Algebra of real functions

Correct answer: A. (0)

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

Correct answer: A. (\frac{2}{(x-2)(x+2)})

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.

Correct answer: A. (\frac{1}{x+2},\ x\ne\pm2)

(\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.

Correct answer: A. Odd function

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.

Correct answer: A. Odd function

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 \(-x\).

Correct answer: A. x²/(x+1), x ≠ 0, −1

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.

Correct answer: A. 4x² − 7

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.

Correct answer: A. \(x\ge4,\ x\ne6\)

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).

Correct answer: A. 2

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.

Correct answer: A. (x^2(x-2)(x+2))

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

Correct answer: A. ([-2,-1]\cup[1,2])

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.

Correct answer: A. It is undefined

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

Correct answer: A. Their formulas match after simplification but domains differ

(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.

Correct answer: A. (2x²+8)/(x²−4), x ≠ ±2

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.

Correct answer: A. 2

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.

Correct answer: A. (\sqrt{x^2+2x-3},\ x\ge1)

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}).

Correct answer: A. (x-1,\ x\ne-1)

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.

Correct answer: A. No minimum value exists

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

Correct answer: A. (\frac{2}{x^2-1},\ x\ne\pm1)

(\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.

Correct answer: A. ([-2,6])

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.

Correct answer: A. ( [2,5] )

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

Correct answer: A. ( \mathbb{R}-{3} )

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

Correct answer: A. ( \mathbb{R}-{-2,2} )

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

Correct answer: A. 2

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.

Correct answer: A. −12

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.