the derivative of the function, apply the chain rule to the function \( f(x) = \ln(x^3 - 1) \). The derivative \( f'(x) \) is given by the formula \( \frac{d}{dx}[\ln u(x)] = \frac{1}{u(x)} \cdot u'(x) \). Here, \( u(x) = x^3 - 1 \).
03

Solve for the Derivative

First, find \( u'(x) \) where \( u(x) = x^3 - 1 \). The derivative \( u'(x) = 3x^2 \). Thus, \( f'(x) = \frac{1}{x^3 - 1} \times 3x^2 = \frac{3x^2}{x^3 - 1} \).
04

Determine the Sign of the Derivative

For the function to be one-to-one, \( f'(x) \) must be positive (or negative) for all \( x \) in its domain. Analyze \( f'(x) = \frac{3x^2}{x^3 - 1} \) to see when it is positive or negative. The critical points of \( f'(x) \) occur when \( x^3 - 1 = 0 \) which gives \( x = 1 \). For values of \( x < 1 \), the denominator is negative and for \( x > 1 \), the denominator is positive.
05

Analyze the Function Behavior

For \( x < 1 \), \( f'(x) < 0 \), and for \( x > 1 \), \( f'(x) > 0 \). The derivative changes sign and \( f(x) \) is strictly decreasing for \( x < 1 \) and strictly increasing for \( x > 1 \), therefore, it's one-to-one across its domain.

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Key Concepts

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

One-to-One Function
Understanding when a function is one-to-one is crucial in mathematics. A function is considered one-to-one when each input (or "x" value) produces a unique output (or "y" value). This means that no two different inputs will result in the same output.

To identify if a function is one-to-one, one can use the horizontal line test on a graph. If no horizontal line intersects the graph at more than one point, then the function is one-to-one.

Another effective approach is to use derivatives. If the derivative of a function is always positive or always negative (except at isolated points), the function is strictly increasing or decreasing, signifying that the function is one-to-one.
Chain Rule
The chain rule is a fundamental technique for differentiating composite functions. A composite function is a function built from two or more functions such as \( f(g(x)) \).

In simpler words, if you have a function within another function, and you need to find the derivative, the chain rule comes to your rescue. Mathematically, if \( y = f(u) \) and \( u = g(x) \), then the derivative of \( y \) with respect to \( x \), expressed in terms of \( y' \), is computed as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

For example, if \( f(x) = \ln(x^3 - 1) \), you would first find the derivative of the inner function \( u = x^3 - 1 \), which is \( 3x^2 \), and then apply the rule: \( f'(x) = \frac{1}{x^3 - 1} \cdot 3x^2 \). This process simplifies the differentiation of complex functions.
Function Domain
The domain of a function refers to all the possible "x" values for which the function is defined. It is crucial to understand the domain to avoid mathematical errors like division by zero or logarithms of negative numbers.

For instance, with the function \( f(x) = \ln(x^3 - 1) \), the expression \( x^3 - 1 \) must be positive because the logarithm of a non-positive number is undefined in real numbers. This means \( x^3 > 1 \), translating to \( x > 1 \) for our domain.

By establishing the correct function domain, you ensure the function behaves consistently and predictably.
Critical Points
Critical points are values of \( x \) where the derivative of a function is zero or undefined. These points indicate where the function's graph may change direction, either rising, falling, or leveling off.

Critical points help identify where a function could reach its minimum or maximum values, or where it changes from increasing to decreasing and vice versa.

In the function \( f(x) = \ln(x^3 - 1) \), the critical point is found when the derivative \( f'(x) = \frac{3x^2}{x^3 - 1} \) equals zero or where the denominator becomes zero (which makes the expression undefined). Here, the critical point occurs at \( x = 1 \). By analyzing the behavior around this point, you can deduce if the function maintains its one-to-one nature.

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Most popular questions from this chapter

A decimal representation of \(e \quad\) Find \(e\) to as many decimal places as your calculator allows by solving the equation \(\ln x=1\) using Newton's method in Section 4.6.

Suppose that a cup of soup cooled from \(90^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) after \(10 \mathrm{min}\) in a room whose temperature was \(20^{\circ} \mathrm{C}\). Use Newton's Law of Cooling to answer the following questions. a. How much longer would it take the soup to cool to \(35^{\circ} \mathrm{C} ?\) b. Instead of being left to stand in the room, the cup of \(90^{\circ} \mathrm{C}\) soup is put in a freezer whose temperature is \(-15^{\circ} \mathrm{C}\). How long will it take the soup to cool from \(90^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C} ?\)

Show that if a function \(f\) is defined on an interval symmetric about the origin (so that \(f\) is defined at \(-x\) whenever it is defined at \(x\) ), then $$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$ Then show that \((f(x)+f(-x)) / 2\) is even and that \((f(x)-\) \(f(-x)) / 2\) is odd.

An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at \(18^{\circ} \mathrm{C}\). After \(10 \mathrm{min}\), the beam warmed to \(2^{\circ} \mathrm{C}\) and after another \(10 \mathrm{min}\) it was \(10^{\circ} \mathrm{C}\). Use Newton's Law of Cooling to estimate the beam's initial temperature.

Find the area of the "triangular" region in the first quadrant that is bounded above by the curve \(y=e^{x / 2},\) below by the curve \(y=e^{-x / 2},\) and on the right by the line \(x=2 \ln 2\)

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