Исследуйте функцию и постройте ее график у=0,5 х^4-4x^2

У=0,5 х^4-4x^2 Substitution x²=u f (x)=0,5u²-4u0,5u²-4u=0u (0,5u-4)=0u₁=0 0,5u-4=0 0,5=4u=8 Resubstitution x²=8x=±√8 Tochki peresichenija aksy X P₁ (0; 0) P₂ (√8; 0) P₃ (-√8; 0) -Teper reshaem gde nahodjatsja Extrema f' (x)=2x³-8x=02x (x²-4)=02x=0x₁=0 x²-4=0 x²=4 x₂,₃=±2Teper prowerjaem eti tochki na maximum ili na minimum, dlja etogo nam nuzhna 2 proizwodnaja f' (x)=6x²-8=06×0-8=-8<0 menshe nolja znachit Maximum6×2²-8=16> bolshe nolja znachit minimum6× (-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) — Teper reshaem tochku peregiba, dlja etogo nam nuzhen f' (x)=0 f' (x)=6x²-8=06x²=8x²=4/3x₁,₂=√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)