Algebra of real functions

(f(1)=1) and (g(1)=-1), so the sum is (0). Before evaluating, check whether (x=1) lies in the domain.

With common denominator ((x-2)(x+2)), numerator is (2x-(x+2)=x-2), so the form is (\frac{1}{x+2}) with (x\ne\pm2). Thus option (C) is the simplification.

(\frac{2x}{(x-2)(x+2)}-\frac{x+2}{(x-2)(x+2)}=\frac{x-2}{(x-2)(x+2)}). Hence the simplified form is (\frac{1}{x+2}), but (x=\pm2) remain excluded.

The direct answer is option A: an odd function. Subtract the functions carefully: \((f-g)(x)=(x^2+3x)-(x^2-3x)=x^2+3x-x^2+3x=6x\). To test parity, replace \(x\) by \(-x\): \(h(-x)=6(-x)=-6x=-h(x)\). Therefore \(h\) is odd. Option A works because \(6x\) changes sign when \(x\) changes sign. Option B is wrong because an even function satisfies \(h(-x)=h(x)\), which does not happen. Option C is wrong because \(6x\) changes with \(x\), so it is not constant. Option D is wrong because the function does satisfy the odd-function rule. A useful cue is that a linear function through the origin, such as \(6x\), is odd.

The direct answer is option A: odd function. Multiply the functions: \((fg)(x)=f(x)g(x)=\frac1x\cdot\frac1{x^2}=\frac1{x^3}\), with domain \(\mathbb{R}\setminus\{0\}\). A function is odd when \(h(-x)=-h(x)\) for every allowed \(x\). Let \(h(x)=1/x^3\). Then \(h(-x)=1/(-x)^3=-1/x^3=-h(x)\), so it is odd. Option A is correct. Option B is wrong because an even function would satisfy \(h(-x)=h(x)\); that does not happen here. Option C is wrong because the sign does change and the oddness condition is satisfied. Option D is wrong because \(1/x^3\) changes with \(x\), so it is not constant. The domain is symmetric about zero, which is necessary for discussing evenness or oddness, and zero itself is simply excluded. Memory cue: an odd power in the denominator gives a sign change when \(x\) becomes \(-x\).

The quotient of two functions is defined by (f/g)(x) = f(x)/g(x), so we must divide x/(x+1) by 1/x. Dividing by a fraction means multiplying by its reciprocal: [x/(x+1)] × x = x²/(x+1). The domain must preserve every restriction from the original functions and must also ensure that the divisor g(x) is not zero. Here f(x) requires x ≠ −1, while g(x) = 1/x requires x ≠ 0; also g(x) never equals zero for any allowed real x. Therefore the domain is all real numbers except 0 and −1. Option A is correct. Option B incorrectly cancels a factor, option C reverses the operation, and option D uses the reciprocal of the final result.

By the definition of operations on functions, (3f − 2g)(x) = 3f(x) − 2g(x). Substituting the given functions gives 3(2x² − 1) − 2(x² + 2) = 6x² − 3 − 2x² − 4 = 4x² − 7. Therefore, option A is correct.

For (f), \(x\ge4\), and for (g), (x\ne6). In \(\frac{f}{g}\), (g(x)\ne0) is also needed, but \(\frac{1}{x-6}\) is never (0).

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \(f(2)=2-1=1\) and \(g(2)=\frac{1}{2-1}=1\), so \((f+g)(2)=1+1=2\). Option B gives the value of only one function, not their sum. Exam tip: evaluate each function separately and then add; although \(g(x)\) is undefined at \(x=1\), \(x=2\) is valid.

(f=x(x-2)) and (g=x(x+2)), so (fg=x^2(x-2)(x+2)). Factoring first is better than long expansion.

The direct answer is option A: \([-2,-1]\cup[1,2]\). For a square root to be real, its expression must be non-negative. For \(f(x)=\sqrt{x^2-1}\), require \(x^2-1\ge0\), so \(x^2\ge1\), giving \(x\le-1\) or \(x\ge1\). Thus the domain of \(f\) is \((-infty,-1]\cup[1,\infty)\). For \(g(x)=\sqrt{4-x^2}\), require \(4-x^2\ge0\), so \(x^2\le4\), giving \(-2\le x\le2\). The domain of \(f+g\) is the common part, or intersection, of both domains: \([-2,-1]\cup[1,2]\). Option A works because every square root is defined there. Option B, \([-1,1]\), contains points where \(x^2-1<0\), so \(f\) is not real. Option C, \([-2,2]\), includes numbers between \(-1\) and \(1\), again invalid for \(f\). Option D is the domain of \(f\) alone and ignores the restriction from \(g\). Memory cue: for a sum, take the intersection of the two domains, not their union.

