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- With what values of parameter a the function f(x)=1/3x^3+a/2x^2+ax is really growing?
- Answer: 0≤a≤4
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- My solution so far:
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- f'(x)=x^2+ax+a
- Zero points of derivation:
- x^2+ax+a=0
- x=(-a±sqrt(a^2-4a))/2
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- Zero points of derivation: assuming f'(x)=x^2+ax+a
- x^2+ax+a=0 |-x²
- ax+a=-x² | (since x² will always be postive we can get the root now) sqrt(-x²)
- x=sqrt(ax+a)
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