Answer :
To find the domain of the function [tex]\( f(x) = 2 \sqrt{-x^2 + 10x} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which the expression under the square root is non-negative. Therefore, we need to solve the inequality:
[tex]\[ -x^2 + 10x \geq 0 \][/tex]
This can be rewritten as:
[tex]\[ x(10 - x) \geq 0 \][/tex]
The critical points occur where the expression [tex]\( x(10 - x) \)[/tex] equals zero, which gives us:
[tex]\[ x = 0 \quad \text{and} \quad x = 10 \][/tex]
To determine the intervals where [tex]\( x(10 - x) \)[/tex] is non-negative, we test values in the intervals [tex]\( (-\infty, 0) \)[/tex], [tex]\( (0, 10) \)[/tex], and [tex]\( (10, +\infty) \)[/tex].
- For [tex]\( x \in (-\infty, 0) \)[/tex], both [tex]\( x \)[/tex] and [tex]\( 10 - x \)[/tex] are of opposite signs, so the product [tex]\( x(10 - x) \)[/tex] is negative.
- For [tex]\( x \in (0, 10) \)[/tex], both [tex]\( x \)[/tex] and [tex]\( 10 - x \)[/tex] are positive, so the product [tex]\( x(10 - x) \)[/tex] is positive.
- For [tex]\( x \in (10, +\infty) \)[/tex], both [tex]\( x \)[/tex] and [tex]\( 10 - x \)[/tex] are of opposite signs again, so the product [tex]\( x(10 - x) \)[/tex] is negative.
Therefore, the expression [tex]\( -x^2 + 10x \)[/tex] is non-negative solely in the interval [tex]\( 0 \leq x \leq 10 \)[/tex].
Thus, the domain of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ 0 \leq x \leq 10 \][/tex]
So, the domain of the function is:
[tex]\[ 0 \leq x \leq 10 \][/tex]
Hence, the correct answer for the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ 0 \leq x \leq 10 \][/tex]
[tex]\[ -x^2 + 10x \geq 0 \][/tex]
This can be rewritten as:
[tex]\[ x(10 - x) \geq 0 \][/tex]
The critical points occur where the expression [tex]\( x(10 - x) \)[/tex] equals zero, which gives us:
[tex]\[ x = 0 \quad \text{and} \quad x = 10 \][/tex]
To determine the intervals where [tex]\( x(10 - x) \)[/tex] is non-negative, we test values in the intervals [tex]\( (-\infty, 0) \)[/tex], [tex]\( (0, 10) \)[/tex], and [tex]\( (10, +\infty) \)[/tex].
- For [tex]\( x \in (-\infty, 0) \)[/tex], both [tex]\( x \)[/tex] and [tex]\( 10 - x \)[/tex] are of opposite signs, so the product [tex]\( x(10 - x) \)[/tex] is negative.
- For [tex]\( x \in (0, 10) \)[/tex], both [tex]\( x \)[/tex] and [tex]\( 10 - x \)[/tex] are positive, so the product [tex]\( x(10 - x) \)[/tex] is positive.
- For [tex]\( x \in (10, +\infty) \)[/tex], both [tex]\( x \)[/tex] and [tex]\( 10 - x \)[/tex] are of opposite signs again, so the product [tex]\( x(10 - x) \)[/tex] is negative.
Therefore, the expression [tex]\( -x^2 + 10x \)[/tex] is non-negative solely in the interval [tex]\( 0 \leq x \leq 10 \)[/tex].
Thus, the domain of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ 0 \leq x \leq 10 \][/tex]
So, the domain of the function is:
[tex]\[ 0 \leq x \leq 10 \][/tex]
Hence, the correct answer for the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ 0 \leq x \leq 10 \][/tex]
