Algebra of real functions

Here (fg)(x) denotes the product f(x)g(x), not composition. Multiply the two functions: (fg)(x) = (x + 2)(x + 3). Applying distribution gives x^2 + 3x + 2x + 6, which simplifies to x^2 + 5x + 6. The constant term is the term independent of x, so it is 6 and option A is correct. There is also a quicker way to identify it: the constant term of a product comes from multiplying the constant terms of the factors, namely 2 × 3 = 6. Option B is the coefficient of x after combining like terms. Options C and D are the individual constant terms of f and g, not the constant term of their product. Recognizing the notation and combining like terms are both essential.

By the definition of the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Hence, \((f-g)(x)=4-(x^2-1)=4-x^2+1=5-x^2\). The brackets are important because the minus sign changes the signs inside \(g(x)\). Exam tip: in \(f-g\), subtract the second function from the first.

(g-f)(x) means g(x)-f(x). Thus, (g-f)(x)=(3x+2)-(x-1)=3x+2-x+1=2x+3. Option D results from multiplying f(x) and g(x), namely (x-1)(3x+2), not from subtracting them. Exam tip: Always follow the order in function subtraction; (g-f) and (f-g) generally give opposite results.

(f−g)(−2)=f(−2)−g(−2)=(-2)²−(-2)=4+2=6. Subtracting the negative value of g(−2) changes the subtraction into addition. Thus, 6 is correct; 4 is only the value of f(−2). Exam tip: evaluate both functions first and then apply the indicated operation.

The sum of functions is defined pointwise: \((f+g+h)(x)=f(x)+g(x)+h(x)\). Therefore, \((x+1)+(x+2)+(x+3)=3x+6\). Option B is incorrect because it counts the term \(x\) only once. Exam tip: for a sum of functions, add their values at the same input.

\(\left(\frac{f}{g}\right)(1)=\frac{f(1)}{g(1)}=\frac{7}{1}=7\). Always check the denominator value in a quotient.

The domain of \(f+g\) is the intersection of the domains of \(f\) and \(g\). For \(f(x)\), \(x-1\neq 0\), so \(x\neq 1\); for \(g(x)\), \(x+1\neq 0\), so \(x\neq -1\). Therefore, both 1 and -1 are excluded. Choosing only one value ignores the restriction from the other denominator. Exam tip: For the sum or difference of functions, exclude every value that makes any relevant denominator zero.

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Thus, \(f(5)=5+4=9\) and \(g(5)=2(5)-3=7\), so \((f+g)(5)=9+7=16\). Exam tip: either add the two function values or first simplify to \((f+g)(x)=3x+1\).

\(f(2)=3(2)-2=4\) and \(g(2)=2+5=7\). Therefore, \((fg)(2)=f(2)\,g(2)=4\times7=28\). Here, \((fg)(x)\) denotes the product of the two function values, not the composition \(f(g(x))\). In an exam, evaluate both functions first and then multiply.

For the sum of two functions, \((f+g)(x)=f(x)+g(x)=x^2+2+2x=x^2+2x+2\). Completing the square gives \(x^2+2x+2=(x^2+2x+1)+1=(x+1)^2+1\), so option A is correct. Option B would produce the term \(-2x\) instead of \(+2x\). Exam tip: for \(x^2+2x+a\), write it as \((x+1)^2+(a-1)\).

The difference of two functions is defined pointwise: (f − g)(x) = f(x) − g(x). Since the two given functions are exactly identical, substitute their expressions to obtain (x − 2) − (x − 2). Removing the brackets gives x − 2 − x + 2, and combining like terms gives 0 for every value of x in their common domain. Thus the difference is the zero function, so option A is correct. Option B results from subtracting the constants incorrectly, option C is the square of x − 2 rather than a difference, and option D simply repeats one function without performing the subtraction. Equality of the two functions is the decisive observation.

The product of functions is defined by \((fg)(x)=f(x)g(x)\). Substituting the given functions gives \((fg)(x)=1\cdot(x^2+2x)=x^2+2x\), so option A is correct.

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Thus, \((f+g)(x)=0+(x+7)=x+7\). Adding the zero function leaves the other function unchanged, so option B is incorrect because \(g(x)\) is not the zero function. Exam tip: For a sum of functions, add their values at the same input \(x\).

