Mohon bantuan.... Nilai maksimum fungsi f(x)= -x'3+12x+3 pada interval -1 <_ x <_ 3 adalah...

f'(x) = -3x^2 + 12

f'(x) = 0
-3x^2 + 12 = 0
-3×(x^2 - 4) =0
-3×(x-2)(x+2) = 0
x = 2 atau x = -2

uju turunan kedua

f''(x) = -6x

x = 2 ==> f''(2) = -6×2 = -12 > 0 [ maksimum]
x = -2 ==> f''(-2) = -6×-2 = 12 < 0 [ minimum ]

nilai maksimumya :
f(2) = -(2)^3 + 12×2 + 3
= -8 + 24 + 3
= 19

f(x) = -x^3 + 12x + 3
f'(x) = -3x^2 + 12 = 0
=> -3(x^2 - 4) = 0
=> -3(x + 2)(x - 2) = 0
=> x = -2 atau x = 2
--- (-2) +++ (2) ---
Karena maksimum maka x = 2 (++ --)
Nilai maksimum = f(2) = -2^3 + 12(2) + 3 = -8 + 24 + 3 = 19