Algebra of real functions

The direct answer is A: \(8x\). First calculate each function sum and difference. \((f+g)(x)=(x+2)+(x-2)=2x\). Next, \((f-g)(x)=(x+2)-(x-2)=x+2-x+2=4\). Their product is therefore \((2x)(4)=8x\). Option A is correct. Option B, \(4x\), leaves out a factor of 2 and would be only half the required product. Option C, \(x^2-4\), is the product \((x+2)(x-2)"), not the requested product of the sum and difference. Option D, \(2x+4\), is just twice \(f(x)\), not the product. Work inside each bracket before multiplying; this prevents confusing \(f+g\) with \(fg\).

For a product \((fg)(x)=f(x)g(x)\), the domain is the intersection of the domains of the two original functions. The function \(f(x)=\frac{1}{x^2-4}\) is undefined when its denominator is zero. Factoring gives \(x^2-4=(x-2)(x+2)\), so \(x=2\) and \(x=-2\) must be excluded. The function \(g(x)=x+2\) is defined for every real number, so the product domain is \(\mathbb R\setminus\{-2,2\}\). Option A is correct.

Although multiplication simplifies formally to \(\frac{x+2}{(x-2)(x+2)}=\frac{1}{x-2}\) for allowed values, the original function was not defined at \(x=-2\). Cancellation does not restore a missing input. Thus both excluded values remain excluded from the domain.

The sum of two functions is defined by (f+g)(x)=f(x)+g(x). Hence, (f+g)(x)=(x²+1)+2x=x²+2x+1. Option B is incorrect because it subtracts 2x instead of adding it. Exam tip: For the sum of functions, add their values at the same input x.

First, f(1)=3(1)-2=1 and g(1)=1+4=5. For the product function, (fg)(1)=f(1)×g(1)=1×5=5, so option A is correct. The value 7 is the sum f(1)+g(1), not their product. In exams, interpret (fg)(x) as f(x)g(x) and multiply the function values.

The direct answer is option A: the simplified expression is \(x+2\), with domain \(x\ne3\). Multiply the functions: \((fg)(x)=\frac{x+2}{x-3}(x-3)\). For \(x\ne3\), the non-zero factor \(x-3\) cancels, giving \(x+2\). However, the original function \(f(x)\) is undefined at \(x=3\), because its denominator becomes zero. Thus the domain remains \(\mathbb R\setminus\{3\}\). Option A is correct. Option B has the right simplified formula but incorrectly includes 3 in the domain. Option C is the wrong factor, not the product. Option D stops before simplifying and is not the requested simplified form, though it shows the original restriction. A cancelled denominator does not restore a forbidden original input. Memory cue: simplify the formula, but preserve every original domain restriction.

By the definition of the difference of functions, (f − g)(x) = f(x) − g(x) = x³ − x. Factoring out x gives x(x² − 1), and the difference-of-squares identity gives x(x − 1)(x + 1). Therefore, option A is correct.

By the definition of the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \((f+g)(x)=(2x+5)+(5-2x)=10\), since \(2x\) and \(-2x\) cancel each other. Option B does not correctly add the corresponding terms, while option C is related to subtraction. In an exam, add the values of the two functions for the same input.

The quotient function is \(\left(\frac{f}{g}\right)(x)=\frac{x^2+6x+9}{x+3}=\frac{(x+3)^2}{x+3}=x+3\), where \(x\ne -3\). Hence, at \(x=0\), its value is \(0+3=3\). The value 9 is only \(f(0)\), not the quotient. Exam tip: simplify the quotient carefully and exclude values that make the denominator zero.

For the algebraic combination of functions, \((3f+g)(x)=3f(x)+g(x)\). Thus, \(3f(x)=3(2x-3)=6x-9\), and \((3f+g)(x)=6x-9+(x^2+1)=x^2+6x-8\). In option B, the coefficient 3 has incorrectly been applied to \(g(x)\) as well. Exam tip: a scalar written before a function multiplies only that function.