For the linear combination, \((f+3g)(x)=f(x)+3g(x)\). Therefore, \(x^2-2x+3x=x^2+x\), so option A is correct. Option B results from subtracting \(3x\) instead of adding it. Exam tip: write the combination rule first, then substitute the functions.

In \(\frac{g}{f}\), the denominator is \(f(x)=x+1\), so \(x=-1\) is excluded. Identify the denominator function carefully.

The product of two functions is defined by \((fg)(x)=f(x)g(x)\). Therefore, \((fg)(x)=(2x+3)(2x-3)\). Using the difference-of-squares identity \((a+b)(a-b)=a^2-b^2\), we get \((2x)^2-3^2=4x^2-9\). Hence, option A is correct. Exam tip: do not confuse the product \((fg)(x)\) with the composition \((f\circ g)(x)\).

Since \(f(x)=g(x)\), \(\frac{f(x)}{g(x)}=1\). Here \(x^2+1\) is never (0), so all real (x) are valid.

The sum of two functions is obtained by adding their values at the same input: \((f+g)(x)=f(x)+g(x)\). Thus, \((x^2+4x)+(2x^2-x)=3x^2+3x\). The quadratic terms and linear terms are combined separately. Exam tip: while adding algebraic functions, combine only like terms.

The governing concept is subtraction of functions, which is defined pointwise: (f − g)(x) = f(x) − g(x). It is not function composition, so the two given expressions should first be evaluated at the same input. At x = 1, f(1) = 5(1) + 2 = 7, while g(1) = 2(1) − 7 = −5. Hence (f − g)(1) = f(1) − g(1) = 7 − (−5) = 7 + 5 = 12. Therefore option A is correct. The negative sign in g(1) is important: subtracting −5 is equivalent to adding 5. Option B could arise from ignoring the negative sign, whereas the other choices do not follow the definition of pointwise function subtraction.

(f(2)=4) and (g(2)=0), so ((fg)(2)=4\cdot 0=0). In multiplication, if one value is (0), the product is (0).

For (\sqrt{x+2}), (x\ge -2), and for (\frac{1}{x-1}), (x\ne 1). Apply both conditions together.

By the definition of the difference of two functions, \((g-f)(x)=g(x)-f(x)\). Hence, \((g-f)(x)=(x^2+1)-(x^2-1)=x^2+1-x^2+1=2\). Option B results from not handling the subtraction of the second bracket correctly. Exam tip: When subtracting a bracket, change the signs of every term inside it.

The governing concept is the product of functions. In this notation, (fg)(x) means f(x)g(x), not composition, because no composition symbol is shown. Evaluate each function at x = 1: f(1) = 1² + 3 = 4 and g(1) = 2(1) = 2. The product is therefore (fg)(1) = f(1)g(1) = 4 × 2 = 8. Hence option A is correct. Option C is only f(1), and option D is only g(1), so each omits one factor. Option B may result from an incomplete or mistaken calculation and is not the product of the two values. The essential procedure is to evaluate both functions at the specified input and then multiply the resulting numbers.

The rule for subtracting functions is (g−f)(x)=g(x)−f(x). Substituting the given expressions carefully gives (g−f)(x)=(x²−4)−(3x+5). The negative sign before the second bracket must be distributed to both terms inside it: x²−4−3x−5. Combining like terms produces x²−3x−9. Thus option A is correct. A common error is to subtract 3x but add 5, which would produce option C; this ignores the effect of the minus sign on the entire expression f(x). Options B and D reverse the sign of the linear term and therefore do not represent g−f. Writing parentheses before simplifying is important because function subtraction means subtracting every term of f from the corresponding value of g.

The governing concept is addition of functions: (f+g)(x)=f(x)+g(x). Add the expressions before substituting x=3: (f+g)(x)=(x^2+2x)+(x^2-2x)=2x^2, because the linear terms +2x and -2x cancel. Therefore, (f+g)(3)=2(3^2)=2×9=18, so option A is correct. A separate calculation confirms it: f(3)=3^2+2(3)=9+6=15 and g(3)=3^2-2(3)=9-6=3; their sum is 15+3=18. Option B or C can result from incomplete addition or omission of a term. Option D may result from squaring the final value incorrectly or multiplying by an unnecessary factor. The cancellation of opposite linear terms is the key step.