Algebra of real functions

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(4)=f(4)+g(4)=\sqrt{4+5}+4=\sqrt{9}+4=3+4=7\). Therefore, the correct answer is 7. Exam tip: write the sum of the functions first and then substitute the given value. The value 6 results from evaluating the square-root term incorrectly.

For an additive inverse, the sign of every corresponding term changes. Thus \(g(x)=-(3x^2-2x+5)=-3x^2+2x-5\). In option B, the sign of \(3x^2\) is unchanged. Exam tip: place a minus sign outside the complete function first.

For the product of two functions, \((fg)(3)=f(3)g(3)\). Here, \(f(3)=\frac{2}{3}\) and \(g(3)=3^2=9\), so \((fg)(3)=\frac{2}{3}\times9=6\). Option C is only \(f(3)\), while option B is only \(g(3)\). Exam tip: interpret \((fg)(a)\) as \(f(a)g(a)\), not as the composition of the functions.

The direct answer is A, \(\frac{5}{2}\). The quotient function is \(\left(\frac{f}{g}\right)(3)=\frac{f(3)}{g(3)}\). Find the numerator: \(f(3)=3+12=15\). Find the denominator: \(g(3)=3+3=6\). Thus \(\frac{15}{6}\) simplifies by dividing numerator and denominator by 3 to \(\frac{5}{2}\). Option A is correct. Option B, 5, is not the value of the fraction. Option C, \(2/5\), is the reciprocal of the correct answer, so numerator and denominator have been reversed. Option D, 3, ignores the quotient operation. Also, the denominator 6 is not zero, so the value is defined. Remember: numerator first, denominator second.

Here, \((fg)(x)=f(x)g(x)=(x+4)(x-4)\). This matches the identity \((a+b)(a-b)=a^2-b^2\), with \(a=x\) and \(b=4\), giving \(x^2-16\). Options B and C represent the sum of squares and \((a-b)^2\), respectively, so they do not apply. Exam tip: whenever conjugate binomials appear, look for the difference-of-squares identity.

By definition, (f + g)(1) = f(1) + g(1). Here, f(1) = 2(1)² + 3 = 5 and g(1) = 1 + 5 = 6, so (f + g)(1) = 5 + 6 = 11. Therefore, option C is correct.

For the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Therefore, \((f-g)(x)=(5x^2+2)-(2x^2-7)=5x^2+2-2x^2+7=3x^2+9\). Option A has the wrong constant-term sign because \(-(-7)=+7\). Exam tip: when a minus sign precedes parentheses, change the sign of every term inside them.

For the product of two functions, \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x-6)x=x^2-6x\). Option B represents \(f(x)+g(x)\), not their product. Exam tip: \((fg)(x)\) denotes the product of functions, whereas \(f\circ g\) denotes function composition.

The sum of functions is defined as \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(x)=(4x+11)+(x-6)=5x+5\), so option A is correct. Option B is obtained from \(f(x)-g(x)\), whereas option D is the product \(f(x)g(x)\). Exam tip: while adding functions, combine like terms and constants separately.

For the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Thus, \((f-g)(x)=8-(x^2-3x)=8-x^2+3x\). Option A reverses the order of subtraction. Exam tip: when a minus sign precedes a bracket, the signs of all terms inside the bracket change.

By definition, (f/g)(x) = f(x)/g(x) = (x² − 25)/(x + 5). Factor the numerator as a difference of squares: x² − 25 = (x − 5)(x + 5). Since x ≠ −5, the denominator is nonzero and the common factor can be cancelled, giving (f/g)(x) = x − 5. Therefore, option A is correct.

\(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-4x+4}{x-2}\). Factoring the numerator as \((x-2)^2\) gives \(\frac{(x-2)^2}{x-2}=x-2\), provided \(x\ne2\). Hence, option A is correct. Option C would be obtained only if the numerator and denominator were identical, which they are not before simplification. Exam tip: factor the numerator first and retain any value excluded by the original denominator.

