Real-Valued Functions, Their Domain and Range

Rewrite the function as \(f(x)=1+\frac{3}{x^2+1}\). Since \(x^2\geq 0\), we have \(x^2+1\geq 1\), so \(0<\frac{3}{x^2+1}\leq 3\). Hence \(1<f(x)\leq 4\), giving the range \((1,4]\). The value \(4\) is attained at \(x=0\), so it is included. The value \(1\) is approached only as \(|x|\to\infty\) and is never attained for real \(x\), so option B is incorrect. Exam tip: always test whether a range endpoint is attained or merely approached.

For a real-valued function, the expression inside each square root must be non-negative. Thus, \(x-2\ge 0\) gives \(x\ge 2\), and \(x-8\ge 0\) gives \(x\ge 8\). Their intersection is \(x\ge 8\), so the domain is \([8,\infty)\). Option D is incorrect because the function is defined at \(x=8\), so 8 must be included. Exam tip: For a function containing multiple radicals, find the domain by taking the intersection of all the resulting conditions.

Two separate restrictions must be applied. The numerator √(x + 2) is real when x + 2 ≥ 0, so x ≥ −2. The denominator √(7 − x) must be real and also nonzero because it is in the denominator. Thus 7 − x > 0, giving x < 7. Intersecting x ≥ −2 with x < 7 produces [−2, 7). The left endpoint is allowed because the numerator becomes √0, which is defined, while the right endpoint is forbidden because the denominator becomes zero. Hence option A is correct. Option B incorrectly includes x = 7, option C incorrectly excludes x = −2, and option D reverses the relevant interval conditions.

The numerator needs (x-3\ge 0), and the denominator needs (x^2-25>0). Intersecting both conditions gives ((5,\infty)).

Here (f(x)=1-\frac{18}{x^2+9}), and (x^2+9\ge 9). So the minimum value is (-1), and (1) is not attained.

The square root is in the denominator, so (|x-2|-3>0) is needed. This gives (|x-2|>3), so the domain is ((-\infty,-1)\cup(5,\infty)).

From (y=\frac{2x+7}{3x-4}), we get (x=\frac{4y+7}{3y-2}). Hence (3y-2\ne 0), so (y\ne \frac{2}{3}).

Both square roots require (x-2\ge0) and (5-x\ge0). In exams take the intersection of all conditions.

The expression is 1 divided by a square root, so two conditions are required. First, x²−9 must be nonnegative for the square root to be real. Second, because the square root is in the denominator, it must not equal zero. Together these conditions require x²−9>0. Factoring gives (x−3)(x+3)>0, which holds outside the interval from −3 to 3.

Therefore x<−3 or x>3, and the domain is (−∞,−3)∪(3,∞), so option A is correct. The endpoints −3 and 3 are excluded because the denominator would be zero there. Option B incorrectly includes them, while option D excludes only the endpoints and still includes values between them, where the square root is not real.

The domain is all real numbers except 2 and 3, so option A is correct. A rational function is defined whenever its denominator is not zero. Here the denominator is x²−5x+6. Factor it: x²−5x+6=(x−2)(x−3), because −2−3=−5 and (−2)(−3)=6. Set the denominator equal to zero: (x−2)(x−3)=0, so x=2 or x=3. These two values must be excluded; every other real number is allowed. Therefore the domain is R\{2,3}. Option A states exactly this. Option B excludes −2 and −3, but those values do not make the denominator zero. Option C, all real numbers, wrongly includes values where division by zero occurs. Option D, {2,3}, gives only the excluded values rather than the allowed values. Exam cue: for a fraction function, factor the denominator and remove its zeroes; the numerator does not restrict the domain unless another condition is given.

Rewrite the function by completing the square: \(f(x)=x^2-4x+7=(x-2)^2+3\). Since \((x-2)^2\geq 0\), we have \(f(x)\geq 3\). At \(x=2\), the squared term is zero, so the minimum value \(3\) is actually attained. Hence, the range is \([3,\infty)\). Option B is incorrect because it excludes the attainable value \(3\). Exam tip: for an upward-opening quadratic, the range begins at the vertex’s \(y\)-coordinate.

Because (x^2+4\ge4), the maximum value is (\frac{1}{4}), and (0) is only approached. In exams do not close an unattained endpoint.

The sum of distances from (-1) and (3) is at least their distance (4). The distance idea is useful for absolute value questions.

From (y=\frac{x-1}{x+2}), we get (x=\frac{1+2y}{1-y}), so (y\ne1). Writing (x) in terms of (y) works well for range.

For a sum of real-valued expressions, every term must be defined at the same input. The radical term requires x + 4 ≥ 0, which gives x ≥ -4. The rational term requires x - 1 ≠ 0, so x ≠ 1. Combining these restrictions means we begin with [-4, ∞) and remove the single point 1. The endpoint -4 is allowed because sqrt(0) is defined, whereas 1 is excluded because it makes the denominator zero. Therefore the domain is [-4, ∞) \ {1}, which is option A. Option B incorrectly excludes -4, C omits all values above 1, and D includes values less than -4 where the square root is not real.

