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Chapter 5: Problem 89

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f\) is an even function, then \(f^{-1}\) exists.

Short Answer

Expert verified
The statement is false. This is because \(\frac{d}{dx}[\arctan(\tan x)]\) is not defined for all \(x\) in its domain, specifically the odd multiples of \(\frac{\pi}{2}\), even though it equals to 1 at all other points in the domain.

Step by step solution

01

Understand and Recall Derivative Formulas

The derivative of \(\arctan(x)\) is \(\frac{1}{1+x^2}\). While taking the derivative of a composed function like \(\arctan(\tan x)\), we will have to use the chain rule. i.e If you have a composite function \(f(g(x))\), the derivative will be \(f'(g(x)) \cdot g'(x)\). In our case, \(f(x) = \arctan(x)\) and \(g(x) = \tan(x)\). The derivative of \(\tan(x)\) is \(\sec^2(x)\).
02

Calculate the Derivative

Using the chain rule and the derivatives we determined in Step 1, we determine the derivative of \(\arctan(\tan x)\) as follows: \( \frac{d}{dx}[\arctan(\tan x)] = \frac{1}{1+(\tan x)^2} \cdot \sec^2(x) = \frac{\sec^2(x)}{1+\tan^2(x)} = \frac{\sec^2(x)}{\sec^2(x)} = 1 \) We see that the derivative is 1, irrespective of x. However, we haven't considered the domain yet.
03

Consider the Domain

The function \(\tan(x)\) is defined for all \(x\) except at odd multiples of \(\frac{\pi}{2}\), where it has vertical asymptotes. These include points such as \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\), etc. However, \(\arctan(\tan x)\) simplifies to \(x\) for \(x\) in the domain of \([- \frac{\pi}{2}, \frac{\pi}{2}]\) and does not equal \(x\) outside this domain due to the periodicity of the \(\tan(x)\) function. So while the derivative of \(\arctan(\tan x)\) is 1 within the specified domain, it is not 1 for all \(x\) in its domain.
04

Formulate the Final Conclusion

The derivative of a function at the discontinuity is undefined (it does not exist). \(\tan(x)\) and hence \(\arctan(\tan x)\) have discontinuities at odd multiples of \(\frac{\pi}{2}\). Thus, the given statement is false because the derivative \(\frac{d}{d x}[\arctan (\tan x)]\) is not defined at these points even though it equals 1 at all other points in the domain.

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