Real-Valued Functions, Their Domain and Range

For the range to be exactly \(\{1,2\}\), every element of \(A\) must map to either 1 or 2, and both values must occur at least once. There are \(2^4=16\) functions from the four-element set \(A\) to \(\{1,2\}\). The two constant functions, mapping every element to 1 or every element to 2, must be excluded. Therefore, the required number is \(2^4-2=14\). Exam tip: the number of onto functions from an \(n\)-element set to a two-element set is \(2^n-2\).

For the domain values, \(f(0)=0\), \(f(1)=2\), \(f(2)=0\), and \(f(3)=2\), since these are the remainders when \(2x\) is divided by 4. Therefore, the set of attained values, or the range, is \(\{0,2\}\). The set \(\{0,1,2,3\}\) is the codomain, not the range. Exam tip: To find the range, evaluate the function for every domain element and list repeated outputs only once.

\(f^{-1}(\{0\})\) is the set of all elements in the domain whose image is \(0\). Among the given domain elements, only \(3\) and \(6\) are divisible by \(3\), so \(f^{-1}(\{0\})=\{3,6\}\). Option A contains the inputs that are not divisible by \(3\), while options C and D incorrectly mix codomain and domain elements. Exam tip: To find \(f^{-1}(\{a\})\), select every domain element whose function value is \(a\).

A function from \(\mathbb{R}\) to \(\mathbb{R}\) must assign exactly one real output to every real input. The given rule covers only \(x<0\) and \(x>0\), so no value is assigned when \(x=0\). Hence its domain is not all of \(\mathbb{R}\). At \(x=1\), only \(1^2=1\) is applicable, and at \(x=-1\), the value is \((-1)+2=1\); therefore options B and C are incorrect. Exam tip: To test whether a relation is a function, check that every input in the stated domain has exactly one output.

For the domain values, \(f(1)=1\), \(f(2)=2\), \(f(3)=3\), and \(f(4)=3\). Hence, the set of distinct values actually attained by the function is \(\{1,2,3\}\), so option A is correct. Option B is the domain, not the range. Exam tip: To find the range, list all outputs and retain only the distinct values.

For the odd inputs, \(f(1)=1\) and \(f(3)=3\). For the even inputs, \(f(2)=5-2=3\) and \(f(4)=5-4=1\). Thus, the distinct output values are only \(1\) and \(3\), so the range is \(\{1,3\}\). Exam tip: the range contains the actual distinct outputs, not necessarily every element of the codomain.

Substituting \(x=0,1,2,3,4\) gives the remainders \(0,3,1,4,2\), respectively. Hence the set of attained values is \(\{0,1,2,3,4\}\), so option A is correct. Option D includes only some even remainders and omits \(1\) and \(3\). Exam tip: To find the range, evaluate the function for every element of the domain; do not simply copy the codomain.

The preimage \(f^{-1}(\{1\})\) consists of all elements of the domain whose image is \(1\). Among the domain elements, only \(4\) and \(8\) are divisible by \(4\), so \(f^{-1}(\{1\})=\{4,8\}\). Option D is incorrect because \(0\) is not in the domain. Exam tip: To find a preimage, select the inputs that map to the specified output value.

For the domain values, the function gives \(f(1)=3\), \(f(2)=2\), \(f(3)=1\), \(f(4)=1\), and \(f(5)=1\). Thus, the set of distinct output values, or the range, is \(\{1,2,3\}\). Option D includes only the repeated output 1 and omits 2 and 3. Exam tip: the range is the set of actual outputs, whereas the codomain may contain values that are not attained.

The denominator must not be zero. Solving \(|x+1|-3=0\) gives \(|x+1|=3\), so \(x=2\) or \(x=-4\). These are the only excluded values, and the domain is therefore \(\mathbb{R}\setminus\{-4,2\}\).

Since \((x-a)^2\ge 0\), the function \(f(x)=(x-a)^2+b\) has minimum value \(b\), attained at \(x=a\). Hence its range is \([b,\infty)\). The given range \([-2,\infty)\) requires \(b=-2\), while \(a\) may be any real number. Therefore, \((5,-2)\) is a possible pair. Exam tip: In this form of a quadratic function, the lower endpoint of the range is the constant term \(b\).

