What is the domain of (f(x)=\sqrt{6x-x^2-5})?

Correct answer: ( [1,5] )

The direct answer is A: [1,5]. For a real square root, the expression inside it must be at least zero. Thus 6x-x^2-5≥0. Rearranging gives -x^2+6x-5≥0, or -(x^2-6x+5)≥0. Factoring gives -(x-1)(x-5)≥0. Since the quadratic opens downward, its value is non-negative between its roots, including the roots themselves. Hence 1≤x≤5, so the domain is [1,5]. Option A is correct because both endpoints make the radicand zero and are allowed. Option B, (-∞,1]∪[5,∞), reverses the sign region; outside the roots the downward quadratic is negative. Option C, (1,5), wrongly excludes x=1 and x=5, although a zero radicand gives √0=0 and is valid. Option D, all real numbers, is wrong because values such as x=0 give -5 inside the square root. A reliable method is to find the roots, test one point between them and one point outside, and remember that zero is allowed under a square root.