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I would like to study the sign of this derivate, but I don't know where to start : http://www.wolframalpha.com/input/?i=derivate+sqrt%28x%5E4-7x%5E2%2B16%29

Pop Flamingo
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2 Answers2

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You want to look at the sign of the following function: $$f(x) = \frac{x(2x^2 - 7))}{\sqrt{ 16 - 7x^2 + x^4 } }$$

The first thing we should do is make sure the expression under the radical is always positive. To do so, we complete the square: $$ x^4 - 7x^2 + 16 = \left( x^4 - 7x^2 + \left(\frac{7}{2}\right)^2 \right) + 16 - \left(\frac{7}{2}\right)^2 = \left( x^2 - \frac{7}{2} \right)^2 + \frac{15}{4}$$

Since this expression is always positive, the denominator of $f(x)$ will be well-defined and positive. So the sign of $f(x)$ will be the same as the sign of $x(2x^2 - 7)$. Can you get the rest from here?

Henry Swanson
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  • Thank you, but I am not getting how you "complete the square"), what is $(\frac{7}{2})^{2}$ ? – Pop Flamingo Sep 19 '14 at 14:41
  • It's the same idea as completing the square for a quadratic. I'll write $y$ instead of $x^2$. We want to turn $y^2 - 7y + 16$ into the form $(y + c)^2 + d$. We know it's not a perfect square, but it is off by a constant. Since $(y + c)^2 = y^2 + 2cy + c^2$, we know $2c$ should be $-7$. Then, to make clear what's going on, we add $c^2 = (7/2)^2$ and subtract it. Then we have the perfect square that we wanted, plus a constant, which turns out positive. – Henry Swanson Sep 19 '14 at 17:20
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Study the sign of the numerator, as the denominator is always positive(try to prove it putting $\ x^2=t$): $\ x>0 \text{ and }2x^2-7>0$ you will get $\ f'(x)>0 \iff -\sqrt{7/2}<x<0, x> \sqrt{7/2}$

Mosk
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