Algebra of real functions

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \((f+g)(4)=\sqrt{4}+(4+1)=2+5=7\). Option B is only the value of \(g(4)\), whereas the values of both functions must be added. Exam tip: For \((f+g)(a)\), find both \(f(a)\) and \(g(a)\), then add them.

For the product of functions, \((fg)(9)=f(9)g(9)=9^2\times\sqrt{9}=81\times3=243\). Therefore, 243 is correct. The value 81 results from calculating only \(f(9)=9^2\) and ignoring \(g(9)\). In an exam, read \((fg)(x)\) as the product of the two functions, not their composition.

For the product of functions, \((fg)(x)=f(x)g(x)\). Therefore, \((fg)(2)=f(2)g(2)=\frac{1}{2}\times2=1\). Option C is only the value of \(f(2)\), whereas the product of both function values is required. In exams, identify \((fg)(x)\) as the product of functions, not their composition.

For quotient functions, \(\left(\frac{f}{g}\right)(0)=\frac{f(0)}{g(0)}\). Here, \(f(0)=6\) and \(g(0)=2\), so \(\frac{6}{2}=3\). Since the denominator is non-zero, the value is defined. Option B gives only \(f(0)\) and ignores the division by \(g(0)\). Exam tip: evaluate both functions first, then divide.

Here, \((fg)(x)=(x-1)(x+1)\). This matches the identity \((a-b)(a+b)=a^2-b^2\), with \(a=x\) and \(b=1\). Hence, \((fg)(x)=x^2-1\). Options B and D are identities for squaring a binomial, while option C is for the sum of two cubes. Exam tip: when two binomials have opposite signs and the same terms, use the difference-of-squares identity.

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(3)=f(3)+g(3)=3^2+(2\times3+1)=9+7=16\). Option B can result from using an incorrect input in one function; in such questions, substitute the same input into both functions.

By definition, \((f-g)(x)=f(x)-g(x)\). Substituting the given functions gives \(2x^2-(x^2+5)=2x^2-x^2-5=x^2-5\), so option A is correct.

For the product of two functions, \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x+3)x=x^2+3x\). Option B, \(2x+3\), is the value of \((f+g)(x)\), not the product. Exam tip: Do not confuse \(fg\) with \(f+g\); multiply the two function values first.

The sum of two functions is defined by \\((f+g)(x)=f(x)+g(x)\\). Hence, \\((2x-3)+(x+4)=3x+1\\). Option B omits the additional \\(2x\\) term, while option D results from multiplying the two functions, not adding them. Exam tip: combine like powers of \\(x\\) and then combine the constant terms separately.

By the definition of the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Hence, \((f-g)(x)=5-x^2\). Option B reverses the order of subtraction, while options C and D represent multiplication and addition, respectively. In an exam, remember to subtract the second function from the first in \((f-g)\).

The direct answer is option A: x−3, with the original restriction x≠−3. The quotient of functions is \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-9}{x+3}\). Factor the numerator as a difference of squares: x^2−9=(x−3)(x+3). Thus \(\frac{(x-3)(x+3)}{x+3}=x-3\), but only when x+3≠0, so x≠−3 remains essential. Option A is correct. Option B, x+3, would result from an incorrect factor cancellation or sign error. Option C, x^2−3, does not come from factoring and simplifying the quotient. Option D, x−9, confuses the constant 9 with the factor x−3; it is not algebraically equivalent. A subtle point is that the simplified rule x−3 cannot be used to say the original quotient is defined at x=−3: the original denominator is zero there. Always carry the excluded value after cancellation. The memory cue is difference of squares: \(x^2-a^2=(x-a)(x+a)\).

By the definition of the quotient of functions, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\). Since \(x^2+2x+1=(x+1)^2\), we get \(\frac{(x+1)^2}{x+1}=x+1\), provided \(x\ne -1\). Option D, \(1\), results from incorrectly cancelling both factors. Exam tip: always exclude the value that makes the denominator zero.

