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I am trying to find definitions range for the function $\sqrt[4]{1-\sqrt[3]{4-\sqrt{25-x²}}}$.

I tried to sloe it like this: Because I know that in real number range under the sqrt I can not have a negative number, I made three equations, where I presumed that every part of the function that is under the root, needs to have solution that is bigger or equal to zero. These are the equations:

$25-x²$ for this equation I got that solution is range form -5 to 5

$4-\sqrt{25-x²}$ for this equation I got that solution is range from -3 to 3

$1-\sqrt[3]{4-\sqrt{25-x²}}$ for this equation I got that solution is range from -4 to 4

I tired to connect these results with "and" statment, and it does not fits, because the solution for this problem is range from -4 to 4, and I got that the solution is range from -3 to 3.

What am I doing wrong???

Thanks!!!

depecheSoul
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    Hint: the cubic root function is defined everywhere on the real line. – DonAntonio Mar 08 '14 at 12:42
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    The "intersection" strategy is good for expressions of the form $f(x)+g(x)$ or $f(x)g(x)$. I would work from outwards in. We want $\sqrt[3]{4-\sqrt{25-x^2}}\le 1$. Equivalently, we want $4-\sqrt{25-x^2} \le 1^3$. Equivalently, we want $\sqrt{25-x^2} \ge 3$. Equivalently, we want $x^2\le 16$. By the way, what you call definitions range is often called domain. – André Nicolas Mar 08 '14 at 12:49

1 Answers1

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The claim

$1-\sqrt[3]{4-\sqrt{25-x²}}$ for this equation I got that solution is range from -4 to 4

is incorrect because the function $\sqrt[3]{x}$ is definet for any real $x$. See for example x=-1 you have: $\sqrt[3]{-1}=-1$ because $(-1)^3=-1$.

Emo
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