If the solve the problem

(b) a minimum at $x=1$

Given: $f(x)=x^{3}+3 x^{2}-9 x+2$

$\Rightarrow f^{\prime}(x)=3 x^{2}+6 x-9$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 3 x^{2}+6 x-9=0$

$\Rightarrow x^{2}+2 x-3=0$

$\Rightarrow(x+3)(x-1)=0$

$\Rightarrow x=-3,1$

Now,

$f^{\prime \prime}(x)=6 x+6$

$\Rightarrow f^{\prime \prime}(1)=6+6=12>0$

So, $x=1$ is a local minima.

Also,

$f^{\prime \prime}(-3)=-18+6=-12<0$

So, $x=-3$ is a local maxima.