Algebra of real functions

The domain of a sum of functions is the intersection of the domains of the individual functions. The polynomial g(x) = x^2 is defined for every real number. The rational function f(x) = 1/(x − 3), however, is undefined when its denominator is zero. Solving x − 3 = 0 gives x = 3, so 3 must be excluded. For every other real x, both f and g are defined and their sum exists. Consequently, the domain of (f + g)(x) is R − {3}, making option A correct. Option B overlooks the denominator restriction, option C excludes the wrong number, and option D lists only the forbidden value instead of all allowed real values.

The quotient function is defined by (f/g)(x) = f(x)/g(x), provided g(x) is not zero. Substituting the given expressions gives (x^2 − 9)/(x + 3). Factor the numerator as a difference of squares: x^2 − 9 = (x − 3)(x + 3). Therefore, for x ≠ −3, (f/g)(x) = [(x − 3)(x + 3)]/(x + 3) = x − 3. Option A is correct. The restriction x ≠ −3 must remain because the original denominator is zero there, even though the common factor cancels algebraically. Option B is the uncancelled denominator, option C uses an incorrect sign in the factorization, and option D does not follow from division.

The domain of a difference of functions is the intersection of the domains of the two functions. The polynomial g(x) = x^2 is defined for every real number. The rational function f(x) = 1/(x + 4) is undefined where its denominator equals zero. Solving x + 4 = 0 gives x = −4, so that single value must be excluded. For all other real numbers, both functions exist and their difference is defined. Therefore, the domain of (f − g)(x) is R − {−4}, which is option A. Option B excludes +4 instead of −4, option C ignores the denominator restriction, and option D includes only positive values greater than 4 without any mathematical reason. The operation of subtraction does not create a new restriction beyond the restrictions already present in the component functions.

The sum of two functions is defined by (f+g)(x)=f(x)+g(x). Therefore, (2x+3)+(x^2−1)=x^2+2x+2. Option B results from adding the constant terms incorrectly. In an exam, combine like terms only after writing the two functions together.

The governing rule for subtracting real-valued functions is (f − g)(x) = f(x) − g(x). Substitute the given expressions without dropping the brackets: (f − g)(x) = (x² + 4x) − (3x − 5). The minus sign applies to every term inside the second bracket, so this becomes x² + 4x − 3x + 5. Combining like terms gives x² + x + 5. Thus option A is correct. Option B fails to change the sign of the constant −5, and option C adds 4x and 3x instead of subtracting them. Option D changes the sign of the x-term incorrectly. The central algebraic precaution is to distribute the outer minus sign across both terms of g(x).

In the algebra of functions, juxtaposition in (fg)(x) denotes the product f(x)g(x), whereas composition would normally be written with a symbol such as f∘g. Therefore, (fg)(x) = (x + 2)(x − 3). Applying the distributive property gives x·x + x(−3) + 2x + 2(−3) = x² − 3x + 2x − 6. Combining like terms, −3x + 2x = −x, so the result is x² − x − 6. Option A is correct. Option B has the wrong sign for the linear term. Options C and D arise from incorrect multiplication or sign handling. Recognizing multiplication rather than composition is the key interpretive step.

\(\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3\), but \(x=3\) is not allowed. Even after cancellation, remember the original denominator.

Both square roots must be defined, so (x-1\geq 0) and (5-x\geq 0), giving (x\in[1,5]). For a sum, take the intersection of domains.

In (f(x)), the denominator (x-2) cannot be zero, so (x\neq 2). The polynomial (g(x)) is defined for all real (x).

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Here, \(f(-1)=(-1)^2=1\) and \(g(-1)=2(-1)+1=-1\), so \((f+g)(-1)=1+(-1)=0\). Option B is only the value of \(f(-1)\), not the sum. In an exam, evaluate both functions separately before adding them.

(g(2)=5) and (f(2)=4), so ((g-f)(2)=5-4=1). Read the order carefully because (g-f) and (f-g) are different.

