Real-Valued Functions, Their Domain and Range

The governing concept is the range of a constant function. A range consists of all output values produced as the input varies over the domain. Here f(x) = 5 for every permitted x, so changing x never changes the output. Consequently, the only value in the range is 5, and the range is the singleton set {5}. It is not all of ℝ, because no value other than 5 is produced. It is not [5, ∞) or (−∞, 5], because those intervals contain many values that the function never takes. The braces are important: they denote a set containing exactly one element. Therefore option A is the only correct answer.

Substitute 3 for x: f(3) = 2(3) − 1 = 6 − 1 = 5. Therefore, the correct answer is 5.

For \(\sqrt{x-2}\) to be real, we need the radicand to be non-negative: \(x-2\ge 0\). Solving gives \(x\ge 2\). Therefore the domain is \([2,\infty)\).

For every real value of \(x\), \(x^2\ge 0\). Hence, \(f(x)=x^2+1\ge 1\). At \(x=0\), \(f(0)=1\), so 1 is included in the range, giving \([1,\infty)\). Option B ignores the effect of the added 1, while option D incorrectly excludes the attainable endpoint 1. Exam tip: For a quadratic of the form \(x^2+c\), find the minimum of \(x^2\) first and then add \(c\).

For every real x, |x| is non-negative, so the function never takes a value below 0. It takes the value 0 when x = 0, and it can take every positive value. Therefore, its range is [0, ∞).

The denominator must not be zero. Since x − 4 = 0 when x = 4, that value is excluded; all other real numbers are allowed. Thus, the domain is ℝ \ {4}.

Substituting \(x=-2\) into the given function gives \(f(-2)=3(-2)=-6\). Therefore, option A is correct. Option B results from missing the negative sign. In such questions, check the sign carefully when multiplying by a negative number.

For a real square root, the expression inside the root cannot be negative. Hence, \(5-x\geq 0\), which gives \(x\leq 5\). Therefore, the domain is \(( -\infty,5]\). Option C wrongly excludes \(x=5\), although \(f(5)=0\) is real. Exam tip: For a square-root function, first set the expression inside the root to be \(\geq 0\).

For every real \(x\), \(x^2\geq 0\), so \(f(x)=2x^2\geq 0\). At \(x=0\), we get \(f(0)=0\). Also, for any \(y>0\), taking \(x=\sqrt{y/2}\) gives \(f(x)=y\). Thus the function attains all and only non-negative real values, so its range is \([0,\infty)\). Option B is the closest incorrect choice because it excludes \(0\). Exam tip: for \(ax^2\) with \(a>0\) and domain \(\mathbb{R}\), the minimum value is \(0\).

The cubic function is defined for every real (x) and can take every real value. Hence both are (\mathbb{R}).

Apply the function to every element of the domain: \(f(1)=2\), \(f(2)=3\), and \(f(3)=4\). Therefore, the range is \(\{2,3,4\}\). Option B is the domain itself, not the range. Exam tip: for a finite domain, evaluate the function at each given input and collect the resulting outputs.

For every real \(x\), \(x^2\ge 0\), so \(x^2+1\ge 1>0\). Thus the denominator is never zero, and the function is defined for every real number. Therefore, its domain is \(\mathbb{R}\). Option B is incorrect because at \(x=0\), the denominator is \(1\), not zero. Exam tip: For a rational function, exclude only those values that make the denominator zero.

For a real square root, the expression under the radical must be zero or positive. Thus, \(x+3\ge 0\), which gives \(x\ge -3\). Therefore, the domain is \([-3,\infty)\). Option B reverses the inequality and includes values less than or equal to \(-3\), which do not make \(x+3\) non-negative except at the endpoint. Exam tip: For a square-root function, set the radicand greater than or equal to zero to find the domain.

An absolute value is always nonnegative, so \(|x-2|\geq 0\). At \(x=2\), the function equals \(|2-2|=0\), so its minimum value is 0.