For the sum of two functions, (f+g)(1)=f(1)+g(1). Here, f(1)=1-1=0 and g(1)=1+10=11, so (f+g)(1)=0+11=11. Although g(1) alone is also 11 in this case, the general rule requires adding the values of both functions. Exam tip: evaluate each function at the given input first, then add the results.

The product of two functions is defined by \\( (fg)(x)=f(x)g(x) \\). Thus, \\(f(3)=15\\) and \\(g(3)=4\\), so \\( (fg)(3)=15\\times4=60\\). The value 45 can result from incomplete or incorrect calculation, whereas both function values must be multiplied. Exam tip: do not confuse \\(fg\\) with composition \\(f\\circ g\\); here it denotes the product of functions.

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Thus, \((f+g)(x)=(x^2-4)+(x^2+4)=2x^2\), since \(-4\) and \(+4\) cancel each other. Option D incorrectly retains an extra constant term. Exam tip: add like terms separately when combining functions.

The difference of two functions is defined as (f-g)(x)=f(x)-g(x). Thus, (f-g)(x)=(4x^2-3)-(x^2+8)=4x^2-3-x^2-8=3x^2-11. In option D, the coefficients of x^2 have been added instead of subtracted. Exam tip: When subtracting a function in parentheses, change the signs of all its terms.

For the product of two functions, \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x-5)\cdot3=3x-15\). Option C is incorrect because multiplying by 3 must also multiply the constant term \(-5\). Exam tip: \((fg)(x)\) denotes the product of functions, whereas \(f(g(x))\) denotes composition.

By the definition of the quotient of two functions, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\). Hence, \(\frac{12x-6}{6}=\frac{12x}{6}-\frac{6}{6}=2x-1\). Therefore, option A is correct. Exam tip: When dividing an algebraic expression by a constant, divide every term in the numerator by that constant.

The sum of two functions is found pointwise: (f+g)(x)=f(x)+g(x)=(2x-11)+(11-2x)=0. The terms 2x and -2x cancel each other, as do -11 and 11. Therefore, option A is correct; option B results from combining the terms with incorrect signs. Exam tip: like terms with opposite signs cancel to zero.

The difference of two functions is defined by \((f-g)(x)=f(x)-g(x)\). Thus, \((f-g)(x)=(3x+8)-(3x-8)=3x+8-3x+8=16\). Therefore, 16 is correct. The expression \(6x+16\) would result from adding the two functions, not subtracting them. Exam tip: when a minus sign precedes parentheses, change the sign of every term inside.

For the product of two functions, \\((fg)(x)=f(x)g(x)\\). Hence, \\((fg)(x)=x^5\cdot x^2=x^{5+2}=x^7\\). Option B, \\(x^{10}\\), can result from composition such as \\(g(f(x))\\), not from multiplication. Exam tip: distinguish \\((fg)(x)\\) from \\(f(g(x))\\).

\(\left(\frac{f}{g}\right)(x)=\frac{x^6}{x^3}=x^3\), where \(x\neq0\). In division, subtract powers of the same base.

Both polynomials are defined on all (\mathbb{R}), so the product domain is (\mathbb{R}). The domain is the set of common allowed values.

The sum \((f+g)(x)\) is defined only where both \(f(x)\) and \(g(x)\) are defined. The function \(f(x)=\sqrt{x}\) requires the quantity inside the square root to be nonnegative, so \(x\ge0\). The function \(g(x)=x+2\) is defined for every real number. Their common domain is therefore \([0,\infty)\), making option A correct.

Indeed, \((f+g)(x)=\sqrt{x}+x+2\). For \(x=0\) and positive values, the square root is real. For a negative value such as \(x=-1\), \(\sqrt{-1}\) is not real in the usual real-number setting, so the sum is not defined. The linear function does not add any further restriction.

In division, \(g(x)\neq0\) is needed, and \(x^2-9=0\) gives \(x=\pm3\). So the domain is \(\mathbb{R}-{-3,3}\).