Question

state whether the function is bounded above, bounded below or bounded. y = \sqrt{1 - x^{2}} a. bounded above b. bounded below c. bounded please select the best answer from the choices provided.

Explanation:

Step1: Analyze domain of function

The function is $y = \sqrt{1 - x^{2}}$. For the square - root to be well - defined, $1 - x^{2}\geq0$, which gives $- 1\leq x\leq1$.

Step2: Find range of function

Let $x=\sin\theta$, where $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$. Then $y = \sqrt{1-\sin^{2}\theta}=\vert\cos\theta\vert$. Since $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$, $y = \cos\theta$. The range of $\cos\theta$ for $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$ is $[0,1]$.

Step3: Determine boundedness

Since the range of $y$ is $[0,1]$, the function is bounded above by $1$ and bounded below by $0$. So it is bounded.

Answer:

C. bounded