Real-Valued Functions, Their Domain and Range

The denominator of a rational function cannot be zero. Here, the denominator is \(x+5\), which becomes zero at \(x=-5\). Therefore, \(-5\) must be excluded from the set of real numbers, giving the domain \(\mathbb{R}\setminus\{-5\}\). Option B incorrectly excludes \(5\) instead of \(-5\). Exam tip: To find the domain of a rational function, set its denominator equal to zero and exclude the resulting value.

To find the range, evaluate the function for every element of the domain: \(f(2)=1\), \(f(4)=2\), and \(f(6)=3\). Hence, the range is \(\{1,2,3\}\). Option A is the domain, not the range, while option C results from multiplying each input by 2. Exam tip: For a finite domain, calculate the function value for every input to obtain the range.

\(2^x\) is defined for every real \(x\), and its value is always greater than 0; hence its range is \((0,\infty)\). In contrast, \(x^2\) also includes 0 in its range. Exam tip: an exponential function never attains 0.

For every real \(x\), \(x^2\ge 0\). Hence, \(f(x)=x^2+9\ge 9\), and equality occurs at \(x=0\). Therefore, the minimum value of the function is \(9\). The value \(-9\) cannot be the minimum because \(x^2+9\) is always at least 9. Exam tip: For a function containing \(x^2\), use \(x^2\ge 0\) to find its minimum quickly.

The direct answer is B: the range is \([9,\infty)\). The range means all possible output values. For every real x, \(x^2\ge 0\). Adding 9 gives \(x^2+9\ge 9\), so no output can be below 9. The value 9 is actually reached when x=0, because f(0)=0^2+9=9. For any value y≥9, choose x=\(\pm\sqrt{y−9}\); then x^2+9=y. Thus every value from 9 upward occurs, giving \([9,\infty)\). Option A, \([0,\infty)\), is wrong because values such as 1 are impossible. Option B is correct. Option C, \((9,\infty)\), wrongly excludes 9, although x=0 produces it. Option D, all real numbers, is wrong because the output can never be less than 9. Memory cue: a square is at least zero; adding 9 shifts the minimum to 9, and the endpoint is included.

For every real number \(x\), \((x-2)^2\geq 0\), so \(5-(x-2)^2\leq 5\). When \(x=2\), the squared term is zero and \(f(2)=5\). Therefore, the maximum value is 5. The distractor 7 is incorrect because the non-negative square is being subtracted. Exam tip: A function of the form \(a-(x-h)^2\) has maximum value \(a\).

Since ((x-2)^2\ge 0), (5-(x-2)^2\le 5). There is no lower bound, so the range is ((-\infty,5]).

For every real \(x\), \(|x-4|\geq 0\). Its minimum value is \(0\), attained when \(x=4\). Therefore, \(f(4)=0+1=1\), so option B is correct. Remember that \(4\) is the value of \(x\) where the minimum occurs, not the minimum value itself.

Since |x − 4| is always non-negative, f(x) is at least 1. It equals 1 when x = 4, and |x − 4| can take every non-negative value. Thus the range is [1, ∞).

Because |x + 2| is always non-negative, f(x) cannot exceed 6. At x = −2, |x + 2| = 0, so f(−2) = 6. Therefore, the maximum value is 6.

The governing concept is the range of an absolute-value function. For every real x, |x + 2| is non-negative, so subtracting it from 6 gives f(x) = 6 − |x + 2| ≤ 6. Equality occurs when x + 2 = 0, or x = −2, so the value 6 is attained and must be included. There is no lower bound: as x moves farther from −2, |x + 2| becomes arbitrarily large, making f(x) arbitrarily negative. Hence every real value less than or equal to 6 occurs, and the range is (−∞, 6]. Option B is correct. Option A reverses the bound, option C adds an unjustified lower bound of zero, and option D incorrectly includes values greater than 6.

Substituting \(x=8\) into the function gives \(f(8)=\sqrt{8+1}=\sqrt{9}=3\). Therefore, the correct value is 3. The value 9 is the radicand before taking the square root, not the function value. Exam tip: simplify the expression inside the square root first, and then evaluate the root.

