tentukan nilai stasioner titik stasioner dan jenis titik stasioner a. f(x)=⅓x³-½x²-2x+3 b.f(x)=3x⁴+4x³-12x³12

f (x) = 1/3x^3 -1/2x^2 -2x +3
f' (x) = x^2 -x -2
stasioner didapat dari f' (x) = 0
x^2 -x -2 = 0
(x + 1)(x -2) = 0
x1 = -1 dan x2 = 2
f (-1) = 1/3(-1)^3 -1/2(-1)^2-2(-1)+3
= -1/3-1/2+2+3
= -5/6+5
= 25/6 ==> nilai stasioner
f (2) = 1/3(2)^3-1/2(2)^2-2(2)+3
= 8/3-2-4+3
= 8/3-3
= -1/3 ==> nilai stasioner
Titik stasioner (-1, 25/6), (2, -1/3)
f" (x) = 2x -1
f" (-1) = 2(-1)-1
f" (-1) = -3. (f"(x) < 0, maka maksimum
f" (2) = 2(2)-1
f" (2) = 3 (f"(x) >0, maka minimum
Jadi (-1, 25/6) adalah puncak maksimum
(2, -1/3) puncak minimum