Algebra of real functions

In addition, like (x) terms are added, so ((f+g)(x)=x+2+3x=4x+2). In exams, open brackets first.

The difference of two functions is defined as (f-g)(x)=f(x)-g(x). Thus, (5x-1)-(2x+4)=5x-1-2x-4=3x-5. Option C has the wrong sign for the constant term because the signs of all terms of the second function change during subtraction. Exam tip: enclose the second function in brackets before subtracting.

For the product of two real functions, \((fg)(x)=f(x)g(x)\). Therefore, \((fg)(x)=(x+1)(x-1)=x^2-1\), using the identity \(a^2-b^2=(a+b)(a-b)\). Option C, \(2x\), results from adding the functions, not multiplying them. In an exam, distinguish the product \((fg)(x)\) from the composition \((f\circ g)(x)\).

By the definition of the quotient of two functions, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\). Hence, \(\frac{2x+6}{2}=x+3\). Option B results from dividing only the constant term by 2, which is incorrect. Exam tip: divide the entire numerator by the denominator and check that the denominator is non-zero; here \(g(x)=2\), so every real \(x\) is allowed.

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(2)=f(2)+g(2)=2^2+4=4+4=8\). Option B gives only \(f(2)\) and does not include \(g(2)\). Exam tip: For \((f+g)(a)\), find the values of both functions at \(a\) and add them.

\((f-g)(1)=f(1)-g(1)\). Here, \(f(1)=3\) and \(g(1)=1+5=6\), so \((f-g)(1)=3-6=-3\). Option B results from reversing the subtraction order. Exam tip: In \((f-g)(a)\), always subtract the value of the second function from that of the first.

(fg)(1)=f(1)×g(1)=(1+2)(1+3)=3×4=12, so the correct answer is 12. The value 7 results from adding f(1) and g(1), but the question asks for their product. Exam tip: Rewrite (fg)(x) as f(x)g(x) before substituting the given value.

The governing concept is the quotient of two real-valued functions. By definition, (f/g)(x) = f(x)/g(x), provided that g(x) is not zero. Evaluate the numerator at x = 3: f(3) = 3^2 − 4 = 9 − 4 = 5. Evaluate the denominator separately: g(3) = 3 − 2 = 1. Therefore (f/g)(3) = f(3)/g(3) = 5/1 = 5, so option A is correct. The denominator is non-zero, so the quotient is defined. Option B is only the denominator value, option C is the unadjusted square 3^2, and option D is not obtained from the quotient rule. Separating numerator and denominator prevents confusion between function operations and ordinary addition or subtraction.

The sum of two functions is defined as \((f+g)(x)=f(x)+g(x)\). Therefore, \((f+g)(x)=(x²+1)+2x=x²+2x+1\), so option A is correct. Option B incorrectly changes the sign of 2x. Exam tip: In function addition, add the function values at the same input x and then combine like terms.

The difference of two functions is defined by \((f-g)(x)=f(x)-g(x)\). Thus, \((f-g)(x)=(4x+7)-(x-3)=4x+7-x+3=3x+10\). In option C, the sign of the constant term has been handled incorrectly. Exam tip: when subtracting a function, change the signs of every term inside its bracket.

For the product of two functions, (fg)(x)=f(x)g(x). Hence, (fg)(x)=(2x)(x^2)=2x^3. Option B, 2x^2, represents f(g(x)), the composition of the functions, not their product. Exam tip: distinguish (fg)(x) from f(g(x)).

\(\left(\frac{f}{g}\right)(x)=\frac{6x^2}{3x}=2x\), because \(x\neq0\). In quotient questions, check the condition first.

Multiplying a function f by 2 gives (2f)(x)=2f(x). Hence, (2f)(x)=2(x+4)=2x+8. Option C is incorrect because it multiplies only the x-term, not the constant term 4. Exam tip: Distribute a scalar multiplier to every term of the function.

