Domain of a Function: Problems with Solutions

Solution:
The square root is defined exactly when the radicand is non-negative:
[tex]23-2x^2\ge 0\iff x^2\le\frac{23}{2}=11.5[/tex]
For an integer [tex]x[/tex] this means [tex]|x|\le 3[/tex], because [tex]3^2=9\le 11.5[/tex], while [tex]4^2=16\gt 11.5[/tex]. Hence the integers in the domain are
[tex]-3, -2, -1, 0, 1, 2, 3[/tex]
that is, [tex]7[/tex] numbers.

Solution:
The function is defined when
[tex]5x^2-x^3=x^2(5-x)\ge 0[/tex]
If [tex]x=0[/tex], the inequality holds. If [tex]x\ne 0[/tex], then [tex]x^2\gt 0[/tex] and the inequality is equivalent to [tex]5-x\ge 0[/tex], i.e. [tex]x\le 5[/tex].
Since [tex]0\le 5[/tex], the domain is [tex]\left(-\infty, 5\right][/tex], and its largest number is [tex]5[/tex]. Indeed,
[tex]f(5)=\sqrt{125-125}=0[/tex]

Solution:
The expression is defined exactly when the denominator is not zero, so we find where it vanishes:
[tex]|x-2|-4+3x=0[/tex]
If [tex]x\ge 2[/tex], then [tex]|x-2|=x-2[/tex] and the equation becomes [tex]4x-6=0[/tex], so [tex]x=\frac{3}{2}[/tex]. This value does not satisfy [tex]x\ge 2[/tex], so it is rejected (indeed, for [tex]x=\frac{3}{2}[/tex] the denominator equals [tex]\frac{1}{2}-4+\frac{9}{2}=1\ne 0[/tex]).
If [tex]x\lt 2[/tex], then [tex]|x-2|=2-x[/tex] and the equation becomes [tex]2x-2=0[/tex], so [tex]x=1[/tex], which satisfies [tex]x\lt 2[/tex].
So the denominator vanishes only at [tex]x=1[/tex], and the admissible values are
[tex]x\in\left(-\infty, 1\right)\cup\left(1, +\infty\right)[/tex]

Solution:
The base of a logarithm must be positive and different from [tex]1[/tex], and its argument must be positive:
[tex]x\gt 0,\qquad x\ne 1,\qquad x^2-3x+2=(x-1)(x-2)\gt 0[/tex]
The last inequality holds for [tex]x\lt 1[/tex] or [tex]x\gt 2[/tex]. Intersecting with the conditions on the base, [tex]x\in\left(0, 1\right)\cup\left(1, +\infty\right)[/tex], we get
[tex]x\in\left(0, 1\right)\cup\left(2, +\infty\right)[/tex]

Solution:
The radicand must be non-negative:
[tex]x^4-x^2=x^2(x-1)(x+1)\ge 0[/tex]
If [tex]x=0[/tex], the left side equals [tex]0[/tex], so the inequality holds and [tex]0[/tex] belongs to the domain.
If [tex]x\ne 0[/tex], then [tex]x^2\gt 0[/tex] and we can divide by it: [tex](x-1)(x+1)\ge 0[/tex], i.e. [tex]x\le -1[/tex] or [tex]x\ge 1[/tex]. Therefore the domain is
[tex]\left(-\infty, -1\right]\cup\left\{0\right\}\cup\left[1, +\infty\right)[/tex]
Note that [tex]0[/tex] is a double root: the factor [tex]x^2[/tex] does not change sign, so the expression is negative on both sides of [tex]0[/tex] (for [tex]0\lt |x|\lt 1[/tex]) and the point [tex]0[/tex] is isolated in the domain.

Solution:
The root of index [tex]112[/tex] has an even index, so its radicand must be non-negative:
[tex]-x^2-4x+5\ge 0\iff x^2+4x-5\le 0\iff (x+5)(x-1)\le 0\iff x\in\left[-5, 1\right][/tex]
The root of index [tex]13[/tex] has an odd index, so it is defined for every real radicand (including negative ones). The only restriction is that it stands in a denominator:
[tex]\sqrt[13]{11-11x}\ne 0\iff 11-11x\ne 0\iff x\ne 1[/tex]
Combining both conditions, the admissible values are
[tex]x\in\left[-5, 1\right)[/tex]

Solution:
Each square root must exist separately, and the denominator must not be zero. For the numerator:
[tex]x^2+4x+3=(x+1)(x+3)\ge 0\iff x\in\left(-\infty, -3\right]\cup\left[-1, +\infty\right)[/tex]
The root in the denominator exists and is non-zero exactly when
[tex]x^2+2x-8=(x+4)(x-2)\gt 0\iff x\in\left(-\infty, -4\right)\cup\left(2, +\infty\right)[/tex]
The domain is the intersection of these two sets:
[tex]x\in\left(-\infty, -4\right)\cup\left(2, +\infty\right)[/tex]
Note that for [tex]x\in\left[-3, -1\right][/tex] the quotient [tex]\frac{x^2+4x+3}{x^2+2x-8}[/tex] is non-negative, but the radicand [tex]x^2+2x-8[/tex] is negative, so these values do not belong to the domain of [tex]f[/tex] (they belong only to the domain of [tex]\sqrt{\frac{x^2+4x+3}{x^2+2x-8}}[/tex]).

