Find value of n such the function has local mimima at x=1. If $f$ is defined by
$$f(x)=(x^2-1)^n(x^2+x-1)$$
then $f$ has a local minimum at $x=1$, when


*
(i) $n=2$


(ii) $n=3$

(iii) $n=4$

 (iv) $n=5$
Multiple options are correct. 

The given answer is $n=2$ and $n=4$.

I tried putting derivative equal to zero and double derivative greater than zero, but both of them were independent of $n$.

$x=1$ is a local minimum if it is a double root for $f$. Since $1$ is not a root of $x^2+x-1$ it must be a even root for $(x^2-1)^n = (x+1)^n(x-1)^n$ and that is when $n$ is even.