The governing logarithm rule is that its argument must be strictly positive. Therefore 7 - 2x > 0. Solving the inequality, subtracting 7 gives -2x > -7, and dividing by -2 reverses the inequality: x < 7/2. The endpoint x = 7/2 is excluded because the logarithm of zero is undefined; it is not enough for the argument merely to be non-negative. The base 2 is valid because it is positive and not equal to 1, but the base does not change the domain restriction here. Hence the domain is (-∞, 7/2), so option A is correct. Option B wrongly includes the endpoint, C reverses the inequality, and D ignores the logarithm condition.

The radicand of a real square root must be non-negative, and the denominator inside the fraction must not be zero. Thus we require (x - 1)/(x + 3) ≥ 0 with x ≠ -3. The critical points are -3, where the fraction is undefined, and 1, where the numerator is zero. A sign check on the intervals shows the quotient is positive for x < -3, negative for -3 < x < 1, and non-negative for x ≥ 1. The point 1 is included because the radicand is zero, while -3 is excluded. Therefore the domain is (-∞, -3) ∪ [1, ∞), option A. The other options either select the negative interval or mishandle the critical endpoints.

From (y=\frac{2x+3}{x-4}), we get (x=\frac{4y+3}{y-2}), so (y\ne2). In such questions find the unavailable (y).

Since (\sqrt{2x-1}\ge0), (3-\sqrt{2x-1}\le3) and is unbounded below. The negative sign reverses the direction of the range.

The only restriction comes from the denominator: it cannot equal zero. Set |x| - 2 = 0, which gives |x| = 2. An absolute-value equation |x| = 2 has two real solutions, x = 2 and x = -2. Both must therefore be removed from the real numbers. There is no square root or logarithm imposing any additional condition, and the absolute-value expression itself is defined for every real x. Hence the domain is R \ {-2, 2}, which is option A. Option B forgets the negative solution, option C retains only the interval between the excluded points even though many outside values are allowed, and option D removes too many valid inputs. Thus A is the unique correct answer.

The key property is that an absolute value is always non-negative: |x - 2| ≥ 0 for every real x. Adding 3 gives f(x) = |x - 2| + 3 ≥ 3. The minimum value is attained, not merely approached, when x - 2 = 0, or x = 2; at that input f(2) = 3. As x becomes very large or very negative, |x - 2| grows without bound, so the function has no upper limit. Therefore the range is [3, ∞), and option A is correct. Option B incorrectly excludes the attainable minimum, option C gives the wrong direction and allows values below 3, while option D ignores the lower bound created by the absolute value.

The direct answer is option A: the domain is [−2,4]. A square root is real only when its radicand is at least zero, so 9−(x−1)²≥0. Rearranging gives (x−1)²≤9, which is equivalent to |x−1|≤3. Therefore −3≤x−1≤3. Adding 1 throughout gives −2≤x≤4, hence the closed interval [−2,4]. The endpoints are included because the radicand becomes zero there, and √0 is defined. Option A is correct. Option B, (−2,4), wrongly removes the endpoints. Option C, [−4,2], shifts the interval incorrectly and does not represent numbers within 3 units of 1. Option D describes values outside the interval, where the radicand is negative and the square root is not real. A helpful picture is a centre 1 and radius 3: all x-values from 1−3 to 1+3 are allowed. Memory cue: for √(9−(x−1)²), stay within 3 units of 1 and include the boundaries.

For a rational function, the denominator must be non-zero. Set x² - 1 = 0 and factor it as (x - 1)(x + 1) = 0. This gives x = 1 or x = -1, so those two inputs are excluded. The numerator x² + 1 is defined for every real x and therefore creates no additional restriction. All other real numbers make the denominator non-zero and give a valid real output. Consequently the domain is R \ {-1, 1}, which is option A. Option B excludes 0 even though the function is defined there, option C keeps only the interval between the forbidden points, and option D keeps only outer intervals while incorrectly including the endpoints -1 and 1, where the denominator is zero.

The governing idea is to determine the least possible value of a function on its stated domain. Since x > 0, both x and 1/x are positive. By AM-GM, x + 1/x ≥ 2√(x·1/x) = 2. Equality is possible when x = 1/x, which gives x² = 1; the positive-domain condition selects x = 1. At this value f(1) = 2. As x tends to 0 from the right or as x tends to infinity, f(x) increases without bound, so every value greater than or equal to 2 is attained. Therefore the range is [2, ∞), making option A correct. Option B incorrectly includes negative values, option C omits values below 2, and option D ignores the lower bound.

For negative (x), the values are (-2) or less, with equality at (x=-1). The sign of the domain changes the range.