Algebra of real functions

The governing concept is that the domain of a difference of functions is the intersection of the domains of the two functions. For f(x)=√(x−2), the radicand must be nonnegative, so x−2≥0 and therefore x≥2. For g(x)=√(x+2), we need x+2≥0, giving x≥−2. Both square roots must be real at the same input, so both conditions must hold simultaneously. Their intersection is x≥2, written in interval notation as [2,∞). Therefore option A is correct. Option B gives only the domain of g and ignores the stricter condition from f. Option C incorrectly adds an upper bound, and option D allows values that would make √(x−2) non-real.

The governing concept is intersection of all domain restrictions, with special care when a square root occurs in a denominator. For f(x)=√(4−x), require 4−x≥0, which gives x≤4. For g(x)=1/√x, the radicand first requires x≥0, but √x is also the denominator, so it cannot be zero. Consequently the stronger condition is x>0. Combining x>0 with x≤4 gives 0<x≤4, or the interval (0,4]. Thus option A is correct. Option B wrongly includes x=0, where g is undefined. Option C wrongly excludes x=4, although f(4)=0 is valid. Option D reverses the restriction from f and includes values greater than 4, where √(4−x) is not real.

The governing concept is subtraction of real-valued functions followed by solving a quadratic inequality. Compute the difference term by term: (f-g)(x)=(2x^2-1)-(x^2+3)=x^2-4. Thus the required condition is x^2-4>0, or (x-2)(x+2)>0. A product of these two factors is positive outside the roots, so x<-2 or x>2. The endpoints x=-2 and x=2 are excluded because the inequality is strict and the expression equals zero there. Option B describes the interval where the difference is negative, option C wrongly includes values between the roots, and option D ignores the sign restriction. Therefore option A is correct.

The sum of two functions is formed by adding their expressions: \\((f+g)(x)=f(x)+g(x)\\). After this addition, the type of function is identified from the simplified expression. A linear function has the form \\(ax+b\\), where the highest power of \\(x\\) is one. A constant function has no variable term, while a quadratic function contains \\(x^2\\).

Substitute the given rules: \\((f+g)(x)=(2x-1)+(5-x)\\). Combine like terms: \\(2x-x=x\\), and \\(-1+5=4\\). Hence \\((f+g)(x)=x+4\\). This has the form \\(ax+b\\) with \\(a=1\\), so it is a linear function. Therefore option A is correct. It is not constant because the term \\(x\\) remains, and it is not quadratic because no \\(x^2\\) term appears.

The governing concept is finding the minimum value of a quadratic by completing the square. Add the two functions pointwise: (f+g)(x)=(x²-1)+2x=x²+2x-1. Now rewrite this expression as x²+2x-1=(x+1)²-2. For every real x, a square is non-negative, so (x+1)²≥0. Consequently, the least possible value of the whole expression is 0-2=-2. This value is attained when x+1=0, or x=-1, so the minimum actually exists. Option A is correct. Option B ignores the contribution of the linear term, while options C and D are larger than the vertex value and therefore cannot be the minimum.

For a product of real-valued functions, the domain consists of all inputs that belong to the domains of both functions. The square-root expression √(x + 2) is real only when x + 2 ≥ 0, giving x ≥ −2. The rational function 1/(x − 1) is defined only when its denominator is nonzero, so x ≠ 1. Combining these conditions gives x ≥ −2 and x ≠ 1, or [−2, ∞) excluding 1. Therefore option A is correct. Option B wrongly excludes −2, although the square root is defined there; option C omits valid values greater than 1, and option D includes values below −2.

The sum of two functions is found by adding their values at the same input. Here both fractions have the common denominator \\(x^2+1\\), so their numerators can be added while the denominator remains unchanged. Thus, the operation is direct and no separate domain problem appears because \\(x^2+1\\) is always positive for every real number \\(x\\). This makes option A the appropriate simplified result.

Using the definitions, \\( (f+g)(x)=f(x)+g(x)=\\frac{1}{x^2+1}+\\frac{x^2}{x^2+1} \\). Combining the numerators gives \\(\\frac{1+x^2}{x^2+1}\\). Since the numerator and denominator are the same expression, the fraction equals \\(1\\). Option B incorrectly adds an extra 1 to the numerator, while option C uses subtraction. Therefore, A follows.

For a quotient of functions, simplify the algebraic expression but preserve every restriction imposed by the original denominator. Factor the numerator: x²-5x+6=(x-2)(x-3). Since g(x)=x-3, we have (f/g)(x)=[(x-2)(x-3)]/(x-3)=x-2, provided x-3≠0. Thus x=3 is excluded from the domain, even though the factor cancels in the simplified expression; the original quotient is undefined there and has a removable hole. Therefore option A is correct. Option B incorrectly includes x=3, option C uses the wrong factor and restriction, and option D leaves the numerator unsimplified.

The governing concept is the quotient of two functions. For every input where g(x) is non-zero, (f/g)(x)=f(x)/g(x). Evaluate the numerator at x=4: f(4)=4²−7(4)+12=16−28+12=0. Next evaluate the denominator: g(4)=4−3=1. Since the denominator is 1, not zero, the quotient is defined. Consequently, (f/g)(4)=0/1=0, so option A is correct. Option B would be obtained by ignoring the zero numerator and treating the denominator as the answer. Option C is not produced by either evaluation. Option D is also incorrect because a quotient is undefined only when its denominator is zero, and g(4)=1. The calculation therefore gives one unambiguous result.

The governing concept is pointwise subtraction of functions: (f−g)(x)=f(x)−g(x). Substitute the given expressions carefully, keeping the minus sign before the second bracket: (f−g)(x)=(x²+1)−(x²−3)=x²+1−x²+3=4. The x² terms cancel exactly, so the output does not depend on x. A function whose value remains 4 for every input in its domain is a constant function. Thus option A is correct. It is not the zero function, because the resulting value is 4 rather than 0. It is not 4x, since no x-term remains after subtraction. It is also not quadratic, because the x² terms cancel completely. The classification follows from the simplified expression, not from the original appearance of the two functions.

The sum of functions is obtained by adding their outputs: (f+g)(x)=f(x)+g(x)=x²+1+2x. Completing the square gives x²+2x+1=(x+1)². Every real square is non-negative, so the expression can never be less than zero. Its minimum value is 0, attained when x+1=0, or x=−1. As x moves away from −1, (x+1)² can take every non-negative value, so no positive values are missed. Hence the range is [0,∞), making option A correct. Option B incorrectly assumes the minimum is 1, while options C and D ignore the square form or reverse its restriction.

The governing concept is addition of rational functions using a common denominator. Start with h(x)=1/(x+1)+1/(x−1). The least common denominator is (x+1)(x−1)=x²−1. Combining the numerators gives h(x)=[(x−1)+(x+1)]/[(x+1)(x−1)]. The numerator simplifies to x−1+x+1=2x, while the denominator remains x²−1. Therefore h(x)=2x/(x²−1). The original expressions are undefined at x=−1 and x=1, so both restrictions must be stated even though the result has been combined into one fraction. Option A includes the correct numerator, denominator, and domain. Option B loses the factor x, option C introduces an incorrect negative sign, and option D loses the factor 2. Thus only option A is mathematically complete.