\((-f)(x)=-f(x)=-(3x-2)=-3x+2\). The negative sign applies to the entire function, so the signs of both terms change. Option B changes only the sign of the constant term and is therefore incorrect. Exam tip: when a minus sign precedes a bracket, change the sign of every term inside it.

The sum of two functions is found pointwise: \\((f+f)(x)=f(x)+f(x)=x^2+x^2=2x^2\\). Hence, option A is correct. Option B results from incorrectly multiplying the terms, but the operation here is addition. Exam tip: when adding like terms, add their coefficients and keep the common power unchanged.

For the product of functions, \((ff)(x)=f(x)\cdot f(x)=(x+1)(x+1)=(x+1)^2=x^2+2x+1\). Hence, option A is correct. Option B, \(f(x)+f(x)=2x+2\), represents the sum of the functions, not their product. In exams, carefully distinguish between \(ff\) and \(f+f\).

The governing concept is pointwise subtraction of functions. For every value of x in the common domain, (f − g)(x) is defined as f(x) − g(x). Since the two given functions are identical, substitute their expressions: (f − g)(x) = (x + 5) − (x + 5). Removing the parentheses gives x + 5 − x − 5, and like terms cancel to produce 0. Thus the difference is the zero function and option A is correct. Option B would be obtained by adding the functions, option C merely repeats one original function, and option D has no basis in the subtraction. The result is zero for every permissible x, not just for one particular input.

\((f+g)(x)=f(x)+g(x)\). So \((f+g)(x)=(2x^2+3)+(2x^2+5x-1)= (2x^2+2x^2)+5x+(3-1)=4x^2+5x-1\).

The governing operation is multiplication of real-valued functions, defined by (fg)(x) = f(x)g(x). Substituting the given functions gives (fg)(x) = 1·(x² − 2x) = x² − 2x. The constant function 1 acts as the multiplicative identity, so it leaves g(x) unchanged for every x in the common domain. Option B would result from adding 1 instead of multiplying; option C confuses the product with f alone; and option D changes the sign of the linear term without justification. Since both functions are defined for every real x, there is no additional domain restriction. Therefore option A is correct.

By the definition of the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Therefore, \((f+g)(x)=(x-2)+(x+2)=2x\), because the constant terms \(-2\) and \(+2\) cancel each other. Option B corresponds to the product \(f(x)g(x)\), while option C considers only the constant terms. Exam tip: In \(f+g\), add the function values; do not multiply them.

The difference of two functions is (f − g)(x) = f(x) − g(x). Substituting the given expressions gives x² + 2x − (x² − x) = x² + 2x − x² + x = 3x. The minus sign changes the sign of each term in g(x).

For the product of two functions, \((fg)(x)=f(x)g(x)\). Thus, \(f(0)=2\) and \(g(0)=1\), so \((fg)(0)=2\times1=2\). Option D is only \(g(0)\), while option C represents the sum of the two values. Exam tip: interpret \((fg)(x)\) as the product of functions, not their composition.

For the difference of two functions, \((f-g)(2)=f(2)-g(2)\). Here, \(f(2)=2^2+3=7\) and \(g(2)=2^2-1=3\), so \((f-g)(2)=7-3=4\). The value 8 results from adding the two function values, so it is incorrect. Exam tip: For \((f-g)(a)\), evaluate both functions at \(a\) and then subtract.

The governing rule for adding functions is (f+g)(x)=f(x)+g(x). We first evaluate both functions at the required input x=−1. Substitution gives f(−1)=2(−1)+1=−2+1=−1. Similarly, g(−1)=−1−4=−5. Adding the two results yields (f+g)(−1)=−1+(−5)=−6. Therefore option C is correct. An equivalent check is to add the functions first: (f+g)(x)=(2x+1)+(x−4)=3x−3, and then substitute x=−1 to obtain −3−3=−6. Option A can arise from a sign or arithmetic error, option B ignores the negative input, and option D does not follow from either evaluation. Careful substitution preserves the signs.

\(\left(\frac{f}{g}\right)(1)=\frac{f(1)}{g(1)}=\frac{5}{2}\). The denominator value is (2), so division is allowed.