已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]的最大值和最小值f(2)=-7,f'(2)=-3
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![已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]的最大值和最小值f(2)=-7,f'(2)=-3](/uploads/image/z/5242950-54-0.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dax%5E3%2Bbx%5E2-3x%2B1%26%2347%3B3%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B-4%2C4%5D%E5%B7%B2%E7%9F%A5f%28x%29%3Dax%5E3%2Bbx%5E2-3x%2B1%2F3%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B-4%2C4%5D%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BCf%282%29%3D-7%2Cf%27%282%29%3D-3)
已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]的最大值和最小值f(2)=-7,f'(2)=-3
已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]
已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]的最大值和最小值
f(2)=-7,f'(2)=-3
已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]已知f(x)=ax^3+bx^2-3x+1/3,求函数f(x)在[-4,4]的最大值和最小值f(2)=-7,f'(2)=-3
f(x)=ax^3+bx^2-3x+1/3,
f(2)=8a+4b-6+1/3=-7,
f'(x)=3ax^+2bx-3,
f'(2)=12a+4b-3=-3,b=-3a,
解得a=1/3,b=-1.
∴f(x)=(1/3)x^3-x^-3x+1/3,
f'(x)=x^-2x-3=(x+1)(x-3),
-1
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f(x)=ax^3+bx^2-3x+1/3,
f(2)=8a+4b-6+1/3=-7,
f'(x)=3ax^+2bx-3,
f'(2)=12a+4b-3=-3,b=-3a,
解得a=1/3,b=-1.
∴f(x)=(1/3)x^3-x^-3x+1/3,
f'(x)=x^-2x-3=(x+1)(x-3),
-1
f(-4)=-25,f(4)=-19/3,f(-1)=2,f(3)=-26/3
∴函数f(x)在[-4,4]的最大值为2,最小值为-25.
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f(2)=-7,f'(2)=-3代入可得a=1/3,b=-1
f(x)=(1/3)x^3-x^2-3x+1/3
f'(x)=x^2-2x-3=(x=1)(x-3)
极大值f(-1)
极小值f(3)
f(x)在(-4,-1)和(3,4)递增
(-1,3)递减
比较f(-4)f(-1)f(3)f(4)就可以了
f(-4)=-25
...
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f(2)=-7,f'(2)=-3代入可得a=1/3,b=-1
f(x)=(1/3)x^3-x^2-3x+1/3
f'(x)=x^2-2x-3=(x=1)(x-3)
极大值f(-1)
极小值f(3)
f(x)在(-4,-1)和(3,4)递增
(-1,3)递减
比较f(-4)f(-1)f(3)f(4)就可以了
f(-4)=-25
f(-1)=2
f(3)=-26/3
f(4)=-19/3
综上所述f(x)MAX=f(-1)=2,f(x)MIN=f(-4)=-25
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