Hallo,
y= 0,2x³+0,2x²-4x
y'= 0.6 x^2 +0.4 x -4 =0
0.6 x^2 +0.4 x -4 =0 |: 0.6
x^2 + (2/3) x -20/3 =0 z.B. pq-Formel
x1.2= -1/3 ± √(1/9 +20/3)
x1.2= -1/3 ± √61/9
x1.2= -1/3 ± (√61)/3
x1= -1/3 + (√61)/3 ≈ 2.27
x2= -1/3 - (√61)/3 ≈ -2.94
dann noch den y-Wert berechnen