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How would I figure out the following


determine the values for which $f'(x)<0$

I know the derivative is using the chain rule is $2(x^2-9)(2x)$ but how would I figure out the rest.

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$f'(x)$ is not $2(x-9)(2x)$. It is $2(x^2-9)(2x)$. – user60725 Feb 14 '13 at 22:14
yes you are right. – Fernando Martinez Feb 14 '13 at 22:15
@FernandoMartinez: Now the you have the correct equation, can you proceed? It sometimes helps to look at a plot to validate your analysis. Regards – Amzoti Feb 14 '13 at 22:16
@amzoti It seems that f(x)<0 is for all real numbers less than zero? – Fernando Martinez Feb 14 '13 at 22:18
@FernandoMartinez: Compare Ross M's nice answer with the plot and what do you get? Regards – Amzoti Feb 14 '13 at 22:19

Expand your derivative to $4x(x-3)(x+3) \lt 0$. It will change sign at $-3, 0 , +3.$ Since the inequality is true when $x$ is very large and negative, the solution is $x \in (-\infty,-3) \cup (0,3)$

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yes I like this answer. – Fernando Martinez Feb 14 '13 at 22:27

This is something you usually do in precalculus, analyzing where a polynomial is negative/positive given its roots.

$4x(x^2-9) < 0 \iff x(x-3)(x+3) < 0.$ Then you analyze $x$ between the zeros.

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$f(x)=(x^2-9)^2$ $$f'(x)=2(x^2-9)2x=4x(x-3)(x+3)<0\Rightarrow x\in(-\infty,-3)\cup (0,3)$$

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thanks for your answer. – Fernando Martinez Feb 14 '13 at 22:29
you are wellcome – Adi Dani Feb 14 '13 at 22:32

Solve the inequality $f'(x)<0$. That is, $$f'(x)=4x(x^2-9)=4x^3-36x<0\implies 4x^3<36x\implies x^3<9x$$ For $x>0$, the final inequality above becomes $$x^2<9\implies x\in(-3,3)$$ so that $x\in(0,3)$. For $x<0$, the same inequality becomes $$x^2>9\implies x\in (-\infty, -3)\cup (3, \infty)$$ so that $x\in(-\infty, -3)$. For $x=0$, the inequality is clearly not satisfied.

So the final solution is $$x\in(-\infty, -3)\cup(0,3).$$

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thanks for your answer. – Fernando Martinez Feb 14 '13 at 22:33

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