Find the values of x, for which the function `f(x)=x^(3)+12x^(2)+36x+6` is increasing.

Explanation

`f(x)=x^(3)+12x^(2)+36x+6`
`:.f'(x)=("d")/("dx")(x^(3)+12x^(2)+36x+6)`
`=3x^(2)+12xx2x+36xx1+0`
`=3x^(2)+24x+36`
f is increasing if `f'(x) gt 0`
i.e. if `3x^(2)+24x+36 gt 0`
i.e. if `x^(2)+8x+12 gt 0`
i.e. if `x^(2)+8x gt -12`
i.e. if `x^(2)+8x+16 gt 16-12`
i.e. if `(x+4)^(2) gt 4`
i.e. if `x+4 gt 2" or "x+4 lt -2`
i.e. if `x gt 2" or "x lt -6`.
`:.` f is increasing, if `x gt 2" or "x lt -6`.
Hence, f is increasing if `x in (-oo, -6)uu(2, oo)`.