Solution:
Both radicands must be non-negative. From [tex]9-x^2\ge 0[/tex] we get [tex]x\in\left[-3, 3\right][/tex], and from [tex]\sin x\ge 0[/tex] we get
[tex]x\in\left[2k\pi, \pi+2k\pi\right],\ k\in\mathbb{Z}[/tex]
We check which of these intervals meet [tex]\left[-3, 3\right][/tex]. For [tex]k=0[/tex] the interval is [tex]\left[0, \pi\right][/tex]; since [tex]\pi\gt 3[/tex], its common part with [tex]\left[-3, 3\right][/tex] is [tex]\left[0, 3\right][/tex].
For [tex]k\le -1[/tex] the intervals end at [tex](2k+1)\pi\le -\pi\lt -3[/tex], and for [tex]k\ge 1[/tex] they start at [tex]2k\pi\ge 2\pi\gt 3[/tex], so they have no common points with [tex]\left[-3, 3\right][/tex]. Therefore the domain is
[tex]x\in\left[0, 3\right][/tex]

Solution:
The argument of [tex]\log_2[/tex] must be positive, the argument of [tex]\log_{\frac{1}{3}}[/tex] must be positive, and the radicand must be non-negative:
[tex]\frac{x^2-7}{x+4}\gt 0,\qquad L=\log_{2}\frac{x^2-7}{x+4}\gt 0,\qquad -\log_{\frac{1}{3}}L\ge 0[/tex]
The last condition means [tex]\log_{\frac{1}{3}}L\le 0=\log_{\frac{1}{3}}1[/tex]. The base is [tex]\frac{1}{3}\lt 1[/tex], so this logarithm is decreasing and the condition is equivalent to [tex]L\ge 1[/tex], i.e.
[tex]\log_{2}\frac{x^2-7}{x+4}\ge 1=\log_2 2\iff\frac{x^2-7}{x+4}\ge 2[/tex]
This inequality already implies the first two conditions. We solve it:
[tex]\frac{x^2-7}{x+4}-2\ge 0\iff\frac{x^2-2x-15}{x+4}\ge 0\iff\frac{(x-5)(x+3)}{x+4}\ge 0[/tex]
By the interval method (critical points [tex]-4[/tex], [tex]-3[/tex], [tex]5[/tex], with [tex]-4[/tex] excluded) the domain is
[tex]x\in\left(-4, -3\right]\cup\left[5, +\infty\right)[/tex]
The interval [tex]\left(-4, -3\right)[/tex] contains no integers, so the smallest integer in the domain is [tex]-3[/tex]. Check:
[tex]\frac{9-7}{-3+4}=2,\qquad \log_2 2=1,\qquad \log_{\frac{1}{3}}1=0,\qquad f(-3)=\sqrt{0}=0[/tex]

Solution:
We need [tex]x\ge 0[/tex] (for the inner root) and [tex]\sin \sqrt{x}\ge 0[/tex] (for the outer root). Put [tex]t=\sqrt{x}\ge 0[/tex]. The inequality [tex]\sin t\ge 0[/tex] holds for
[tex]t\in\left[2k\pi, (2k+1)\pi\right],\ k\in\mathbb{Z}[/tex]
For [tex]k\le -1[/tex] the whole interval is negative (its right end is [tex](2k+1)\pi\le -\pi\lt 0[/tex]), so it contains no admissible [tex]t[/tex]; hence [tex]k\in\left\{0, 1, 2, \ldots\right\}[/tex]. For [tex]t\ge 0[/tex] the function [tex]x=t^2[/tex] is increasing, so
[tex]2k\pi\le\sqrt{x}\le(2k+1)\pi\iff 4k^2\pi^2\le x\le(2k+1)^2\pi^2[/tex]
Therefore the domain is the union of the intervals [tex]\left[4k^2\pi^2, (2k+1)^2\pi^2\right][/tex], [tex]k\in\left\{0, 1, 2, \ldots\right\}[/tex]:
[tex]\left[0, \pi^2\right]\cup\left[4\pi^2, 9\pi^2\right]\cup\left[16\pi^2, 25\pi^2\right]\cup\ldots[/tex]