state whether the function is bounded above, bounded below or bounded. y = \\sqrt{1 - x^{2}} a. bounded…

Explanation:

Step1: Determine domain

For $y = \sqrt{1 - x^{2}}$, we need $1 - x^{2}\geq0$. Solving $1 - x^{2}\geq0$ gives $- 1\leq x\leq1$.

Step2: Find range

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

Step3: Analyze bounds

The function has a maximum value of $y = 1$ and a minimum value of $y = 0$. So it is bounded above by $1$ and bounded below by $0$, which means it is bounded.

Answer:

C. bounded