\(\frac{f}{g}=\frac{(x+2)^2}{x+2}\), but at \(x=-2\) the original denominator is (0). Cancellation does not bring that point back into the domain.

(f(x)=x+3) only for (x\ne3), while (g) is defined for all real (x). The same algebraic form does not always mean the same function.

To add rational functions, use a common denominator rather than adding numerators directly. The denominators are x−2 and x+2, so their common denominator is (x−2)(x+2) = x²−4. Thus (f+g)(x) = [(x+2)² + (x−2)²]/[(x−2)(x+2)]. Expanding gives (x²+4x+4) + (x²−4x+4) = 2x²+8; the middle terms cancel. Hence the result is (2x²+8)/(x²−4). The original functions are undefined at x = 2 and x = −2, so these values must remain excluded even if later cancellation were possible. Option A is correct. Option B omits the square terms, C omits a factor of 2, and D incorrectly treats the reciprocal expressions as exact additive inverses.

For a quotient function, the denominator must be non-zero. Since \(f(x)=x^2-6x+9=(x-3)^2\), we get \(\left(\frac{f}{g}\right)(x)=\frac{(x-3)^2}{x-3}=x-3\), for \(x\neq 3\). Therefore, \(\left(\frac{f}{g}\right)(5)=5-3=2\). The value 4 is only \(f(5)\), not the quotient. Exam tip: evaluate the numerator and denominator separately, then divide.

For both square roots, (x+3\ge0) and (x-1\ge0), so (x\ge1). The product is (\sqrt{(x+3)(x-1)}=\sqrt{x^2+2x-3}).

The direct answer is option A: \(x-1\), with \(x\ne-1\). Compute the difference: \((f-g)(x)=\frac{x^2}{x+1}-\frac{1}{x+1}=\frac{x^2-1}{x+1}\). Factor the numerator: \(x^2-1=(x-1)(x+1)\). Cancelling gives \(x-1\), but the original denominator still shows that \(x=-1\) is forbidden. Option A is correct. Option B, \(x+1\), comes from incorrect cancellation or factoring. Option C is not fully simplified and also gives the wrong excluded value \(x\ne1\); the original denominator excludes \(-1\). Option D fails because it does not cancel the common factor and is not equal to the difference. Always retain restrictions from the original expressions.

((fg)(x)=2x^3+4x), which tends to (-\infty) as (x\to-\infty). Therefore no real minimum value exists.

(\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x^2-1}=\frac{2}{x^2-1}). The zeros (x=\pm1) of the denominator are excluded.

For both square roots, \(x+2\ge0\) and \(6-x\ge0\), so \(-2\le x\le6\). The denominator (f(x)+g(x)) is never (0) on this interval, so no point is removed.

The common domain is ( [2,\infty)\cap(-\infty,5]=[2,5] ). For addition, always take the intersection of domains.

For product, the domain is the intersection of original domains, so only (x=3) is excluded. Check the original domain before simplifying.

We need (g(x)\neq0), and (x^2-4=0) gives (x=\pm2). In quotient questions, always exclude zeros of the denominator.

The sum of the functions is \((f+g)(x)=f(x)+g(x)=(2x-5)+(3-x)=x-2\). Therefore, \((f+g)(4)=4-2=2\), so option A is correct. Option B results from an incorrect simplification of the sum. In an exam, first form the sum of the functions and then substitute the given value.

The difference of two functions is defined pointwise: (f−g)(x) = f(x) − g(x). Substituting the given expressions gives (x²+2x) − (x²−2x). The minus sign applies to every term in the second parentheses, so this becomes x²+2x−x²+2x = 4x. Now evaluate at x = −3: (f−g)(−3) = 4(−3) = −12. Hence option A is correct. A direct check gives f(−3) = 9−6 = 3 and g(−3) = 9+6 = 15, so f(−3)−g(−3) = 3−15 = −12 as well. Option B results from reversing the subtraction, C from incorrectly cancelling the entire difference, and D from incomplete substitution. The governing concept is pointwise subtraction of functions.