The domain of the function $f(x) = \sqrt{25-x^2}+\sqrt{x-2} + \frac{1}{\sqrt{1 - x}}$ is an interval of what width?
The function $\sqrt{25-x^2}$ is defined for $-5\leq x\leq 5$, since $25-x^2$ must be nonnegative. The function $\sqrt{x-2}$ is defined for $x\geq 2$, since $x-2$ must be nonnegative. Lastly, the function $\frac{1}{\sqrt{1-x}}$ is defined for $x<1$, since $1-x$ must be positive. Therefore, the function $f(x) = \sqrt{25-x^2}+\sqrt{x-2} + \frac{1}{\sqrt{1 - x}}$ is defined for $x<1$ and $x\geq2$.
The width of this interval is $\boxed{\infty}$.
To determine the domain of the function $f(x) = \sqrt{25-x^2}+\sqrt{x-2} + \frac{1}{\sqrt{1 - x}}$, we need to consider the restrictions on the square roots and division.
1. Square Roots: The expression $\sqrt{25-x^2}$ requires the quantity under the square root to be non-negative, which means $25-x^2\geq 0$. Solving this inequality, we find that $-5 \leq x \leq 5$.
2. Square Roots: The expression $\sqrt{x-2}$ requires $x-2\geq 0$. Solving this inequality, we find that $x \geq 2$.
3. Division: The expression $\frac{1}{\sqrt{1 - x}}$ requires the denominator $\sqrt{1 - x}$ to be non-zero. Thus, $1 - x \neq 0$, which implies $x \neq 1$.
Putting all the restrictions together, we have $-5 \leq x \leq 5$ and $x \geq 2$, with $x \neq 1$. The common interval that satisfies all the conditions is $[2, 5)$, which means the domain of the function is an interval of width $5 - 2 = \boxed{3}$.
To find the domain of a function, we need to determine the values of $x$ for which the function is defined. In this case, there are three parts to consider: $\sqrt{25-x^2}$, $\sqrt{x-2}$, and $\frac{1}{\sqrt{1-x}}$.
For $\sqrt{25-x^2}$ to be defined, the value inside the square root must be non-negative. So we have $25-x^2 \geq 0$. This inequality implies that $-5 \leq x \leq 5$.
For $\sqrt{x-2}$ to be defined, we need $x-2 \geq 0$. Solving this inequality gives $x \geq 2$.
For $\frac{1}{\sqrt{1-x}}$ to be defined, we require $\sqrt{1-x} \neq 0$, which means $1-x > 0$. This inequality implies that $x < 1$.
To find the overlapping interval from these three conditions, we take the intersection of the intervals $-5 \leq x \leq 5$, $x \geq 2$, and $x < 1$. The only interval that satisfies all three conditions is $2 \leq x < 1$, which means the domain of the function is the interval $(2, 1]$.
The width of an interval is defined as the difference between the endpoints. Therefore, the width of the interval $(2, 1]$ is $2 - 1 = 1$. Hence, the width of the domain is 1.