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Find the domain of the function $${\sqrt{\log_{0.4} (x-x^2)}}$$ Where $0.4$ is the base $0.5$ is the power on the whole bracket.

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  • $\begingroup$ @lulu Your expression is also hard to read. $\endgroup$ – callculus Apr 29 '16 at 9:23
  • $\begingroup$ @callculus Indeed it was. Let me try again: $\sqrt {{\log_{.4} (x-x^2)}}$ $\endgroup$ – lulu Apr 29 '16 at 9:23
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First, we want to make sure that the term inside the logarithm is always positive, as $\log{x}$ is defined only for $x > 0$.

$$x-x^2 = x(1-x) > 0 \implies 0 < x < 1$$

Next, we need the term inside the square root to be positive or equal to 0, so we get:

$$\begin{align*} \log_{0.4}(x-x^2) &\geq 0\iff\\\frac{\ln(x-x^2)}{\ln{0.4}} &\geq 0\iff\\\ln(x-x^2) &\leq 0\iff \\x-x^2 &\leq 1 \iff \\x^2 - x + 1 &\geq 0 \iff \\ \left(x-\frac12\right)^2 + \frac34 &\geq 0\end{align*}$$

And this statement is always true. Hence, our domain is $$0<x<1$$

We have used:

$(x-a)(x-b) < 0,\, a < b,\,\implies a < x < b$ and $\log_{a}x = \frac{\ln{x}}{\ln{a}} $

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You have here two conditions on the domain: $$ x-x^2 >0 \tag{1}$$ because $\log_{0.4}(t)$ is defined only for $t>0$, and $$\log_{0.4}(x-x^2)\geq 0 \tag{2}$$ because $\sqrt{t}$ is defined only for $t\geq 0$.

From $(1)$ we want that $x>x^2$ and so $x\in (0,1)$ because $x^2 \geq 0$ and $x^2\geq x$ for $x\geq 1$.

For condition $(2)$, we know that for $s\in (0,1)$ we have $\log_{s}(x)\geq 0$ if and only if $x\leq 1$ so we want that $x-x^2\leq 1$ which is true whenever $x\leq x^2+1$. This is satisfied for every $x\in (0,1)$.

It follows that the domain is $(0,1)$.

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