Real-Valued Functions, Their Domain and Range

The original denominator is (x-3), so (x=3) is not allowed. Even after simplification, do not add the removed point to the domain.

For both square roots to be real, each radicand must be non-negative. From \(x+2\ge 0\), we get \(x\ge -2\), and from \(8-x\ge 0\), we get \(x\le 8\). Hence, the intersection is \([-2,8]\). At \(x=-2\) and \(x=8\), one square root becomes zero, so both endpoints are included; therefore, option B is incorrect. Exam tip: For radical functions, require every radicand to be \(\ge 0\) and intersect all resulting conditions.

For real values, \(x+2\ge 0\) and \(8-x\ge 0\), so the domain is \(-2\le x\le 8\). By the Cauchy–Schwarz inequality, \((\sqrt{x+2}+\sqrt{8-x})^2\le 2[(x+2)+(8-x)]=20\). Hence \(f(x)\le 2\sqrt{5}\). Equality occurs when \(\sqrt{x+2}=\sqrt{8-x}\), which gives \(x=3\). Therefore, the maximum value is \(2\sqrt{5}\). Exam tip: first determine the domain of radical expressions, then check the equality condition for the inequality used.

The square roots require x+2≥0 and 8−x≥0, so the domain is the closed interval [−2,8]. Let a=x+2 and b=8−x. Then a,b are nonnegative and a+b=10. By the inequality (√a+√b)²=a+b+2√(ab)≥a+b=10, we obtain f(x)≥√10. Equality occurs when ab=0, which happens at x=−2 or x=8. Therefore the minimum is √10.

Option A is correct. The value 2√5 is larger than √10 and is actually the maximum, reached when the two radicands are equal. The value 0 is impossible because both square roots cannot vanish at the same x. The supplied explanation gives the correct endpoint idea, although the inequality makes the minimum argument more explicit.

(|2x|\le x^2+1) because ((x-1)^2\ge 0) and ((x+1)^2\ge 0). Hence the values lie from (-1) to (1), and both occur.

Since \(|3x|\le \frac{x^2+9}{2}\), \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\). Equality occurs at \(x=\pm 3\).

Since \(\sqrt{x^2-4}\) is in the denominator, its radicand must be non-negative and the denominator must also be nonzero. Hence, \(x^2-4>0\), so \(x^2>4\), giving \(x<-2\) or \(x>2\). Therefore, the domain is \((-\infty,-2)\cup(2,\infty)\). Option B is incorrect because it includes \(x=\pm2\), where the denominator becomes zero. Exam tip: when a square root occurs in the denominator, require the expression under the root to be strictly positive.

The radicand must satisfy \(9-x^2\ge 0\), giving \(-3\le x\le 3\). The denominator must also be nonzero, so exclude \(x=-1\). Thus the domain is \([-3,3]\setminus\{-1\}\).

Since (\sqrt{x}\ge 0), the denominator is at least (1). At (x=0), the value is (1), and as (x\to\infty), it approaches (0).

Complete the square: \(x^2+2x+2=(x+1)^2+1\), so the logarithm's argument has minimum value \(1\), attained at \(x=-1\), and is unbounded above. Since \(\log_2\) is increasing, the function has minimum value \(\log_2 1=0\) and grows without bound. Its range is \([0,\infty)\).

Since \(x\in[0,2)\), we have \(2x\in[0,4)\). Therefore, the possible integer values of \(\lfloor 2x\rfloor\) are \(0,1,2,3\), and each of them is attained. For example, the value is \(0\) when \(x\in[0,\tfrac12)\), \(1\) when \(x\in[\tfrac12,1)\), and so on. The endpoint \(x=2\) is excluded, so \(2x=4\) and the value \(4\) cannot occur; hence option B is incorrect. Thus, the range is \(\{0,1,2,3\}\). Exam tip: For the greatest-integer function, check carefully whether the endpoints of the input interval are included.

