The function `f(x)=sqrt(a x^3+b x^2+c x+d)` has its non-zero local minimum and local maximum values at `x=-2 and x = 2`, respectively. If `a` is a root of `x^2-x-6=0`, then find a,b,c and d.

Answer

`ararrs;brarrp;crarrq;drarrr`

Explanation

Since f(X) is minimum at x=-2 and maximum at x=2 let
`g(X)=ax^(3)+bx^(2)+cx+d`
Thus g(X) is also minimum at x=-2 and maximum at x=2 thus `alt0`
since a is a root of `x^(2)-x-6=0 i.e x=3,-2` we get a=-2
Then `g(X) =-2x^(3)+bvx^(2)+cx+d`
`therefore g(X)=-6x^(2)+2bx+c=-6(x+2)(x-2)`
on comparing we get

Since minimum and maximum values are positive
`g(-2)gt0 rarr 16-48+dgt0 rarr dgt32`
and `g(x2)gt0 rarr -16+48dgt0 rarrdgt-32`
It is clear that `dgt32`
Hence `a=-2,b=0,c=24,dlt32`