If the function `f(x)=x^4+b x^2+8x+1` has a horizontal tangent and a point of inflection for the same value of `x` then the value of `b` is equal to `-1` (b) 1 (c) 6 (d) `-6`

Options

A. `-2`
B. `-6`
C. 6
D. 3

Explanation

According to the question we must have
`f'(x)=0 and f''(x)=0` for the same `x=x_(0)`
now `f'(x)=4x^(3)+2bx+8`
`therefore" "f'(x_(0))=2[2x_(0)^(3)+bx_(0)+4]=0" (i)"`
`"and "f''(x_(0))=2[6x_(0)^(2)+b]=0" (ii)"`
From (ii) `b=-6x_(0)^(2)`
Substituting this value of b in (i)
`2x_(0)^(3)+(-6x_(0)^(2))+4=0 rArr 4x_(0)^(3)=4`
`rArr" "x_(0)=1.`
`rArr" "b=-6`