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Chapter 4: Problem 31

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=\tan ^{-1} x$$

Short Answer

Expert verified
Based on the analysis and solution provided, the function \(f(x) = \tan^{-1}(x)\) is always increasing on its entire domain, which is \((-\infty, \infty)\). There are no decreasing intervals for this function.

Step by step solution

01

Find the derivative of the function

First, we will find the derivative of the given function \(f(x) = \tan^{-1}(x)\) using the chain rule. The derivative of \(\tan^{-1}(x)\) is already a well-known formula, which is: $$f'(x) = \frac{1}{1+x^2}$$ Now that we have the derivative, we can proceed to check the intervals where it is positive and negative.
02

Determine the intervals for the derivative to be positive or negative

The derivative \(f'(x) = \frac{1}{1+x^2}\) is a fraction, and its numerator is always positive (1), and its denominator is always positive as well (since \(1+x^2 > 0\) for all x). Therefore, the entire fraction \(f'(x)\) is positive for all values of \(x\). Since the first derivative is positive for all x, the function \(f(x) = \tan^{-1}(x)\) is increasing on the entire domain.
03

Write down the final increasing and decreasing intervals

The function \(f(x) = \tan^{-1}(x)\) has an increasing interval on its entire domain: - Increasing interval: \((-\infty, \infty)\) - Since the function is always increasing, there are no decreasing intervals.

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Key Concepts

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

Derivative
Understanding derivatives is vital in calculus since they reveal how functions change. The derivative of a function measures its rate of change or the slope of the tangent line at any point. For the function given, \(f(x) = \tan^{-1}(x)\), its derivative is derived from a known calculus formula. By differentiating \(\tan^{-1}(x)\), you obtain:
  • \(f'(x) = \frac{1}{1+x^2}\)
This derivative highlights important characteristics about the function's behavior. Since the formula \(\frac{1}{1+x^2}\) involves a fraction, let鈥檚 break down why it鈥檚 crucial:
  • The numerator is 1, a constant value.
  • The denominator, \(1+x^2\), is positive for all real numbers \(x\) because squares of numbers are non-negative.
Thus, the derivative \(f'(x)\) is positive over the entire domain, which signifies a monotonically increasing function.
Chain Rule
The chain rule is a powerful tool in calculus used for finding derivatives of composite functions. A composite function is one where a function is applied inside another function, such as \(h(x) = g(f(x))\). To differentiate such functions, the chain rule provides a structured approach:
  • Identify the outer function \(g(u)\) and the inner function \(f(x)\).
  • Differentiate the outer function \(g(u)\) with respect to the inner function \(u\).
  • Differentiated the inner function \(f(x)\) with respect to \(x\).
This exercise used the chain rule implicitly, even though \(\tan^{-1}(x)\) is straightforward. Learning to recognize when to apply it is crucial for more complex derivatives.
Increasing and Decreasing Intervals
In calculus, identifying where a function increases or decreases is essential for graphing and understanding function behavior. To determine these intervals, we inspect the derivative:
  • If the derivative \(f'(x) > 0\), the function is increasing on that interval.
  • If the derivative \(f'(x) < 0\), the function is decreasing on that interval.
For \(f(x) = \tan^{-1}(x)\), since \(f'(x) = \frac{1}{1+x^2}\) is positive for all \(x\), the function is continuously increasing on its entire domain. Therefore, it is increasing for:
  • Interval: \((-\infty, \infty)\)
Because the derivative never changes sign (it remains positive), there are no intervals where the function is decreasing.

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