Question

How do you find the implied range and domain of $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right] $ ?

Answer 1

Hint: We explain the function $ \arccos \left( x \right) $ . We express the inverse function of cos in the form of $ \arccos \left( x \right)={{\cos }^{-1}}x $ . We find the range and domain for $ \arccos \left( x \right)={{\cos }^{-1}}x $ . From there we replace the values to find the range and domain for $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right] $ .

Complete step-by-step answer:
The given expression is the inverse function of trigonometric ratio cos.
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $ 2\pi $ .
The general solution for that value where
 $ \cos \alpha =x $ will be $ 2n\pi \pm \alpha ,n\in \mathbb{Z} $ .
But for $ \arccos \left( x \right) $ , we won’t find the general solution. We use the principal value. For ratio cos we have $ 0\le \arccos \left( x \right)\le \pi $ .
Therefore, the range for function $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right] $ is $ 2n\pi \pm \theta ,n\in \mathbb{Z} $ where $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right]=\theta $ .
Eventually the range becomes the whole real space. So, the actual range is $ \left( -\infty ,\infty \right) $ .
Now we try to find the domain for $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right] $ .
We know that the principal domain for $ \arccos \left( x \right)={{\cos }^{-1}}x $ is $ \left[ -1,1 \right] $ . This gives $ x\in \left[ -1,1 \right] $ .
Replacing the value for $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right] $ , we get $ {{\left( x-1 \right)}^{2}}\in \left[ -1,1 \right] $ .
Now we know that square value is always greater than or equal to 0.
This gives $ {{\left( x-1 \right)}^{2}}\in \left[ 0,1 \right] $ . The simplified form is $ \left( x-1 \right)\in \left[ -1,1 \right] $ .
Now we add 1 to the equation and get $ x\in \left[ 0,2 \right] $ .
Therefore, the domain of $ \arccos \left[ {{\left( x-1 \right)}^{2}} \right] $ is $ \left[ 0,2 \right] $
So, the correct answer is “ Domain is $ \left[ 0,2 \right] $ and Range is $ \left( -\infty ,\infty \right) $ ”.

Note: If we are finding an $ \arccos \left( x \right) $ of a positive value, the answer is between $ 0\le \arccos \left( x \right)\le \dfrac{\pi }{2} $ . If we are finding the $ \arccos \left( x \right) $ of a negative value, the answer is between $ \dfrac{\pi }{2}\le \arccos \left( x \right)\le \pi $ .