To determine the domain of the function [tex]\( f(x) = 2 \sqrt{-x^2 + 10x} \)[/tex], we need to ensure that the expression inside the square root is non-negative. This is because the square root of a negative number is not defined in the real number system.
1. Start with the expression inside the square root: [tex]\(-x^2 + 10x \geq 0\)[/tex].
2. Factor the quadratic expression: [tex]\(-x^2 + 10x = 10x - x^2 = x(10 - x)\)[/tex].
3. Set up the inequality: [tex]\(x(10 - x) \geq 0\)[/tex].
4. Determine the values of [tex]\(x\)[/tex] that satisfy this inequality. The expression [tex]\(x(10 - x)\)[/tex] will be zero at [tex]\(x = 0\)[/tex] and [tex]\(x = 10\)[/tex].
5. Analyze the sign of the expression [tex]\(x(10 - x)\)[/tex] within the interval [tex]\([0, 10]\)[/tex]:
- For [tex]\(x\)[/tex] in [tex]\(0 \leq x \leq 10\)[/tex], the expression [tex]\(x(10 - x)\)[/tex] is non-negative because it represents the product of two non-negative numbers within this interval.
- Outside this interval, the expression [tex]\(x(10 - x)\)[/tex] becomes negative because one or both factors would be negative.
Thus, the function [tex]\( f(x) \)[/tex] is defined for [tex]\( x \)[/tex] in the interval [tex]\([0, 10]\)[/tex].
Hence, the domain of the function is [tex]\( 0 \leq x \leq 10 \)[/tex].
