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Chapter 1: Problem 44

Find the range of the function \(f(x)=\cos ^{-1}\left(\frac{x^{2}}{1+x^{2}}\right)\)

Short Answer

Expert verified
The range of the function \(f(x)=\cos^{-1}\left(\frac{x^{2}}{1+x^{2}}\right)\) is \([0, \frac{\pi}{2}]\)

Step by step solution

01

Determine the range of the function inside the cosine inverse

We want to find the range of the function \( \frac{x^2}{1 + x^2} \). To do this, let's note that \(x^2\) is always non-negative. In other words, \(x^2 \geq 0\). Therefore, the smallest possible value of the function \( \frac{x^2}{1 + x^2} \) is when \(x^2 = 0\), which gives us 0. On the other hand, as \(x^2\) becomes very large, the function tends towards 1. Therefore, the range of the function \( \frac{x^2}{1 + x^2} \) is \([0, 1]\).
02

Apply the range to the cosine inverse function

The function \( \cos ^{-1}(x) \) is defined for all real numbers \(x\) such that \(-1 \leq x \leq 1\). It outputs an angle in the interval \([0, \pi]\). Because we have found that \( \frac{x^2}{1 + x^2} \) takes values in \([0, 1]\), \(\cos ^{-1}\left(\frac{x^2}{1 + x^2}\right)\) will vary in the interval \([0, \frac{\pi}{2}]\).
03

Conclusion

Combining Steps 1 and 2, we see that the range of \(f(x) = \cos^{-1}\left(\frac{x^2}{1 + x^2}\right)\) is \([0, \frac{\pi}{2}]\). This means that for any value of \(x\), the output of this function will always fall within this range.

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