у=0,5х^4-4x^2 Substitution x²=u
f(x)=0,5u²-4u
0,5u²-4u=0
u(0,5u-4)=0
u₁=0 0,5u-4=0
0,5=4
u=8
Resubstitution x²=8
x=±√8 Tochki peresichenija aksy X P₁(0;0) P₂(√8;0) P₃(-√8;0)
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Teper reshaem gde nahodjatsja Extrema
f'(x)=2x³-8x=0
2x(x²-4)=0
2x=0
x₁=0 x²-4=0
x²=4
x₂,₃=±2
Teper prowerjaem eti tochki na maximum ili na minimum, dlja etogo nam nuzhna 2 proizwodnaja
f''(x)=6x²-8=0
6×0-8=-8<0 menshe nolja znachit Maximum</p>
6×2²-8=16> bolshe nolja znachit minimum
6×(-2)²-8=16> bolshe nolja znachit minimum
i eshe my delaem wywod chto parabala semetrichna k x-osi
teper reshaem znachenie y, dlja etogo wstawljaem 0, 2, -2 w f(x)=0
f(x)=0,5×0⁴-4×0²=0 Pmax(0;0)
f(x)=0,5×2⁴-4×2²=-8 Pmin(2;-8)
f(x)=0,5×(-2)⁴-4×(-2)²=8 Pin(-2;8)
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Teper reshaem tochku peregiba, dlja etogo nam nuzhen f''(x)=0
f''(x)=6x²-8=0
6x²=8
x²=4/3
x₁,₂=√4/3≈1,3333333
f(x)=0,5×(√4/3)⁴-4(√4/3)²=-40/9≈-4,444444 Tochka peregiba P(√4/3;-40/9)
f(x)=0,5×(-√4/3)⁴-4(-√4/3)²=-40/9≈-4,444444 Tochka peregiba P(√-4/3;-40/9)