# Let` f(x) = ax^3 + 6x^2 + bx + 4` . Determine the constants a and b so that f has a relative minimum at x = -1 and a relative maximum at x = 2.

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`f(x)=ax^3+6x^2+bx+4`

Differentiating with respect to x,

`f'(x)=3ax^2+12x+b`

Differentiating again with respect to x,

`f''(x)=6ax+12`

f has a relative minimum at x=-1 and a relative minimum at x=2,

For a function to have a relative extremum (maximum/minimum) at a particular point, the first derivative at this point must be equal to zero and the second derivative should be less than zero for maximum and greater than zero for minimum.

Put the first derivatives at x=-1 and also at x=2 equal to zero to get two equations:

`f’(-1)=3a(-1)^2+12(-1)+b=0`

`rArr 3a+b-12=0`

`rArr 3a+b=12` --- (i)

and

`f’(2)=3a(2)^2+12(2)+b`

`rArr 12a+b+24=0`

`rArr 12a+b=-24` --- (ii)

(ii)-(i),

`9a=-24-12=-36`

`rArr a=-36/9=-4`

Put this value of a in eq. (i),

`3(-4)+b=12`

`rArr -12+b=12`

`rArr b=24`

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Check the values of the second derivative at these two points:

At x=-1, `f’’(-1)=6*(-4)*(-1)+12`

=36,

>0

At x=2, `f’’(2)=6*(-4)*(2)+12`

=-36,

<0

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Therefore, for **a=-4 and b=24**, the function `f(x)=ax^3+6x^2+bx+4` has a relative minimum at x=-1 and a relative maximum at x=2.