Differentiation and Critical Points

[Hoslerd]

This is the equation: f(x)=x(x[sup:2xgcvpvu]2[/sup:2xgcvpvu]-7)[sup:2xgcvpvu]3[/sup:2xgcvpvu]
The first part is to find f'(x) and I got 7x[sup:2xgcvpvu]6[/sup:2xgcvpvu]-105x[sup:2xgcvpvu]4[/sup:2xgcvpvu]+441x[sup:2xgcvpvu]2[/sup:2xgcvpvu]-343
The next step of the problem is to find the critical points of the function. So far I have -1, and 1 but there's supposed to be more and I can't figure them out.
Then i need to find f"(x) and I got 42x[sup:2xgcvpvu]5[/sup:2xgcvpvu]-420x[sup:2xgcvpvu]3[/sup:2xgcvpvu]+882x
Finally, I needed to find the inflection points by setting f"(x) equal to zero and solving. The only point I got was 0 but again, there should be more.
Please tell me if I'm on the right track and then help me find the remaining critical points and inflection points. The help is much appreciated.

 

[galactus]

Your derivative may be best expressed factored as:

\(\displaystyle f'(x)=7(x^{2}-7)^{2}(x^{2}-1)\)

Now, you should be able to see the other zeros.

\(\displaystyle f''(x)=42x(x^{2}-7)(x^{2}-3)\)

Now, the inflection points.