For the sum of functions, add their values at the same input: \((f+g+h)(x)=f(x)+g(x)+h(x)=(x+1)+(x-1)+x^2\). The constants \(+1\) and \(-1\) cancel, leaving \(x^2+2x\). In an exam, write every term with its sign before combining like terms.

The scalar multiple of a function is defined by \((3f)(x)=3f(x)\). Hence, \((3f)(x)=3(2x^2-1)=6x^2-3\). Option B is incorrect because multiplying the entire function by 3 changes \(-1\) to \(-3\). Exam tip: do not confuse \((3f)(x)\) with \(f(3x)\); the former multiplies the whole function by 3.

In a quotient, the denominator \(g(x)=x\) must not be zero, so \(x\neq 0\). Apply the restriction before simplifying.

The domain of a product is the intersection of both domains, and in (f(x)), (x\neq 4). Since (g(x)) is a polynomial, it adds no new restriction.

\(g(-2)=0\), so the quotient \(\left(\frac{f}{g}\right)(-2)\) is not defined. Even if the numerator is (0), a zero denominator is not allowed.

The governing concept is pointwise subtraction of functions: (f − g)(a) = f(a) − g(a). The modulus function gives the non-negative distance of a number from zero, so f(−3) = |−3| = 3. Since g(x) = x is the identity function, g(−3) = −3. Substituting these values, (f − g)(−3) = 3 − (−3) = 3 + 3 = 6. Therefore option A is correct. Option D is only f(−3), not the requested difference. Option B would incorrectly treat the two function values as equal, while option C results from failing to apply the rule that subtracting a negative number changes subtraction into addition. The modulus sign must be evaluated before performing the subtraction.

((f+g)(x)=2x+5+5-2x=10), so it is a constant function. When opposite terms cancel, write the remaining constant carefully.

The direct answer is A, the zero function \(0\). The sum of functions is found point by point: \((f+g)(x)=f(x)+g(x)\). Substitute the definitions: \((f+g)(x)=x^2+(-x^2)=x^2-x^2=0\) for every x. Since the output is always zero, this is the zero function. Option A is correct. Option B, the identity function x, would require output x, not 0. Option C, the constant function 1, always outputs 1, not 0. Option D, the quadratic function \(x^2\), would be the result only if the second function were not added as its negative. The opposite signs cancel exactly. Memory cue: a function plus its additive opposite gives zero.

The coefficients apply to the entire functions: \\((2f-3g)(x)=2f(x)-3g(x)=2(x+5)-3(2x-1)=2x+10-6x+3=-4x+13\\)。Hence, option A is correct. In option B, the sign of the \\(x\\)-term is handled incorrectly. In an exam, expand the brackets first and then combine like terms carefully.

By the definition of the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Hence, \((f-g)(x)=(x^2+1)-(x^2-1)=x^2+1-x^2+1=2\). Option B results from adding the functions, not subtracting them. Exam tip: When a bracket is preceded by a minus sign, the sign of every term inside the bracket changes.

((fg)(x)=(x+1)\cdot\frac{1}{x+1}=1), but (x=-1) is not allowed. Always check the denominator in a reciprocal function.

For (\sqrt{x+2}), (x\geq -2), and for (\frac{1}{x-1}), (x\neq 1). Their intersection gives ([-2,\infty)-{1}).

\(\frac{x^2-2x}{x-2}=\frac{x(x-2)}{x-2}=x\), but \(x\neq 2\). A cancelled factor does not remove the original restriction.

The governing concept is addition of two real-valued functions. By definition, (g+f)(x)=g(x)+f(x), so the corresponding expressions must be added term by term: (x²+2x)+(x−4)=x²+(2x+x)−4=x²+3x−4. Therefore option A is correct. Option B would result from incorrectly keeping only one x-term or adding the coefficients improperly. Option C omits the contribution of x from f(x), while option D changes the sign of the constant −4 without justification. Function addition is commutative, so (g+f)(x) and (f+g)(x) give the same result, but it is still necessary to include every term accurately.