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I've got some trouble: I don't know how to determine the sign of : $$\frac{4t^2- \sqrt t}{t}$$ for $t > 0$.

Thank you in advance

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  • $\begingroup$ What values can $t$ possibly take? Assuming that you are talking about real numbers, note that $\sqrt{t}$ is only defined for $t \ge 0$. For the rest, note that it's not too hard to find the points where $4x^4 - x < 0$. $\endgroup$ – Johannes Kloos Jan 20 '13 at 15:45
  • $\begingroup$ @Dan1: Is that a correct edit? $\endgroup$ – Dennis Gulko Jan 20 '13 at 15:45
  • $\begingroup$ Yes thanks. In fact : t is higher than 0 (but not 0). Sorry, I don't know how to write it. $\endgroup$ – Dan1 Jan 20 '13 at 16:00
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Since $\sqrt{t}$ is only defined when $t \ge 0$, we can only consider inputs from that range.

Now we have to consider the sign of both the numerator and the denominator. The fraction will be positive whenever the sign of $4t^2 - \sqrt{t}$ matches that of $t$. Clearly though, $t$ is only positive when $t \ge 0$, so we see that the sign of the numerator determines the sign of the fraction.

Consider the equation $4t^2 - \sqrt{t} = 0.$ We can solve this as follows: $$4t^2 = \sqrt{t}$$ $$(4t^2)^2 = t$$ $$16t^4 = t$$ $$16t^4 - t = 0$$ $$t(16t^3 - 1) = 0.$$

So the numerator equals zero whenever $t = 0$ (which is not on our interval) or $16t^3 - 1 = 0$. Solving this last equation we see that the numerator equals zero when $t = (\frac{1}{16})^\frac{1}{3}$.

We now pick a point larger than this value, say t = 1. Plugging in, we see that $4(1)^2 - \sqrt{1} = 4 - 1 = 3 > 0.$ Thus, whenever $t > (\frac{1}{16})^\frac{1}{3}$, the numerator, and the fraction, are positive. Picking a point between $0$ and $(\frac{1}{16})^\frac{1}{3}$ will reveal that the numerator is negative on the interval $(0, (\frac{1}{16})^\frac{1}{3}).$ For instance, $t = \sqrt{\frac{1}{16}}.$ Plugging in this value to the numerator gives $$4(\sqrt{\frac{1}{16}})^2 - \sqrt{\sqrt{\frac{1}{16}}} = $$ $$4(\frac{1}{16}) - (\frac{1}{16})^{\frac{1}{4}} = $$ $$ \frac{4}{16} - \frac{1}{2} < 0.$$

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First, we note we have $\sqrt{t}$, so $t \ge 0$. $t$ is in the denominator of the fraction, so the function is undefined at $t = 0$. Now, we have $t > 0$. Let's say we want to find where the function is negative.

$$\frac{4t^2- \sqrt t}{t} < 0$$ $$4t^2- \sqrt{t} < 0$$

Solving, we get:

$$ 4t^2 < \sqrt{t} $$ $$ 16t^4 < t $$ $$ 16t^4 - t < 0 $$ $$ t(16t^3 - 1) < 0 $$

Finally, we get:

$$ 0 < t < \frac{1}{2\sqrt[3]{2}} $$

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