Algebra of real functions

Correct answer: A. 3x-2

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(x)=(2x+3)+(x-5)=3x-2\). Option B results from handling the constant terms incorrectly. In exams, remember to add the values of both functions at the same input.

Correct answer: B. 5x − 13

By definition, (f − g)(x) = f(x) − g(x). Substituting the given functions gives (7x − 4) − (2x + 9) = 7x − 4 − 2x − 9 = 5x − 13. The minus sign changes the signs of both terms in g(x).

Correct answer: A. x² + 4x − 12

The product of two real functions is defined by (fg)(x) = f(x)g(x). Hence, (fg)(x) = (x + 6)(x − 2) = x² + 4x − 12. Option C has an incorrect coefficient for the middle term. Exam tip: distinguish (fg)(x), which denotes the product, from f(g(x)), which denotes composition.

Correct answer: A. (x+2)

\(\frac{4x^2+8x}{4x}=\frac{4x(x+2)}{4x}=x+2\), where \(x\neq0\). Check the denominator condition before cancellation.

Correct answer: C. 11

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \(f(2)=2^2+2=6\) and \(g(2)=3(2)-1=5\). Therefore, \((f+g)(2)=6+5=11\), so option C is correct. In an exam, calculate the values of the two functions separately before adding them to avoid substitution errors.

Correct answer: A. 8

The difference of two functions is defined by \((f-g)(x)=f(x)-g(x)\). Thus, \(f(3)=5(3)+2=17\) and \(g(3)=3^2=9\), so \((f-g)(3)=17-9=8\). Reversing the order, as in \((g-f)(3)\), would give \(-8\). Exam tip: Evaluate both functions first and subtract them in the stated order.

Correct answer: B. 18

For the product of functions, \((fg)(x)=f(x)g(x)\). Thus, \(f(2)=2(2)-1=3\) and \(g(2)=2+4=6\), so \((fg)(2)=3\times6=18\). An answer such as 21 results from an arithmetic mistake; in the exam, evaluate both functions first and then multiply their values.

Correct answer: C. (9)

The direct answer is option C: 9. By definition, \(\left(\frac{f}{g}\right)(5)=\frac{f(5)}{g(5)}\). Compute f(5)=5^2−16=25−16=9. Compute g(5)=5−4=1. Therefore the quotient is 9/1=9. The denominator is nonzero, so the value is defined. Option A, 5, is wrong because it does not equal the quotient after substitution. Option B, 7, may result from an arithmetic error such as 25−16 being treated as 7; the correct difference is 9. Option C, 9, is correct. Option D, 11, also does not follow from either the numerator or the quotient. Another valid check is to factor x^2−16=(x−4)(x+4), so for x≠4 the quotient simplifies to x+4; at x=5 this is 9. The condition matters because at x=4 the denominator would be zero, but x=5 is allowed. Exam cue: evaluate numerator and denominator separately, then divide.

Correct answer: A. \(5x^2+1\)

The sum of two functions is found pointwise: \((f+g)(x)=f(x)+g(x)=3x^2+(2x^2+1)=5x^2+1\). Hence, option A is correct. Option B incorrectly combines the expressions as if they were multiplied and also changes the powers. Exam tip: when adding like terms, add their coefficients and keep the common power unchanged.

Correct answer: A. \(5x+5\)

The difference of two functions is defined as \((f-g)(x)=f(x)-g(x)\). Thus, \((f-g)(x)=(9x-2)-(4x-7)=9x-2-4x+7=5x+5\). The key step is that \(-(4x-7)=-4x+7\), so both signs inside the brackets change. Exam tip: when subtracting a bracketed expression, reverse the sign of every term inside it.

Correct answer: C. \(2x^4\)

The product of two real functions is defined by \((fg)(x)=f(x)g(x)\). Thus, \((fg)(x)=x^3\cdot 2x=2x^{3+1}=2x^4\), so option C is correct. Option B misses the coefficient 2. Exam tip: when multiplying powers with the same base, add their exponents.

Correct answer: A. \(2x^2\)

\(\left(\frac{f}{g}\right)(x)=\frac{10x^3}{5x}=2x^2\), where \(x\neq0\). In division, powers are subtracted.

Correct answer: C. 3x − 24

The expression 3f(x) means 3 times the entire function value. Substituting f(x) = x − 8 gives 3f(x) = 3(x − 8) = 3x − 24, so option C is correct.

