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f(x)=(1/3x+2)³+12x
![1) (x^3/27+2*x^2/3+16x+8)+12x= (16x)'+(x^3/27)'+(2*x^2/3)+(8)'+(12x)'= 1) (x^3/27+2*x^2/3+16x+8)+12x= (16x)'+(x^3/27)'+(2*x^2/3)+(8)'+(12x)'=](https://tex.z-dn.net/?f=1%29+%28x%5E3%2F27%2B2%2Ax%5E2%2F3%2B16x%2B8%29%2B12x%3D+%2816x%29%27%2B%28x%5E3%2F27%29%27%2B%282%2Ax%5E2%2F3%29%2B%288%29%27%2B%2812x%29%27%3D)
![=16+x^2/9+4x/3+12 =16+x^2/9+4x/3+12](https://tex.z-dn.net/?f=%3D16%2Bx%5E2%2F9%2B4x%2F3%2B12)
2)![x^{2}/9+4x/3+16+12=(x/3+2)^2+12 x^{2}/9+4x/3+16+12=(x/3+2)^2+12](https://tex.z-dn.net/?f=+x%5E%7B2%7D%2F9%2B4x%2F3%2B16%2B12%3D%28x%2F3%2B2%29%5E2%2B12)
3)![f'(-3)=(-3/3+2)^2+12=(-1+2)^2+12=1-4+4+12=13 f'(-3)=(-3/3+2)^2+12=(-1+2)^2+12=1-4+4+12=13](https://tex.z-dn.net/?f=f%27%28-3%29%3D%28-3%2F3%2B2%29%5E2%2B12%3D%28-1%2B2%29%5E2%2B12%3D1-4%2B4%2B12%3D13)
Ответ: 13
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Ответ: 13
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