How many critical points does the function f(x) = (x+2)^5 (x^2-1)^4 have?

Apr 16, 2015

5 critical points 1, -1, -2 , $\frac{- 8 \pm \sqrt{129}}{13}$

Find f'(x) = $5 {\left(x + 2\right)}^{4} {\left({x}^{2} - 1\right)}^{4} + {\left(x + 2\right)}^{5} 4 {\left({x}^{2} - 1\right)}^{3} 2 x$

= ${\left(x + 2\right)}^{4} {\left({x}^{2} - 1\right)}^{3} \left(5 {x}^{2} - 5 + 8 {x}^{2} + 16 x\right)$

=${\left(x + 2\right)}^{4} {\left({x}^{2} - 1\right)}^{3} \left(13 {x}^{2} + 16 x - 5\right)$

Critical points would be that make f '(x) =0

These would be x=1,-1, -2,$\frac{- 8 \pm \sqrt{129}}{13}$