Correct answer: A. 2x + 3

The notation (1/2 f)(x) means that the function f is multiplied by the scalar 1/2; it does not mean that the input x is replaced by x/2. Therefore, apply 1/2 to the complete function value: (1/2 f)(x) = (1/2)f(x) = (1/2)(4x + 6). Distributing the scalar gives (4x)/2 + 6/2 = 2x + 3. Hence option A is correct. Option B divides only the constant term, option C leaves the constant unchanged, and option D doubles both terms instead of multiplying them by one-half. The governing concept is scalar multiplication of a real-valued function, where every term in the output is multiplied by the same scalar.

Correct answer: A. \(-2x^2+5\)

\((-f)(x)=-f(x)=-(2x^2-5)=-2x^2+5\). The negative sign applies to every term inside the function, so the sign of \(-5\) changes to \(+5\). Option D is the original function \(f(x)\), while option B fails to change the sign of \(2x^2\). Exam tip: when a minus sign precedes a bracket, reverse the sign of every term while expanding it.

Correct answer: B. \(2x^2+8x\)

By the definition of addition of functions, \((f+f)(x)=f(x)+f(x)=2f(x)\). Hence, \((f+f)(x)=2(x^2+4x)=2x^2+8x\), so option B is correct. In option C, only the \(4x\) term has effectively been doubled, whereas adding two identical functions doubles every term. Exam tip: Rewrite \(f+f\) as \(2f\) before substituting the given function.

Correct answer: C. \(x^2-6x+9\)

By the definition of the product of functions, \((ff)(x)=f(x)\cdot f(x)\). Hence, \((ff)(x)=(x-3)(x-3)=(x-3)^2=x^2-6x+9\), so option C is correct. Option A results from adding \(f(x)\) to itself, not from multiplying the functions. Exam tip: interpret \((fg)(x)\) as \(f(x)g(x)\), whereas \((f+g)(x)\) means \(f(x)+g(x)\).

Correct answer: A. (0)

The difference of equal functions is (0), so ((f-g)(x)=0). In exams, first notice if both functions are equal.

Correct answer: B. (0)

The direct answer is option B: \((fg)(x)=0\). The product of two functions is defined pointwise: \((fg)(x)=f(x)g(x)\). Substitute the given values: \((fg)(x)=0\cdot(3x^2-2)=0\). This is true for every allowed \(x\). Option A is only \(g(x)\), not the product, because it forgets the factor \(f(x)=0\). Option B is correct because multiplying any expression by zero gives zero. Option C, \(3x^2\), drops the constant term and also ignores the zero factor. Option D, \(2\), is not the product and has no valid calculation leading to it. The zero function outputs zero everywhere, so multiplying by it makes the entire product the zero function. Exam cue: for a product of functions, multiply their outputs, not their names or formulas selectively.

Correct answer: A. \(5x-5\)

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(x)=1+(5x-6)=5x-5\). Option B results from forgetting to add the constant \(1\). In an exam, first write the function-addition rule and then combine like terms.

Correct answer: B. 18

The sum of two functions is defined pointwise: (f + g)(x) = f(x) + g(x). Therefore, (f + g)(x) = (x + 9) + (9 - x) = x - x + 18 = 18. Hence, option B is correct. Exam tip: the terms x and -x cancel each other, so the result is not 2x.

Correct answer: B. \(3x\)

The difference of two functions is defined as \((f-g)(x)=f(x)-g(x)\). Therefore, \((f-g)(x)=(x^2+5x)-(x^2+2x)=3x\). The like terms \(x^2\) cancel each other. Exam tip: when subtracting functions, change the sign of every term in the second function before combining like terms.

Correct answer: A. 8

The product of two functions is defined by (fg)(x)=f(x)g(x). Here, f(1)=1+7=8 and g(1)=3(1)-2=1. Therefore, (fg)(1)=8×1=8. Exam tip: Do not confuse (fg)(x), the product of functions, with f(g(x)), their composition.

Correct answer: C. 6

First, \(f(1)=1^2+10=11\) and \(g(1)=1^2+4=5\). Hence, \((f-g)(1)=f(1)-g(1)=11-5=6\). Equivalently, \(f(x)-g(x)=(x^2+10)-(x^2+4)=6\), since the \(x^2\) terms cancel. Exam tip: For \((f-g)(a)\), always calculate \(f(a)-g(a)\), not their sum.

Correct answer: A. \(\frac{1}{3}\)

The quotient function is \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\). Here, \(f(0)=1\) and \(g(0)=3\), so \(\left(\frac{f}{g}\right)(0)=\frac{1}{3}\). Option B gives only the denominator value, not the quotient. Exam tip: evaluate both functions at the given input and divide, ensuring that the denominator is not zero.