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In Exercises \(23-28\), use the derivative to determine whether the function is strictly monotonic on its entire domain and therefore has an inverse function. \(f(x)=\ln (x-3)\)

Short Answer

Expert verified
The function \(f(x)=\ln (x-3)\) is strictly monotonic and has an inverse function.

Step by step solution

01

Compute the derivative

The derivative of the function \(f(x)=\ln (x-3)\) is given by using the chain rule of derivatives as follows: \(f'(x)=\frac{1}{x-3}\)
02

Determine the sign of the derivative

The derivative \(f'(x)=\frac{1}{x-3}\) is greater than 0 for all \(x > 3\) in the domain of \(f(x)\). This indicates that the function \(f(x)=\ln (x-3)\) is increasing in its entire domain.
03

Determine the monotonicity and the existence of an inverse function

Since the function \(f(x)=\ln (x-3)\) is increasing in its entire domain (i.e., it's strictly monotonic), it can be concluded that it has an inverse function.

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Key Concepts

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

Strictly Monotonic Functions
When we talk about strictly monotonic functions, we're describing functions that move consistently in one direction. These functions neither plateau nor change directions at any point in their domain. They will either keep increasing or keep decreasing.
  • A strictly increasing function means that as you move from left to right along the x-axis, the output of the function continuously gets larger.
  • Conversely, a strictly decreasing function does the opposite, continually getting smaller as you move from left to right.
This continuous movement in one direction ensures that each input corresponds to a unique output, which is a crucial characteristic for a function to have an inverse. If a function is strictly monotonic across its entire domain, it means there's no chance of the function turning back or duplicating any output value, making it invertible.
Derivatives
Derivatives are like the speedometer of mathematics, showing us how fast a function's output is changing at any given point. When we calculate the derivative of a function, we're essentially finding a new function that tells us the rate of change.
For example, if we take the derivative of our original function, \(f(x) = \ln (x-3)\), we use the rules of differentiation to get \(f'(x) = \frac{1}{x-3}\). This represents the slope or steepness of the function at any point where \(x > 3\).
  • If the derivative is positive (\(f'(x) > 0\)), the function is increasing.
  • If the derivative is negative (\(f'(x) < 0\)), the function is decreasing.
  • And if the derivative is zero, the function could be at a plateau, indicating a possible maximum, minimum, or inflection point.
Derivatives are vital for analyzing the behavior of functions, helping us determine if they are consistently moving up or down, which is key in identifying strictly monotonic functions.
Chain Rule
The chain rule is a powerful tool in calculus for dealing with the derivatives of composed functions. It helps us to differentiate a function that is nested inside another function. Let's say we have a composite function \(f(g(x))\), where \(f\) and \(g\) are two different functions.
Here's how the chain rule works:
  • First, differentiate the outer function \(f\), but keep the inner function \(g\) unchanged.
  • Then multiply this result by the derivative of the inner function \(g\).
For our example, \(f(x) = \ln (x-3)\), the chain rule tells us to recognize \((x-3)\) as an inner function. Applying the chain rule gives us \(f'(x) = \frac{1}{x-3}\), simplifying \(\frac{d}{dx}[\ln u] = \frac{1}{u} \cdot \frac{du}{dx}\), where \(u = x-3\) and \(\frac{du}{dx} = 1\).
This approach allows us to handle complex differentiations step by step, ensuring that we accurately capture how the nested changes impact the overall function.

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