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Chapter 2: Problem 16

Find the range of \(f(x)=\cot ^{-1}\left(2 x-x^{2}\right)\).

Short Answer

Expert verified
The range of the function \(f(x)=\cot^{-1}(2x-x^{2})\) is \((0, \pi)\).

Step by step solution

01

Determine the Domain of the Function

The domain of \(f(x) = \cot^{-1}(2x - x^{2})\) includes all real numbers because the quadratic function \(2x - x^{2}\) is defined for all real numbers. You won't find any undefined x-values, such as division by zero or taking negative square roots, in this expression.
02

Observe the Function \(2x - x^2\)

It's a downward-opening parabola that takes all real values as \(x\) varies over \(-\infty\) to \(\infty\). Since \(2x - x^{2}\) takes all real values, there'll be no problem with the domain when it comes to the inverse cotangent function.
03

Calculate the Range of Inverse Cotangent

The range of \(\cot^{-1}(x)\) is from 0 (exclusive) to \(\pi\) (exclusive). Thus, the range will be \((0, \pi)\).
04

Combine the Information

\(\cot^{-1}\) is defined for all real numbers, and its range is \((0, \pi)\). The function within it, \(2x - x^{2}\), is also defined for all real numbers. So, there's no restriction that could limit the typical range of the \(\cot^{-1}\) function. Therefore, the range of \(f(x) = \cot^{-1}(2x-x^{2})\) is \((0, \pi)\). This includes all possible values of \(f(x)\), apart from 0 and \(\pi\).

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Most popular questions from this chapter

Prove that: $$ \tan ^{-1}\left(\frac{1-x}{1+x}\right)-\tan ^{-1}\left(\frac{1-y}{1+y}\right)=\sin ^{-1}\left(\frac{y-x}{\sqrt{\left(1+x^{2}\right)\left(1+y^{2}\right)}}\right) $$

Prove that: Let \(m=\frac{\left(\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3\right)}{\left(\cot ^{-1} 1+\cot ^{-1} 2+\cot ^{-1} 3\right)}\), then prove that \((m+2)^{m+1}=64\)

Find the values of: $$ \sin ^{-1}(\sin 1)+\sin ^{-1}(\sin 2)+\sin ^{-1}(\sin 3) $$

Find the integral values of \(x\) satisfying the inequality, \(x^{2}-3 x<\sin ^{-1}(\sin 2)\)

Prove that: If \(\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}\), then prove that \(x+y+x y=1\).
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