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f'x=6x² -12 x =0 ..... x=0 dan x=2
f'' x= 12 x -12
x=0 ... f''x= -12 <0 ... maksimum
x=2 .... f"x=12>0 .... minimum
2. Tentukan titik stasioner dari fungsi f(x) = x³ + 6x² - 15x + 3 dan jenisnya
f'x=3x² + 12x -15=0 .... (3x -1)(x+5)=0
x=1/3 dan x= -5
f"x=6x+12
x=1/3..... 6/3+12 = 14>0 minimu
x= -5 ..... 6(-5) +12 = - 18 <0 mqksimu