Real-Valued Functions, Their Domain and Range

For each given value of \(x\), calculating \(y=x^2\) gives \(4,1,0,1,4\). The range contains only distinct \(y\)-values, so it is \(\{0,1,4\}\). Option B is the set of given \(x\)-values, which is the domain, not the range. Exam tip: Apply the rule to every domain value and list repeated output values only once.

Substitute each domain value into \(y=\frac{x+1}{2}\): for \(x=1\), \(y=1\); for \(x=3\), \(y=2\); for \(x=5\), \(y=3\); and for \(x=7\), \(y=4\). Hence, the range is \(\{1,2,3,4\}\). Option B is the domain, not the range. Exam tip: to find a function’s range, list all distinct output values obtained from the given inputs.

The range of a function is the set of all actual output values, כלומר the second components of its ordered pairs. Here, \(1\mapsto a\), \(2\mapsto b\), and \(3\mapsto b\), so the range is \(\{a,b\}\). The set \(\{a,b,c,d\}\) is the codomain, but \(c\) and \(d\) are not attained by the function. Exam tip: To find the range, collect the second components of all ordered pairs and list repeated values only once.

The domain of a function is the set of first components of all its ordered pairs. The first components here are \(a,b,c,d\), so \(A=\{a,b,c,d\}\). Option B is only the range, option C is the complete set of ordered pairs, and option D incorrectly combines domain and range elements. Exam tip: identify first components for the domain and second components for the range.

By definition, the even inputs \(2\) and \(4\) give \(f(x)=0\), while the odd inputs \(1\) and \(3\) give \(f(x)=1\). Hence, the set of all output values is \(\{0,1\}\). Option D is the domain, not the range. Exam tip: To find a range, evaluate the function at every domain element and list only the distinct output values.

The denominator must be non-zero: \(x^2-1\ne0\), so \(x\ne\pm1\). Hence, the domain is \(\mathbb{R}\setminus\{-1,1\}\). For the range, let \(y=\frac{x^2+1}{x^2-1}\). Then \(x^2=\frac{y+1}{y-1}\). Since \(x^2\ge0\) and \(y\ne1\), we must have \(y\le-1\) or \(y>1\). The value \(y=-1\) occurs at \(x=0\), whereas \(y=1\) is never attained. Option B incorrectly treats excluded domain values as excluded range values. Exam tip: for rational functions, find denominator restrictions first, then determine range separately.

The odd elements of \(A\), namely \(1\) and \(3\), give \((-1)^x=-1\). The even elements, \(2\) and \(4\), give \((-1)^x=1\). Therefore, the set of all output values is \(\{-1,1\}\). The choices \(\{1\}\) and \(\{-1\}\) omit the outputs from odd and even inputs, respectively. Exam tip: \((-1)^x\) equals \(-1\) for an odd exponent and \(1\) for an even exponent.

Dividing \(0,1,2,3\) by 3 gives the remainders \(0,1,2,0\), respectively. Hence the set of actual values of the function, or its range, is \(\{0,1,2\}\). The equality \(f(0)=f(3)=0\) does not violate the definition of a function; different inputs may have the same output, provided each input has exactly one output. Exam tip: To find the range, list the actual outputs rather than simply copying the codomain.

The governing concept is the range of a real-valued function. Put u = x², so u ≥ 0 for every real x. Then f(x) = (u + 4)/(u + 1) = 1 + 3/(u + 1). Since u + 1 ≥ 1, the added term is positive and at most 3. Its maximum 3 occurs when u = 0, that is, x = 0, giving f(0) = 4. As u grows without bound, 3/(u + 1) approaches 0 but never equals 0 for any finite real x. Hence f(x) approaches 1 without attaining it, and every value between 1 and 4 is obtained. Therefore the range is (1, 4], so option A is correct. Option B incorrectly includes 1, while C and D do not represent the actual set of values.

The direct answer is option A: the domain is x>2, or (2,∞). For a square root to be real, its radicand must be at least zero. Here x−2≥0 would be needed for the root itself, but the root is in the denominator, and a denominator cannot equal zero. Therefore we need √(x−2)>0, which is equivalent to x−2>0. Adding 2 gives x>2. Option A is correct. Option B, x≥2, incorrectly includes x=2; at x=2 the denominator is √0=0, so division is undefined. Option C, x<2, makes x−2 negative, so the square root is not real. Option D, x≠2, is too broad because values such as x=1 also make the radicand negative. Thus both restrictions matter: the radicand must be nonnegative and the denominator must be nonzero. Memory cue: a square root in a denominator requires a strictly positive radicand.

For a square-root function to have a real value, the expression inside the square root must be non-negative. In this function the radicand is 5−2x, so the required condition is \\(5-2x\\ge0\\). Solving it gives \\(-2x\\ge-5\\). When we divide by the negative number −2, the inequality sign reverses, giving \\(x\\le5/2\\).

The endpoint x=5/2 is included because the radicand then equals zero, and \\(\\sqrt{0}\\) is defined. Values smaller than 5/2 make the radicand positive, while values larger make it negative and produce no real function value. Therefore the domain is all real x satisfying \\(x\\le5/2\\), so option A is correct. The strict inequality in option C wrongly excludes the valid endpoint.

The denominator (x^2-x-6=(x-3)(x+2)) cannot be zero. Values that make the denominator zero are excluded from the domain.

