Algebra of real functions

The domain of a sum is the intersection of both domains, so (x\ge 2). In exams, keep the radicand of a square root non-negative.

Using a common denominator gives numerator ((x+1)+(x-3)=2x-2). While simplifying, remember original restrictions (x\ne3,-1).

The domain of a product is the intersection of both domains, so (5-x\ge0) and (x+1\ge0). Hence (-1\le x\le5).

\(\frac{x^2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2\), but \(x=-2\) remains excluded. Do not forget the original denominator restriction after cancellation.

((2f-3g)(x)=2(2x-1)-3(3-x)=7x-11), so at (x=4) the value is (17). Always simplify before substituting.

((2f-3g)(x)=4x-2-9+3x=7x-11), hence ((2f-3g)(4)=17). In linear combinations, pay close attention to signs.

The equal (x^2) terms cancel and (2x-(-2x)=4x). In subtraction, the sign of the whole second function changes.

Although the product becomes (1), the original functions exclude (x=0) and (x=1). It is safer to decide the domain before simplification.

For (\sqrt{x^2-9}), (x^2-9\ge0), and for the fraction (x\ne4). Thus (x\le-3) or (x\ge3), but (x=4) is removed.

The direct answer is A: \(f+g\) is an odd function. First add the functions: \((f+g)(x)=x^3+2x\). To test oddness, replace \(x\) by \(-x\): \((f+g)(-x)=(-x)^3+2(-x)=-x^3-2x=-(x^3+2x)=-(f+g)(x)\). Therefore it satisfies the definition of an odd function. Option A is correct. Option B is wrong because an even function would satisfy \(h(-x)=h(x)\), which does not happen here. Option C is wrong because the exact oddness test succeeds. Option D is wrong because the expression changes with \(x\); it is not a fixed constant. Both \(x^3\) and \(2x\) are odd, and the sum of odd functions is odd. Memory cue: every term with an odd power and no constant term changes sign when \(x\) changes to \(-x\).

Both (x^2) and (|x|) are even functions, so their sum is even. To test evenness, replace (x) by (-x).

\((fg)(x)=f(x)g(x)=x^2\cdot x=x^3\). Since \((fg)(-x)=(-x)^3=-x^3=-(fg)(x)\), the product is an odd function. It is not even because an even function must satisfy \(h(-x)=h(x)\). Exam tip: the product of an even and an odd function is odd, provided the domain is symmetric about zero.

\(x^2-5x+6=(x-2)(x-3)\), so the quotient is (x-3) and \(x=2\) is excluded. Keep the original denominator restriction after factorization.

(f(0)=\frac{1}{-1}=-1) and (g(0)=\frac{-1}{1}=-1), so the sum is (-2). First check whether the given (x) lies in the domain.

((f-g)(x)=x^2-2x+1=(x-1)^2), which is (0) at (x=1). Therefore (>0) holds for all (x\ne1); none of the broad options is exact.

((f-g)(x)=(x-1)^2), which is (0) at (x=1) and positive elsewhere. A square form gives a quick clue in inequalities.

The direct answer is option A: \((f+g)(-2)=18\). Addition of functions means \((f+g)(x)=f(x)+g(x)\). First combine the formulas: \((3x^2-2)+(x^2+4)=4x^2+2\). Now substitute \(x=-2\): \((-2)^2=4\), so \(4(4)+2=16+2=18\). Option A is correct. Option B, 14, may result from an arithmetic mistake involving the constants; the constants actually give \(-2+4=2\). Option C, 20, does not follow from the correct substitution. Option D, 10, also comes from an incorrect calculation and is not the function sum. Remember that squaring a negative number gives a positive result: \((-2)^2=4\), not \(-4\). Memory cue: add the functions first, then substitute, and use parentheses around negative inputs.

The direct answer is option A: all real numbers except \(x=-1\) and \(x=1\). The domain contains inputs for which both functions are defined. The function \(g(x)=x+5\) is a polynomial, so it is defined for every real number. For \(f(x)=\frac{1}{x^2-1}\), the denominator cannot be zero. Thus \(x^2-1\ne0\), so \(x^2\ne1\), giving \(x\ne-1,1\). Therefore the domain is \(\mathbb{R}\setminus\{-1,1\}\). Option A is correct because it states exactly these exclusions. Option B wrongly excludes only zero; zero actually gives a valid denominator of \(-1\). Option C includes only numbers greater than 1, although many values less than 1 are also allowed. Option D incorrectly includes the two values that make the denominator zero. Memory cue: for a reciprocal, set the denominator unequal to zero.

The direct answer is option A: \(x>-2\). For \(f(x)=x^2-4\), every real number is allowed. For \(g(x)=\sqrt{x+2}\), the radicand must satisfy \(x+2\ge0\), so \(x\ge-2\). However, in \((f/g)(x)\), this square root is in the denominator, so it must not equal zero. Thus \(\sqrt{x+2}>0\), which requires \(x+2>0\), or \(x>-2\). Option A is correct. Option B includes \(x=-2\), but at that point the denominator is zero. Option C excludes 2 for no valid reason; the numerator may be zero there and the quotient is still defined. Option D includes values below -2, where the square root is not real, and also includes -2. The key difference is that a square root in a numerator may be zero, but a square root in a denominator may not. Memory cue: denominator square root means strictly positive radicand.

((fg)(x)=(2x+3)(x-5)=2x^2-7x-15), so the coefficient of (x^2) is (2). In multiplication, check the leading terms first.

From (x^2+x+1=x^2(x+1)), the equation is (x^3-x-1=0). The listed integer pairs are not exact solutions.

(2x+1=x^2+x) gives (x^2-x-1=0). The quadratic formula gives (x=\frac{1\pm\sqrt{5}}{2}), so the listed option pattern would be wrong.

((f+g)(x)=2x+1) and ((fg)(x)=x(x+1)), so (x^2-x-1=0). The quadratic formula gives (\frac{1\pm\sqrt{5}}{2}).

\(\frac{f}{g}=\frac{x-2}{x+2}\cdot\frac{x-2}{x+2}=\frac{(x-2)^2}{(x+2)^2}\). The restrictions \(x\ne\pm2\) from the original functions remain.

For the difference of two functions, (g − f)(x) = g(x) − f(x) = (x² + 1) − (x² − 1) = 2. The x² terms cancel, so the result remains 2 for every x; hence it is a constant positive function. Option B has the wrong sign, while option C results from adding rather than subtracting the functions. Exam tip: In (g − f)(x), subtract the entire first function from the second and change the signs inside its parentheses carefully.