diketahui Fungsi g(x) = -1/4 x^3 + A^2x+2, A konstanta. jika f(x) = g(3-x) dan fungsi f(x) turun pada interval 1

Nilai minimum

g(x)= - 1/4 x³ + a² x + 2

f(x) =  g(3-x)
f(x) = -1/4 (3-x)³ + a²(3-x) + 2
f(x) turun jika f' (x) < 0

3(-1/4)(3-x)²(-1) - a² < 0
3/4(3-x)² - a²  < 0
3(3-x)² - 4a²  < 0
3(9-6x + x²) - 4a² <0
3x² -18x + 27 - 4a² < 0
x² - 6x + (9 - 4/3 a²)  =  (x - 1)(x -5)
9 - 4/3 a² = 5
27 - 4a² = 15
-4a² = - 12
a² = 3
a = √3

g(x)= -1/4 x³ + a² x + 2
g(x)= - 1/4 x³ + 3x + 2

g '(x)= 0
-3/4 x² + 3 = 0
-3/4 x² = - 3
x² = 4
x = 2  atau x = - 2
g(x)= -1/4 x³ + 3x + 2
g(2) = -1/4 (2)³ + 3(2) + 2 =  -2 + 6 + 2 = 6
g(-2) = -1/4(-2)³ + 3(-2) + 2 = 2 - 6 + 2 = -2

nilai minimum g(x) = -2 , untuk x = - 2