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Thus, \(f(5)=5+2=7\) and \(g(5)=5-5=0\), so \((f+g)(5)=7+0=7\). The value 12 results from an incorrect evaluation of the two functions. Exam tip: evaluate each function at the given input first, then add the results.

Here, \((fg)(x)=f(x)g(x)\), so \(f(2)=8\) and \(g(2)=4\). Therefore, \((fg)(2)=8\times4=32\). The value 16 represents \(f(g(2))\), which is composition rather than multiplication. Exam tip: distinguish \((fg)(x)=f(x)g(x)\) from \((f\circ g)(x)\), which denotes composition.

The sum of two functions is obtained by adding their values at the same input: \((f+g)(x)=f(x)+g(x)=(x^2-1)+(x^2+1)=2x^2\). The constant terms \(-1\) and \(+1\) cancel each other. Option C represents the product \(f(x)g(x)\), not the sum. Exam tip: For function addition, add the two expressions term by term.

By definition, \((f-g)(x)=f(x)-g(x)\). Thus, \((f-g)(x)=(3x^2+1)-(x^2-4)=3x^2+1-x^2+4=2x^2+5\). Option C results from keeping the constant term as \(-4\) instead of changing it to \(+4\). Exam tip: when removing a bracket preceded by a minus sign, change the sign of every term inside it.

By the definition of the product of functions, \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x+1)\cdot2=2x+2\), so option A is correct. Option B represents the sum \(f(x)+g(x)\), not the product. In exams, distinguish \(fg\), which denotes a product here, from \(f\circ g\), which denotes composition.

By the definition of the quotient of two functions, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\). Thus, \(\frac{8x+4}{4}=\frac{8x}{4}+\frac{4}{4}=2x+1\). In option B, the term \(8x\) has not been divided correctly by 4, while option C has an incorrect constant term. Exam tip: when dividing a polynomial by a nonzero constant, divide every term by that constant.

By the definition of the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Therefore, \((f+g)(x)=(x-7)+(7-x)=0\), so option A is correct. Remember that opposite terms cancel in addition; \(x^2-49\) would arise from multiplication, not addition.

The difference of two functions is defined as (f−g)(x)=f(x)−g(x). Hence, (f−g)(x)=(2x+5)−(2x−5)=2x+5−2x+5=10. The terms containing x cancel each other. Option B results from effectively adding the functions instead of subtracting them. In an exam, remember to change the signs of every term when removing the parentheses after a minus sign.

By definition of the product of two functions, \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=x^2\cdot x^3=x^{2+3}=x^5\). Option B, \(x^6\), would result from the composition \(f(g(x))\), not from multiplication. Exam tip: distinguish \((fg)(x)\) from \(f(g(x))\).

\(\left(\frac{f}{g}\right)(x)=\frac{x^4}{x^2}=x^2\), where \(x\neq0\). In division with the same base, powers are subtracted.

Both linear functions are defined on all (\mathbb{R}), so the domain of the sum is also (\mathbb{R}). Check common allowed values for domain.

The direct answer is A: the domain is \(\mathbb{R}\setminus\{0\}\). The domain means all input values for which a function is defined. Here, \(f(x)=1/x\) is undefined when \(x=0\), because division by zero is not allowed. The function \(g(x)=x+1\) is defined for every real number. For a sum \((f+g)(x)=f(x)+g(x)=1/x+x+1\), both parts must be defined at the same time. Therefore every real number except zero is allowed. Option A is correct because it gives \(\mathbb{R}\setminus\{0\}\). Option B, all real numbers, wrongly includes zero. Option C removes \(-1\), but \(x=-1\) is allowed: only \(1/x\) causes a restriction. Option D contains only zero, which is precisely the forbidden input and also leaves out all valid nonzero real numbers. Memory cue: for a sum, take the common domain, and always exclude values that make a denominator zero.

In division, the denominator \(g(x)=x-3\) must not be zero, so \(x\neq3\). Hence the domain is \(\mathbb{R}-{3}\).