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Find the relative extrema of the function by applying the first-derivative test:

$$f(x)=x^5-5x^3-20x-2$$

So I found the $f'(x)$

$$f'(x) = 5x^4-15x^2-20$$

Now, I'm trying to find the critical values, which $x=0$ or undefined, so I can apply the first-derivative test. However, I can't simpliy this. How can I find the relative extrema now? Thanks.

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2 Answers 2

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To find the critical values, set $f'(x)=0$
This function certainly can be simplified:$f'(x)=5x^4−15x^2−20=5(x^4-3x^2-4)=5(x^2-4)(x^2+1)=5(x+2)(x-2)(x^2+1)$

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Hint. If one set $X=x^2$ then one has to solve a classic quadratic equation $$ 5X^2-15X-20=0. $$

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