Real-Valued Functions, Their Domain and Range

The direct answer is option A: ℝ\{5}. A domain includes all real inputs that make the formula meaningful. In f(x)=1/(x−5)+2, the only possible problem is division by zero. The denominator is x−5, so set x−5=0; this gives x=5. Therefore every real number except 5 is allowed, and the domain is ℝ\{5}. Option A is correct. Option B, ℝ\{2}, is wrong because 2 is added after the fraction and does not make the denominator zero; at x=2 the expression is defined. Option C, (5,∞), wrongly removes all values less than 5, even though they work; for example, x=4 gives 1/(-1)+2=1. It also does not include 5, but it is still incomplete. Option D, [2,∞), is wrong for the same reason and includes x=5, where division by zero occurs. The added constant 2 changes output values, not the allowed inputs. Remember: for rational expressions, inspect the denominator only and exclude its zeros.

(\frac{1}{x-5}) is never (0), so the output cannot be (2). In exams a vertical shift of a reciprocal changes the excluded range value.

Because (x^2+4>0) for every real (x). In exams if the square root expression is always positive, the domain is (\mathbb{R}).

The minimum value of (x^2+4) is (4), so (\sqrt{x^2+4}\ge 2). In exams first find the minimum value inside.

In (\frac{1}{x}), the denominator is (x), so (x=0) is not allowed. In exams do not let the denominator become zero.

(x^2) is never negative and can take all values from (0) upward. In exams remember the range of the square function as ([0,\infty)).

Substituting \(x=16\) in the given function gives \(f(16)=\sqrt{16}=4\). The principal square root is taken as the positive value, so the correct answer is 4. In such questions, first substitute the given input into the function and then simplify.

Substituting \(x=-8\) in the function gives \(f(-8)=|-8+5|=|-3|=3\). Therefore, the correct answer is 3. Option B is incorrect because the modulus of a number cannot be negative. Exam tip: substitute the given value first, then evaluate the modulus as a non-negative quantity.

The two conditions (x-1\ge 0) and (6-x\ge 0) together give (1\le x\le 6). In exams take the intersection of conditions for two square roots.

For \(f(x)=\frac{1}{(x-1)(x+2)}\), the denominator \((x-1)(x+2)\) cannot be zero.

So \(x-1\neq 0\Rightarrow x\neq 1\) and \(x+2\neq 0\Rightarrow x\neq -2\).

Therefore the domain is \(\mathbb{R}\setminus\{1,-2\}\).

Since ((x+2)^2\ge 0), (3-(x+2)^2\le 3). In exams the maximum occurs when the squared part is (0).

Since ((x-4)^2\ge 0), the minimum value is (6). In exams set the squared part to (0) to get the minimum value.

For the square root, (1-x^2\ge 0), so (x^2\le 1) and (-1\le x\le 1). In exams (x^2\le a^2) gives a closed interval.

Since (0\le 9-x^2\le 9), we get (0\le \sqrt{9-x^2}\le 3). In exams first check the maximum and minimum values inside.

The denominator (x^2+4) has minimum value (4), so the maximum output is (\frac{1}{4}) and (0) is not reached. In exams (0) is often not included in reciprocal ranges.

Both square roots need (x-2\ge 0) and (8-x\ge 0), so (2\le x\le 8). In exams take the intersection of conditions for the combined domain.

The square root needs (x+2\ge 0) and the denominator needs (x-3\ne 0). So (x\ge -2) and (x\ne 3); in exams take the intersection of all conditions.

For a function with a finite domain, the range is found by evaluating the rule at every element of the domain and collecting the distinct outputs. At x=−3, f(−3)=9−1=8. At x=−1, f(−1)=1−1=0. At x=2, f(2)=4−1=3. Thus the range is {8,0,3}, conventionally written in increasing order as {0,3,8}; option A is correct. Option B is the domain itself, not the range. Option C includes 4, which forgets to subtract 1 at x=2, and option D contains incorrect evaluations. Since all three outputs are different, no repeated value needs to be removed from the set.

The function f(x)=5x−4 is linear and has positive slope 5, so it is strictly increasing on the entire domain. Therefore its output values follow the same endpoint order as the input interval. At the lower boundary, x approaches −3 from the right. The corresponding limiting output is 5(−3)−4=−19, but x=−3 is excluded, so −19 is not included in the range. At the upper boundary, x=2 is included and gives f(2)=5(2)−4=6, so 6 is included. Hence the range is (−19,6], which is option A. Option B incorrectly includes the output −19; option C merely repeats the domain, and option D reverses or miscalculates the endpoint values.

The defining condition for a real square root is that its radicand is nonnegative. Thus x+b≥0, which can be rewritten as x≥−b. The domain begins at the boundary x=−b. Since the given domain is [-6,∞), its boundary must satisfy −b=−6. Multiplying by −1 gives b=6, so option A is correct. The left endpoint is included because the radicand is zero there: at x=−6, x+b=−6+6=0. Option B has the sign reversed and would give the domain [6,∞). Option C would give [0,∞), while option D would give [−12,∞); neither matches the stated domain. Comparing the endpoint is therefore sufficient and decisive.

Substitute x = -2 into f(x) = x³: f(-2) = (-2)³ = (-2) × (-2) × (-2) = -8. Thus, the correct answer is -8; a negative number raised to an odd power remains negative.

For every real x, x² ≥ 0, and x = 0 gives the value 0. Every positive value y is also attained by taking x = √y. Therefore, the range is [0, ∞).

For every real number \(y\), \(x=\sqrt[3]{y}\) is real and gives \(f(x)=x^3=y\). Hence, \(x^3\) can produce every real value, so its range is \(( -\infty,\infty)\). Option A excludes negative values, while option C excludes positive values. Exam tip: an odd-power function such as \(x^3\) has all real numbers as its range.

The polynomial \(x^2-3x+1\) is defined for every real \(x\), so its domain is \(\mathbb{R}\). In contrast, \(\sqrt{x-1}\) requires \(x\ge1\). Exam tip: first check restrictions from roots, denominators and logs.

This is a constant function: for every input in its domain, the output is 2. Thus, its range consists of the single value {2}.