\(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-5x+6}{x-2}\). Factoring the numerator gives \(x^2-5x+6=(x-2)(x-3)\), so for \(x\ne2\), the quotient is \(x-3\). Option C is only the denominator, while option B has an incorrect sign. Exam tip: factor the quadratic before dividing, and exclude the value that makes the original denominator zero.

The product is defined by (fg)(x) = f(x)g(x). Substituting the given functions gives (x + 1)·[1/(x + 1)] = 1, but this cancellation is valid only when x + 1 ≠ 0. Thus x cannot equal −1, because g(x) has a zero denominator there and is not defined. The correct statement is (fg)(x) = 1 for x ≠ −1, so option A is correct. Option B incorrectly includes the excluded input. Option C gives only one factor rather than the product, and option D treats an undefined value as zero. The simplified formula is constant, but its original domain restriction must still be retained.

By the definition of subtraction of functions, \((g-f)(x)=g(x)-f(x)\). Therefore, \((g-f)(x)=(2x^2-1)-(x^2+2)=2x^2-1-x^2-2=x^2-3\), so option A is correct. Option C results from mishandling the signs of the constants, while option B is the result of addition, \((f+g)(x)\). Exam tip: in \(g-f\), write \(g\) first and subtract the entire expression for \(f\).

\((fg)(x)=f(x)g(x)\) is defined only when both fractions are defined.
For \(f(x)=\frac{x-2}{x+3}\), denominator \(x+3\neq 0\Rightarrow x\neq -3\).
For \(g(x)=\frac{x+3}{x-2}\), denominator \(x-2\neq 0\Rightarrow x\neq 2\).
So the domain is \(\mathbb{R}\setminus\{-3,2\}\).

The sum of two functions is defined pointwise: \((f+g)(x)=f(x)+g(x)\). Therefore, \((x^2+x)+(x^2-x)=2x^2\), because the terms \(+x\) and \(-x\) cancel each other. Hence, option A is correct. Exam tip: combine like terms carefully, especially when their signs are opposite.

The sum (f + g)(x) is √(4 − x) + 1/x. The square-root condition is 4 − x ≥ 0, which gives x ≤ 4. The reciprocal function requires x ≠ 0 because division by zero is undefined. Both conditions must hold simultaneously, so the domain is (−∞, 4] with 0 removed. Equivalently, it can be written as (−∞, 0) ∪ (0, 4]. Therefore option A is correct. Option B forgets the denominator restriction, option C reverses the square-root inequality, and option D includes values greater than 4 where √(4 − x) is not real. Addition does not cancel either restriction.

The direct answer is A: \(1-x\), for \(x\ne1\). Start with \(\left(\frac{f}{g}\right)(x)=\frac{x^2-2x+1}{1-x}\). The numerator is \((x-1)^2\). The denominator is \(1-x=-(x-1)\). Hence the quotient is \(\frac{(x-1)^2}{-(x-1)}=-(x-1)=1-x\), provided \(x\ne1\). Option A is correct. Option B gives \(x-1\), which has the opposite sign; it forgets that \(1-x=-(x-1)\). Option C, \(x+1\), does not follow from either factorisation. Option D, \(x^2-1"), is not the result after cancellation and has unrelated factors. At \(x=1\), the original denominator is zero, so that value is excluded even though the simplified expression can be evaluated there. Sign check: reverse the order in a subtraction and a minus sign appears.

Here (f(2)=1) and (g(2)=5), so (h(2)=2\cdot1-5=-3). In exams, first find individual function values, then apply the algebraic operation.

For (f(x)), (x\geq1), and for (g(x)), (x\neq3), so the domain of (fg) is ([1,\infty)-{3}). For product domains, take the intersection of the domains of both functions.