For every real \(x\), \(|x-2|\) can never be negative, so \(f(x)\ge 0\). At \(x=2\), \(f(2)=0\), and every non-negative value is obtained by taking \(x\) suitably far from 2. Therefore, the range is \([0,\infty)\). The option \([2,\infty)\) incorrectly treats the horizontal shift in the input as a shift in the range. Exam tip: \(|x-a|\) has minimum value 0 when \(x=a\) lies in the domain.

The denominator of a fractional function cannot be zero. Here, the denominator \(x+1\) is zero when \(x=-1\), so every real number except \(-1\) belongs to the domain. Therefore, the domain is \(\mathbb{R}\setminus\{-1\}\). Option B incorrectly excludes \(1\), although the function is defined at \(x=1\). Exam tip: To find the domain of a rational function, set its denominator equal to zero and exclude the resulting values.

A square root \(\sqrt{A}\) is real when \(A\ge 0\). Here \(A=9-x^2\). So the correct condition for the real domain is \(9-x^2\ge 0\).

For the square root to be real, its radicand must be non-negative: 9 − x² ≥ 0. This gives x² ≤ 9, so −3 ≤ x ≤ 3. The endpoints are included because at x = 3 and x = −3 the radicand equals zero, and √0 is defined. Hence the real domain is [−3, 3], making option A correct.

For every real number \\(x\\), \\(x^2\ge 0\\). Hence, \\(f(x)=x^2-4\ge -4\\), and equality occurs at \\(x=0\\), giving \\(f(0)=-4\\). Therefore, the minimum value is \\(-4\\). Exam tip: For a function of the form \\(x^2+c\\), the minimum value is \\(c\\) at \\(x=0\\).

The correct answer is option A: the range is [-4,∞). The range is the set of all output values that f(x) can produce. Since x is real, x^2 is never negative: its smallest value is 0, reached when x=0. Therefore f(x)=x^2-4 has smallest value 0-4=-4. For every larger value, let y=x^2-4; then x^2=y+4 can be obtained whenever y≥-4. Hence the range is [-4,∞), including -4. Option B, [0,∞), forgets the subtraction of 4 and is the range of x^2, not of x^2-4. Option C, (-∞,-4], incorrectly says outputs are only below -4, whereas x=0 gives -4 and x=2 gives 0. Option D, all real numbers, is wrong because values less than -4 cannot occur. The graph is an upward-opening parabola with vertex (0,-4), so its lowest height is -4. Exam cue: find the minimum of x^2 first, then apply the vertical shift.

Evaluating the function at each element gives f(−1) = (−1)² = 1, f(0) = 0² = 0, and f(1) = 1² = 1. Thus, the set of distinct output values is {0, 1}, which is the range. Option B is the original set A, the domain, not the range. Exam tip: list each output value only once when forming the range.

The denominator is \(x^2\), which equals zero only when \(x=0\). Since a fraction cannot have a zero denominator, every real number except zero belongs to the domain. Hence, the domain is \(\mathbb{R}\setminus\{0\}\). Remember that negative real numbers are also included, so options C and D are incomplete.

Since \(x^2\) is defined and real for every real x, \(x^2+1\) is also real for all real inputs. In contrast, \(\sqrt{x-1}\) needs x≥1. Exam tip: check input restrictions such as roots, denominators and logs.

For \(x^2+1\), squaring and adding 1 are valid for every real \(x\), so its domain is \(\mathbb{R}\). In contrast, \(\sqrt{x}\) requires \(x\ge0\). Exam tip: always exclude values that make a denominator zero.

For every real \(x\), \(x^2\ge 0\), so \(f(x)=x^2+4\ge 4\). At \(x=0\), \(f(0)=4\), so 4 is included in the range. Therefore, the range is \([4,\infty)\). Option B is the range of \(x^2\) before adding 4, while option C incorrectly excludes the endpoint 4. Exam tip: For a quadratic involving a square, first identify its minimum value.