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Chapter 2: Problem 61

Find the range of $$ f(x)=\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x $$

Short Answer

Expert verified
The range of the given function \( f(x) = \sin^{-1}x + \cos^{-1}x + \tan^{-1}x \) is \(-\frac{\pi}{2} < y < \frac{\pi}{2}\).

Step by step solution

01

Identify Constants in the Function

Analyze the function \( f(x) = \sin^{-1}x + \cos^{-1}x + \tan^{-1}x \) and realize that the sum of \(\sin^{-1}x\) and \(\cos^{-1}x\) is a constant, namely \(\frac{\pi}{2}\). Thus, the function can be simplified to \( f(x) = \frac{\pi}{2} + \tan^{-1}x \).
02

Determine the Range of the Remaining Function

Next, we need to find the range of \(\tan^{-1}x\). The inverse tangent function, \(\tan^{-1}x\), is defined for all real numbers, and its range is \(-\frac{\pi}{2} < y < \frac{\pi}{2}\).
03

Add Constant to the Range

Now, to find the range of the whole function, we add our constant from Step 1 (\(\frac{\pi}{2}\)) to all parts of the range from Step 2: \(\frac{\pi}{2} - \frac{\pi}{2} < y + \frac{\pi}{2} < \frac{\pi}{2} + \frac{\pi}{2} \). The inequality simplifies to \(0 < y + \frac{\pi}{2} < \pi\).
04

Solve for y

Finally, to solve for 'y' in the inequality, subtract \(\frac{\pi}{2}\) from all sections: \(0 - \frac{\pi}{2} < y < \pi - \frac{\pi}{2}\), simplifying this inequality gives us \(-\frac{\pi}{2} < y < \frac{\pi}{2}\). Thus, this is our range for the function \( f(x)\).

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Most popular questions from this chapter

Prove that: If \(\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi\), then prove that $$ x^{2}+y^{2}+z^{2}+2 x y z=1 $$

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