For a real square root, the expression inside the root cannot be negative. Hence, \(x+10\ge 0\), so \(x\ge -10\). Therefore, the real domain is \([-10,\infty)\). Option C excludes \(-10\), but at \(x=-10\), \(f(x)=\sqrt{0}=0\) is defined. Exam tip: For a square-root function, set the radicand \(\ge 0\) before solving the inequality.

For a real square root, the expression inside the root must be non-negative. Thus, \(12-3x\geq 0\). Solving gives \(-3x\geq -12\), hence \(x\leq 4\). Therefore, the real domain is \(( -\infty,4]\). Option C incorrectly excludes 4, although \(f(4)=\sqrt{0}=0\) is defined. Exam tip: for real square-root functions, set the radicand greater than or equal to zero.

The denominator of a rational function cannot be zero. Here, \(x^2-9=0\), so \(x^2=9\) and \(x=\pm3\). Therefore, \(-3\) and \(3\) are excluded from the domain. Option A incorrectly uses \(-9\) and \(9\); the denominator is not zero at those values. Exam tip: To find the domain of a rational function, set its denominator equal to zero and exclude the resulting values.

The denominator must not be (0), and (x^2-9=0) gives (x=-3,3). Hence the domain is (\mathbb{R}-{-3,3}).

For a rational function, the denominator must not be zero. Here, \(x^2+6\geq 6>0\) for every real \(x\), so the denominator never vanishes and all real numbers belong to the domain. Therefore, the domain is \(\mathbb{R}\). Exam tip: Set the denominator equal to zero first to identify any excluded values.

The governing idea is that a positive fraction with fixed numerator becomes largest when its positive denominator is smallest. For every real x, x² ≥ 0, so x² + 6 ≥ 6. The smallest denominator, 6, occurs exactly when x² = 0, that is, when x = 0. Consequently f(0) = 1/6, which is the maximum value. At x = 1 the value is 1/7, and at x = 6 it is 1/42, both smaller than 1/6. The maximum is actually attained because x = 0 belongs to the real domain. Therefore option A is correct. Option D is false, while options B and C identify inputs that produce smaller values rather than the greatest one.

The minimum denominator is (6), so the maximum value is (\frac{1}{6}). The value (0) is never reached but can be approached.

For the square root to be real, its radicand must be non-negative: 16 − x² ≥ 0. This gives x² ≤ 16, or −4 ≤ x ≤ 4. The endpoints are included because the function equals 0 at x = −4 and x = 4. Therefore, the domain is [−4, 4].

For real values, \(16-x^2\ge 0\), and \(x^2\) is smallest when \(x=0\). Thus, \(f(0)=\sqrt{16}=4\), so the maximum value is 4. The value 16 is the maximum radicand, not the value of the function after taking the square root. Exam tip: For a square-root function, first maximize the expression inside the radical and then take its square root.

A square-root value is always non-negative, so \(f(x)\geq 0\). At \(x=0\), \(f(x)=4\), and at \(x=\pm4\), \(f(x)=0\). Therefore, the range is \([0,4]\). Option B represents the domain, not the range. In exams, check whether the minimum and maximum values are actually attained before choosing open or closed endpoints.

For every real \(x\), \((x+5)^2\geq 0\). The minimum value is \(0\), attained when \(x=-5\), and arbitrarily large values are possible. Hence the range is \([0,\infty)\). \([5,\infty)\) is incorrect because the \(+5\) is inside the square; it shifts the vertex horizontally and does not make the minimum function value 5. Exam tip: For a squared expression, find the input that makes the squared part zero to identify the minimum value.

Since \((x-3)^2\geq 0\), we have \(f(x)=(x-3)^2+2\geq 2\). At \(x=3\), the function attains the value 2, so its range is \([2,\infty)\). Option B gives 3, which is the x-coordinate of the vertex, not the minimum function value. Exam tip: the range of \((x-a)^2+b\) is \([b,\infty)\).

For every real \(x\), \((x+1)^2\ge 0\), so \(f(x)=2-(x+1)^2\le 2\). The value 2 is attained at \(x=-1\), and as \(|x+1|\) grows, the function decreases without bound. Thus the range is \(( -\infty,2]\).