Lisantiom

2022-10-03

Determine the natural domain of $f(x)=\frac{\sqrt{N+3-{x}^{2}}}{\mathrm{log}(x)}$

falwsay

Beginner2022-10-04Added 8 answers

For the square root to yield real numbers it must hold that

$${x}^{2}\le N+3,$$

meaning

$$x\in [-\sqrt{N+3};\sqrt{N+3}].$$

Now consider the denominator. First of all, the natural logarithm must take in positive values of x, so $x>0$. Besides that, the denominator cannot be 0. The logarithm assumes that value at $x=1$

Now what you do is the intersection of all of these constraints to obtain your domain.

$$x\in [-\sqrt{N+3};\sqrt{N+3}]\cap (0;1)\cap (1;\mathrm{\infty}).$$

Finally

$$x\in (0;\sqrt{N+3}]\setminus \{1\}.$$

If N is such that the square root stays below 1, the set substraction operation will just yield the interval $(0;\sqrt{N+3}]$

$${x}^{2}\le N+3,$$

meaning

$$x\in [-\sqrt{N+3};\sqrt{N+3}].$$

Now consider the denominator. First of all, the natural logarithm must take in positive values of x, so $x>0$. Besides that, the denominator cannot be 0. The logarithm assumes that value at $x=1$

Now what you do is the intersection of all of these constraints to obtain your domain.

$$x\in [-\sqrt{N+3};\sqrt{N+3}]\cap (0;1)\cap (1;\mathrm{\infty}).$$

Finally

$$x\in (0;\sqrt{N+3}]\setminus \{1\}.$$

If N is such that the square root stays below 1, the set substraction operation will just yield the interval $(0;\sqrt{N+3}]$

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