If (f(x)=\sqrt{x+2}+\sqrt{8-x}), what is the minimum value of (f)?

Correct answer: ( \sqrt{10} )

The square roots require x+2≥0 and 8−x≥0, so the domain is the closed interval [−2,8]. Let a=x+2 and b=8−x. Then a,b are nonnegative and a+b=10. By the inequality (√a+√b)²=a+b+2√(ab)≥a+b=10, we obtain f(x)≥√10. Equality occurs when ab=0, which happens at x=−2 or x=8. Therefore the minimum is √10.

Option A is correct. The value 2√5 is larger than √10 and is actually the maximum, reached when the two radicands are equal. The value 0 is impossible because both square roots cannot vanish at the same x. The supplied explanation gives the correct endpoint idea, although the inequality makes the minimum argument more explicit.