F(x) = 3/√(-1/4х²+х +3) [-1; 3]
решение
ОДЗ
(-1/4х²+х +3) > 0
корни -2 и 6
ОДЗ х∈(-2;6)
f(x) = 3/√(-1/4х²+х +3) = (-1/4х²+х +3)^-1/2
f' (x) = -3/2(-1/4х²+х +3)^-3/2= -3/2*1/√ (-1/4х²+х +3)^3 * (-1/2 x +1)
-3/2*1/√ (-1/4х²+х +3)^3 * (-1/2 x +1) = 0
(-1/2 x +1) = 0
-1/ х = -1
х = 2
а) х = -1
f(-1) = 3/√(3 -1 -1/4) = 6/√7 = 6√7/7
б)х = 2
f(2) = 3/√(3 +2 -1) =6/√4 = 3
в) х = 3
f(3) = 3/√(3 +3 -9/4) = 6/√15 = 6√15/15 = 2√15/5
min f(x) = f(2) = 3
[-1;3]