A cubic function `f(x)` vanishes at `x=-2` and has relative minimum/maximum at `x=-1a n dx=1/3ifint_(-1)^1f(x)dx=(14)/3dot` Find the cubic function `f(x)dot`

Answer

`f(x)=x^(3)+x^(2)-x+2`

Explanation

Since , f(x) is a cubic polynomial . Therefore , f ' (x) is a quadratic polynomial and f(x) has relative maximum and minimum at `x=(1)/(3)andx=-1` respectively, therefore , `-1 and 1//3 ` are the roots of f ' (x) =0.
`:. f' (x) =a (x+1)(x(1)/(3))=a(x^(2)-(1)/(3)x+x-(1)/(3))`
`=a(x^(2)+(2)/(3)x-(1)/(3))`
Now , integrating w . r . t. x , we get
`f(x) =a ((x^(3))/(3)+(x^(2))/(3)-(x)/(3))+c`
where , c is constant of integration. Again , `f(-2)=0`
`:. f (-2)=a(-(8)/(3)+(4)/(3)+(2)/(3))+c`
`rArr 0=a ((-8+4+2)/(3))+c`
`rArr 0= (-2a)/(3)+crArrc=(2a)/(3)`
`:. f(x) =a ((x^(3))/(3)+(x^(2))/(3)-(x)/(3))+(2a)/(3)=(a)/(3)(x^(3)+x^(2)=x+2)`
Again , ` int _(-1)^(1)f(x)dx=(14)/(3)`
`rArrint_(-1)^(1)(a)/(3)(x^(3)+x^(2)-x+2) dx = (14)/(3)`
`rArr int_(-1)^(1)(a)/(3)(0+x^(2)-0+2)dx=(14)/(3)`
`[:' y = x^(3) and y =- x` are odd functions]
` rArr (a)/(3)[2int_(0)^(1)x^(2)dx +4int_(0)^(1)1 dx ]=(14)/(3)`
`rArr (a)/(3)[((2x^(3))/(3)+4x)]_(0)^(1)=(14)/(3)`
`rArr(a)/(3)((2)/(3)+4)=(14)/(3)rArr(a)/(3)((14)/(3))=(14)/(3)`
`rArr a= 3`
Hence , `f(x)=x^(3)+x^(2)-x+2`