1. решите уравнение:
f'(x)= f (1), если
f (x)= x^3-2x^2+3x+1
2.f (x)=12-3x^4+5x^6
f (x)= x^3+6 (корень)x
f(x)=(4x-3)(3+4x)
Ответы
Ответ дал:
0
=========== 1 ===========
![f(x)=x^3-2x^2+3x+1\
f'(x)=3x^2-4x+3\
f(1)=1^3-2cdot1^2+3cdot1+1=1-2+3+1=3\
f'(x)=f(1)\
3x^2-4x+3=3\
3x^2-4x=3-3\
3x^2-4x=0\
x(3x-4)=0\
x_1=0\
3x-4=0\
3x=4\
x_2= frac{4}{3} =1 frac{1}{3} f(x)=x^3-2x^2+3x+1\
f'(x)=3x^2-4x+3\
f(1)=1^3-2cdot1^2+3cdot1+1=1-2+3+1=3\
f'(x)=f(1)\
3x^2-4x+3=3\
3x^2-4x=3-3\
3x^2-4x=0\
x(3x-4)=0\
x_1=0\
3x-4=0\
3x=4\
x_2= frac{4}{3} =1 frac{1}{3}](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3-2x%5E2%2B3x%2B1%5C%0Af%27%28x%29%3D3x%5E2-4x%2B3%5C%0Af%281%29%3D1%5E3-2cdot1%5E2%2B3cdot1%2B1%3D1-2%2B3%2B1%3D3%5C%0Af%27%28x%29%3Df%281%29%5C%0A3x%5E2-4x%2B3%3D3%5C%0A3x%5E2-4x%3D3-3%5C%0A3x%5E2-4x%3D0%5C%0Ax%283x-4%29%3D0%5C%0Ax_1%3D0%5C%0A3x-4%3D0%5C%0A3x%3D4%5C%0Ax_2%3D+frac%7B4%7D%7B3%7D+%3D1+frac%7B1%7D%7B3%7D+)
=========== 2 ===========
![f(x)=12-3x^4+5x^6\
f'(x)=-12x^3+30x^5 f(x)=12-3x^4+5x^6\
f'(x)=-12x^3+30x^5](https://tex.z-dn.net/?f=f%28x%29%3D12-3x%5E4%2B5x%5E6%5C%0Af%27%28x%29%3D-12x%5E3%2B30x%5E5)
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![f(x)=x^3+6 sqrt{x} \
f'(x)=3x^2+ frac{6}{2 sqrt{x} } =3x^2+ frac{3}{ sqrt{x} } f(x)=x^3+6 sqrt{x} \
f'(x)=3x^2+ frac{6}{2 sqrt{x} } =3x^2+ frac{3}{ sqrt{x} }](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3%2B6+sqrt%7Bx%7D+%5C%0Af%27%28x%29%3D3x%5E2%2B+frac%7B6%7D%7B2+sqrt%7Bx%7D+%7D+%3D3x%5E2%2B+frac%7B3%7D%7B+sqrt%7Bx%7D+%7D)
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![f(x)=(4x-3)(3+4x)=(4x-3)(4x+3)=16x^2-9\
f'(x)=32x f(x)=(4x-3)(3+4x)=(4x-3)(4x+3)=16x^2-9\
f'(x)=32x](https://tex.z-dn.net/?f=f%28x%29%3D%284x-3%29%283%2B4x%29%3D%284x-3%29%284x%2B3%29%3D16x%5E2-9%5C%0Af%27%28x%29%3D32x)
=========== 2 ===========
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