For every real \(x\), \(x^2+4x+5=(x+2)^2+1>0\), so the domain is \(\mathbb{R}\). Put \(t=(x+2)^2\ge0\). Then \(f(x)=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). At \(t=0\), the value \(4\) is attained. As \(t\to\infty\), the function approaches \(1\), but it never equals \(1\). Hence the range is \((1,4]\). Option B incorrectly includes \(1\), while option C excludes the attained value \(4\). Exam tip: For such rational functions, complete the square and determine the allowed range of the substituted variable.

Put (t=(x-1)^2\ge 0), then (f=\frac{t+4}{t+1}=1+\frac{3}{t+1}). The maximum is (4), and (1) is only a limiting value.

Each radicand must be non-negative: x − 1 ≥ 0 requires x ≥ 1, and 9 − x ≥ 0 requires x ≤ 9. Their intersection is [1, 9]. The endpoints are included because a square root of zero is defined.

At the endpoints the value is (2\sqrt{2}), and at the midpoint (x=5), the maximum is (4). For symmetric radicals, check both endpoints and the midpoint.

The direct answer is option A: the domain is [2,\infty)\setminus\{7\}. A real square root is defined only when its radicand is non-negative, so x-2\ge0, giving x\ge2. Also, a denominator cannot be zero; therefore x-7\ne0, giving x\ne7. Combining both conditions gives [2,\infty) with 7 removed. Option A is correct because it includes 2 and all larger numbers except 7. Option B wrongly removes 2, although \sqrt{0} is defined. Option C includes numbers below 2, where the square root is not real. Option D wrongly excludes every x greater than 7, even though such values are allowed. Memory cue: square root means radicand \ge0; denominator means denominator \ne0.

For a real square root, the expression inside the root must be non-negative. Thus, \(x^2-6x+5\ge 0\). Factoring gives \((x-1)(x-5)\ge 0\), which is true for \(x\le 1\) or \(x\ge 5\). Therefore, the domain is \(( -\infty,1]\cup[5,\infty)\). Option \([1,5]\) is incorrect because the quadratic is negative between its roots. Exam tip: for any square-root function, set the radicand greater than or equal to zero first.

On the domain, the inside expression starts at (0) and goes to infinity. Hence the square root range is ([0,\infty)).

Completing the square gives \(f(x)=x^2+2x+3=(x+1)^2+2\). Since \((x+1)^2\geq 0\), the function has minimum value \(2\), attained at \(x=-1\). Hence its range is \([2,\\infty)\), so \(m=2\). Exam tip: For a quadratic written as \(a(x-h)^2+k\) with \(a>0\), the minimum value is \(k\). The constant term \(3\) is not the minimum value.

The square root needs (\log_{\frac{1}{2}}(x-4)\ge 0) and (x-4>0). Since the base (\frac{1}{2}<1), (0<x-4\le 1), so the domain is ((4,5]).

The expression inside the square root must satisfy (\frac{x-5}{x-2}\ge 0) and (x\ne 2). A sign chart gives (x<2) or (x\ge 5).

From (y=\frac{7x+4}{2x-3}), (x=\frac{3y+4}{2y-7}), so (y\ne \frac{7}{2}). In a linear fractional function, remove the impossible (y).

Since \(12-x-x^2=\frac{49}{4}-\left(x+\frac{1}{2}\right)^2\), the inside maximum is \(\frac{49}{4}\). Taking square root gives maximum \(\frac{7}{2}\).

Put (t=x^2\ge 0), then (f=\frac{t+5}{t+2}=1+\frac{3}{t+2}). The maximum is (\frac{5}{2}), and (1) is only a limiting value.

The argument of a logarithm must be strictly positive, so \(x^2-9>0\). Factoring gives \((x-3)(x+3)>0\), which holds for \(x<-3\) or \(x>3\). Thus the domain is \((-\infty,-3)\cup(3,\infty)\); the endpoints are excluded because the logarithm’s argument is zero there.