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$$f'(x)= (x+1)^2(x-4)^5(x-2)^4 $$

For the critical points, I got $x = -1,2,$ and $4$. According to my professor's answer key, the interval is increasing on $(4, \infty)$ and decreasing on $(-\infty,4)$. I got increasing on $(-1,2)$ and $(4,\infty)$ and decreasing on $(-\infty,1)$ and $(2,4)$. Why are my answers wrong and why doesn't my professor consider the $-1$ and $2$ when determining where the function increases/decreases?

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Observe that $$(x+1)^2(x-2)^4(x-4)^4\ge0$$ for real $x$

So, the sign of $f'(x)$ will be dictated by that of $x-4$ unless $(x+1)^2(x-2)^4\ne0$

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  • $\begingroup$ Would you care to elaborate? $\endgroup$ – iii Nov 21 '13 at 12:04
  • $\begingroup$ @igknighton, So, $f'(x)>0$ if $x<-1,-1<x<2,2<x<4, x>4$ excluding the critical points $\endgroup$ – lab bhattacharjee Nov 21 '13 at 12:05
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$x^{2k} \geq 0$ for every $x \in \mathbb{R}$ and for every $k \in \mathbb{N}$. Hence you should study $(x-4)^5 \geq 0$ only.

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You can make a table of the values:

  • if $x<-1$, then $f'(x)<0$

  • $f'(-1)=0$

  • if $-1<x<2$, then $f'(x)<0$

  • $f(2)=0$

  • if $2<x<4$, then $f'(x)<0$

  • $f(4)=0$

  • if $x>4$, then $f'(x)>0$

The checks are straigthforward.

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