Completing the square gives \(f(x)=2(x-2)^2+3\). Since \((x-2)^2\ge 0\), the function is at least 3, and it attains 3 when \(x=2\). Thus its range is \([3,\infty)\).

For x = 0,1,2,3,4,5, the remainders obtained on dividing 4x by 6 are 0, 4, 2, 0, 4, 2, respectively. Thus, the distinct output values are {0,2,4}, which is the range of f. Option B is the codomain, not the set of values actually attained by the function. Exam tip: To find the range of a finite function, evaluate every domain element and list the distinct outputs.

The preimage of a set consists of all elements of the domain whose images belong to that set. Here, \(f(n)=1\) exactly when \(n\) is divisible by \(3\). Among the given domain elements, these are \(3,6,9\), so \(f^{-1}(\{1\})=\{3,6,9\}\). Option A contains numbers not divisible by \(3\), while option D incorrectly includes \(0\), which is not in the domain. Exam tip: To find a preimage, check which domain elements map to the specified target value or set.

Because a > 0, the squared term is nonnegative and the parabola opens upward. Its minimum occurs at x = 2, where f(2) = b, so its range is [b, ∞). For this range to be [5, ∞), b must equal 5; any positive a is then possible. Option A has a = 3 and b = 5.

For both square roots, (x-2\ge 0) and (5-x\ge 0) must hold. In exams, take the intersection of all conditions.

From (y=\frac{x+1}{x-2}), (x=\frac{2y+1}{y-1}), so (y\ne 1). For a linear fractional function, isolate the impossible value.

The sum of distances from (-1) and (3) is minimized at (4). On the interval ([-1,3]), the function stays constant.

The denominator (1+|x|\ge 1), so the maximum value is (1). As (x) grows, the value approaches (0) but never becomes (0).

Complete the square inside the square root: \[x^2+4x+7=(x+2)^2+3\] Since \((x+2)^2\ge 0\), we get \((x+2)^2+3\ge 3\). Therefore \[f(x)=\sqrt{(x+2)^2+3}\ge \sqrt{3}\] The minimum value \(\sqrt{3}\) occurs when \((x+2)^2=0\Rightarrow x=-2\). As \((x+2)^2\) increases without bound, \(f(x)\) increases without bound. So the range is \([\sqrt{3},\infty)\).

The square root needs (x\ge -1), and the denominator needs (x\ne \pm 2). Since (-2) is already outside the domain, only (2) is removed.

For a real-valued function, the expression inside each square root must be non-negative. Thus, \(x-1\ge 0\) gives \(x\ge 1\), while \(x+2\ge 0\) gives \(x\ge -2\). The intersection of these conditions is \(x\ge 1\), so the domain is \([1,\infty)\). Option B considers only the condition for \(\sqrt{x+2}\) and ignores the stricter condition from \(\sqrt{x-1}\). Exam tip: For a function with multiple square roots, impose every radicand condition and take their intersection.

From (y=\frac{2x-3}{x+4}), (x=\frac{-3-4y}{y-2}), so (y=2) is impossible. In such fractions, the ratio of leading coefficients often gives the missing value.

Rewrite the quadratic by completing the square: x²−6x+11=(x−3)²+2. Since a square is always nonnegative, the smallest possible value is 2, attained at x=3, which lies in the interval [1,5]. To find the largest value on this interval, compare the endpoints because the parabola opens upward and its vertex gives the minimum. We get f(1)=1−6+11=6 and f(5)=25−30+11=6.

Consequently, the range is [2,6], so option A is correct. The endpoint values are both 6, while every value between 2 and 6 is attained as x moves through the interval. The interval [1,5] is the input domain, not the output range. The supplied explanation correctly identifies the minimum and maximum.

The difference of distances from (2) and (-2) stays between (-4) and (4). Removing signs interval-wise is the safe method.