Let `f(x)=x^3+3x^2-9x+2` . Then, `f(x)` has a maximum at `x=1` (b) a minimum at `x=1` (c) neither a maximum nor a minimum at `x=-3` (d) none of these

Explanation

`f(x)=x^3+3x^2−9x+2`

`f′(x)=3(x^2+2x−3)=3(x+3)(x−1)=0`

`\Rightarrow x=−3,1`


`f′′(x)=6(x+1)`

now` f′′(−3)<0`

and `f′′(1)>0`

Therefore `f(x)` is maximum at `x=−3 `and minimum at `x=1`