The governing concept is piecewise evaluation of an absolute-value function. We need (f−g)(x)=|x−2|−(x−2)=0, which is equivalent to |x−2|=x−2. If x≥2, then x−2 is non-negative, so |x−2|=x−2 and the difference is zero for every value in this interval. If x<2, then |x−2|=−(x−2)=2−x, and the difference becomes (2−x)−(x−2)=4−2x. For x<2 this expression is positive, not zero. The boundary value x=2 is included in x≥2 and gives f(2)=g(2)=0. Hence the complete solution set is x≥2, making option A correct. Option B excludes all valid values above 2, option C includes only the boundary point, and option D incorrectly includes the negative-side interval.

The governing concept is the parity of a product of functions. Here (fg)(x) means f(x)g(x), so h(x)=(fg)(x)=|x|x². Both factors are even because |−x|=|x| and (−x)²=x². Direct substitution confirms this: h(−x)=|−x|(−x)²=|x|x²=h(x), which is precisely the definition of an even function. It is not odd, since an odd function must satisfy h(−x)=−h(x). Option D is not the correct classification: although h(x) is non-negative, it is zero at x=0, so it is not positive for every real x. Thus option A is correct.

The governing concept is determining the parity of a function after performing algebraic operations on functions. Add the expressions term by term: (f+g)(x)=(x²+3x)+(x²−3x)=2x², because the linear terms +3x and −3x cancel. Let h(x)=2x². Then h(−x)=2(−x)²=2x²=h(x), which satisfies the definition of an even function. It is not odd, since an odd function would require h(−x)=−h(x); for example, h(1)=2 and h(−1)=2 rather than −2. It is not constant because h(0)=0 while h(1)=2. Therefore option A is the only correct classification. The cancellation of the opposite linear terms is the key algebraic step.

Answer: A, −8. The notation (f−g)(x) means f(x)−g(x), so the order must remain unchanged. Substitute x=4 into f: f(4)=|4−3|=|1|=1. Substitute into g: g(4)=2(4)+1=9. Therefore (f−g)(4)=1−9=−8. Option A is correct. B, 8, comes from reversing the subtraction and calculating g(4)−f(4), which is not what was asked. C, −6, results from an arithmetic or substitution error. D, 10, is the sum 1+9 rather than the difference. A useful check is that f(4)=1 is smaller than g(4)=9, so their difference in the stated order must be negative; the gap is 8. Memory cue: evaluate each function first, then subtract in the displayed order.

The governing concept is that the domain of a product of functions is the intersection of the domains of the individual functions. For f(x), the denominator x−1 cannot be zero, so x ≠ 1. For g(x), the denominator x+1 cannot be zero, so x ≠ −1. Both restrictions must remain even though the algebraic product simplifies to [(x+1)(x−1)]/[(x−1)(x+1)] = 1 wherever it is defined. Hence the domain is ℝ excluding −1 and 1, making option A correct. Options B and C omit one necessary restriction; option D incorrectly permits zero denominators.

The governing concept is addition of rational functions by using a common denominator while retaining domain restrictions. Start with (f+g)(x)=1/(x+1)+1/(x−1). The common denominator is (x+1)(x−1)=x²−1. Combining the numerators gives (x−1)+(x+1)=2x, so the sum becomes 2x/(x²−1). The original denominators show that x cannot be −1 or 1; these exclusions remain even if a later algebraic expression appears simplified. Therefore option A is correct. Option B omits x from the numerator, option C omits the factor 2, and option D uses the false identity (x+1)(x−1)=x²+1 and incorrectly allows all real x.

The governing concept is transforming a function equation into a polynomial equation without losing the original domain restriction. Since f(x)+g(x)=x²+1/x, the given condition is x²+1/x=2, where x≠0 because g is undefined at zero. Multiply the equation by x, which is legitimate under that restriction: x³+1=2x. Bringing every term to the left gives x³−2x+1=0. Hence option A is correct. Option B does not have the cubic term produced when the fraction is cleared. Option C changes the sign of 2x, and option D changes the sign of the constant 1. The condition x≠0 must also be retained because multiplying can otherwise conceal the original domain issue.