Q. Let `f(x) = ax^3+bx^2+x+d` has local extrema at `x=alpha` and `beta`

such that `alpha,beta<0 , f(alpha).f(beta)>0` ;Then the equation `f(x)=0` has

A) 3 distinct real roots

B) has only one real root if `af(alpha)>0`

C) has only one real root if `af(beta)>0`

D) has 3 equal real roots

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Given `f(x)=ax^3+bx^2+x+d` with extrema at `x=alpha,beta` and `alpha,beta<0` and `f(alpha),f(beta)>0` :

A) The graph cannot have 3 distinct real roots. There are two turning points (extrema), and both of these are positive. Thus the graph can only change from positive to negative (or negative to positive) once. (A cubic can have at most two relative extrema.)

If you have calculus, the sign of the derivative can change at most twice.

D) The graph cannot have 3 equal real roots, as this implies that it has no extrema.

B) C) Both of these would be correct. If `af(alpha)>0` , then `f(alpha)>0==> a>0` . If a>0, then `af(beta)>0` .

If you have calculus:

`f'(x)=3ax^2+2bx+1`

Use the quadratic formula to find the zeros of f'(x) -- these are the critical points, or the x values where the extrema occur.

`f'(x)=0 ==> x=-b/(3a)+-sqrt(b^2-3a)/(3a)`

Suppose a<0; then `-b/(3a)-sqrt(b^2-3a)/(3a)>0` . But the two extrema occured when x<0, so a must be greater than zero.

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Both B and C are correct.

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An example graph is `f(x)=3x^3+5x^2+x+3`

Then `alpha=-1,beta=-1/9,f(alpha)=4,f(beta)~~2.947` so the requirements are satisfied. The graph:

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