A fraction is undefined wherever its denominator is zero. Factor the denominator: x^2+4x+4=(x+2)^2. This square equals zero only when x+2=0, which gives x=-2. Every other real number makes the denominator positive and therefore gives a defined function value. Thus the domain is all real numbers except -2, matching option A.

The excluded value must not be confused with 2 or -4. Substituting x=2 gives a denominator of 16, and substituting x=-4 also gives 4, so both are allowed. At x=-2, however, the denominator is 0 and the expression would require division by zero. In set notation the domain is \\(\mathbb{R}\setminus\{-2\}\\.

Because the square root appears in the denominator, its radicand must be strictly positive, not merely non-negative. Thus x²-4>0. Factoring gives (x-2)(x+2)>0, which holds when both factors are positive or both are negative: x>2 or x<-2. Therefore the domain is (-∞,-2)∪(2,∞), option A. The endpoints -2 and 2 are excluded because x²-4 becomes zero there, making the denominator zero. Option B incorrectly includes those endpoints, option C contains values for which the radicand is negative, and D removes only the zero points without excluding the interval where the square root is non-real.

For the square root to be real, its radicand must be non-negative: (x − 1)(x + 4) ≥ 0. The zeros are −4 and 1, and the product is non-negative outside the interval between them, including the zeros. Thus the domain is (−∞, −4] ∪ [1, ∞).

For a square root to be real, its radicand must be nonnegative. Here that condition is 9-x^2 greater than or equal to 0. However, the square root is in the denominator, so it cannot itself equal zero. Therefore the stronger condition is 9-x^2>0. Solving it gives x^2<9, hence -3<x<3, which is option A.

The endpoints -3 and 3 are excluded because at either endpoint the denominator becomes \\(\sqrt{0}=0\\, making the fraction undefined. Values between the endpoints make the radicand positive and are allowed. Values outside the interval make the radicand negative, so the square root is not real. Thus neither the closed interval [-3,3] nor all real numbers is correct; the domain is (-3,3).

The direct answer is A, \\(\\mathbb{R}\\setminus\\{-3,3\\}\\). A rational function is defined only when its denominator is nonzero. Here the denominator is \\(|x|-3\\), so require \\(|x|-3\\ne0\\), or \\(|x|\\ne3\\). The equation \\(|x|=3\\) has two solutions, x=3 and x=-3. Remove both from all real numbers. Option A is correct. Option B removes only 3 and wrongly keeps -3, where the denominator is zero. Option C removes only -3 and wrongly keeps 3. Option D keeps only numbers between -3 and 3, although many outside values, such as 4 or -5, are perfectly allowed. The absolute-value equation always requires both signs.

The radicand needs (|x|-2\ge 0), so (|x|\ge 2). View absolute value inequalities on a number line.

\((x+1)^2\) is always \(\ge 0\) for real \(x\).
So the minimum of \((x+1)^2\) is \(0\), which occurs when \(x+1=0\Rightarrow x=-1\).

Therefore, the minimum value of \(f(x)=(x+1)^2-9\) is \(0-9=-9\). Since \((x+1)^2\) can grow without bound, \(f(x)\) increases without bound too.

Hence the range is \([ -9,\infty)\).

For every real x, x² ≥ 0, so x² + 1 ≥ 1 and therefore 0 < f(x) ≤ 1. The value 1 is attained at x = 0. The function approaches 0 as |x| grows, but never equals 0, since its numerator is 1. Hence the range is (0, 1].

Here (f(x)=1-\frac{1}{x^2+1}), so values start at (0) and stay below (1). Check limiting values and attained values separately.

From (y=\frac{x+2}{x-3}), (x=\frac{3y+2}{y-1}), so (y\ne 1). The range excludes the (y)-value that makes (x) undefined.

Let y=(2x−5)/(x+4), where x≠−4 because the original denominator cannot be zero. To determine which y-values are possible, solve for x: y(x+4)=2x−5, so yx+4y=2x−5 and x(y−2)=−5−4y. Hence x=(−5−4y)/(y−2), provided y≠2. For every real y other than 2, this formula gives a real x, and that x is not −4; substituting x=−4 into the rearranged equation would produce a contradiction. The value y=2 is impossible because it would require 2(x+4)=2x−5, giving 8=−5. Thus the only excluded output is 2, and the range is R\{2}. Option A is correct. Option B confuses an excluded domain value with an excluded range value; C and D have no valid algebraic basis.

The absolute value |x-4| measures a distance, so it can never be negative. Its smallest possible value is 0, reached when x=4. Adding 2 to this nonnegative quantity gives f(x)=|x-4|+2, whose least value is 2. Since the absolute value can become arbitrarily large as x moves away from 4, the function has no upper bound.

The value 2 is actually attained at x=4, so the lower endpoint is included. For any number y greater than or equal to 2, choose x=4+(y-2); then |x-4|+2=y. Hence every value from 2 onward occurs, and the range is [2,∞), option A. The values below 2 cannot occur because they would require a negative absolute value.

For option A, put \(t=x^2\ge0\); then \(f=\frac{t}{1+t}\). It starts at 0 and approaches 1 without attaining it, so the range is \([0,1)\). Option C instead has range \((0,1]\). Exam tip: first check whether any denominator can be zero.