Algebra of real functions

Both denominators must be non-zero, so (x\neq -3) and (x\neq 1). For rational functions, check all denominators.

For the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Thus, \(f(0)=0^2=0\) and \(g(0)=0+1=1\), so \((f-g)(0)=0-1=-1\). Option B results from reversing the order of subtraction. Exam tip: For \((f-g)(a)\), calculate \(f(a)\) first and subtract \(g(a)\) from it.

The sum of two functions is defined by (f + g)(x) = f(x) + g(x). At x = 3, f(3) = 2(3) − 7 = −1 and g(3) = 3 + 5 = 8. Therefore, (f + g)(3) = −1 + 8 = 7.

((x^2-1)(x^2+1)=x^4-1) by difference of squares. Recognizing the formula makes calculation faster.

\(g(x)=(x+2)^2\), so \(\frac{g(x)}{f(x)}=x+2\), but \(x\neq -2\). In a quotient, remove the zero value of the denominator.

(\sqrt{x}) needs (x\geq 0), and (g(x)) is defined for all real (x). Therefore, the product domain is ([0,\infty)).

The direct answer is option A, \\(x+2)^2-2\\). First add the functions: \\( (f+g)(x)=f(x)+g(x)=(x^2+2)+4x=x^2+4x+2\\). To complete the square, take half of the coefficient of x, namely 4 divided by 2, which is 2. Add and subtract its square, 4: \\(x^2+4x+2=x^2+4x+4-4+2=(x+2)^2-2\\). Thus option A is correct. Option B gives \\(x^2-4x+2\\), so its middle term has the wrong sign. Option C expands to \\(x^2+4x+6\\), not the required expression. Option D expands to \\(x^2-8x+18\\), so both the middle term and constant are wrong. The safe method is: half the x-coefficient, square it, add and subtract that square.

In (\frac{1}{x}), the denominator becomes zero at (x=0). Therefore, the domain of the sum is (\mathbb{R}-{0}).

\((f-g)(x)=f(x)-g(x)=x^3-x\). Taking \(x\) as the common factor gives \(x(x^2-1)\). Using the difference-of-squares identity \(x^2-1=(x-1)(x+1)\), the factorized form is \(x(x-1)(x+1)\). Option C is incorrect because \((x-1)\) is squared, so it does not equal \(x^3-x\). Exam tip: take out the common factor first, then apply \(a^2-b^2=(a-b)(a+b)\).

Here (fg)(x) denotes the product of the two functions, so (fg)(0)=f(0)g(0). First evaluate each function at the given input: f(0)=0+3=3 and g(0)=2(0)−6=−6. Hence (fg)(0)=3×(−6)=−18, making option A correct. Option B has the correct magnitude but loses the negative sign. Option C could arise from using only part of one expression or from an incorrect arithmetic step, and option D would be the result of mistakenly treating the product as a sum or assuming one factor is zero. The important rule is that function multiplication is evaluated by finding both function values at the same input and then multiplying those values.

\(x^2+5x+6=(x+2)(x+3)\), so the quotient is (x+3), but \(x\neq -2\). Write the restriction along with the factorization.

The sum of the functions is \\( (f+g)(x)=f(x)+g(x)=(2x+1)+(x-4)=3x-3 \\). Setting it equal to zero gives \\(3x-3=0\\), so \\(3x=3\\) and hence \\(x=1\\). Therefore, option A is correct. In an exam, form the sum first and then solve the resulting linear equation; setting only \\(f(x)\\) or \\(g(x)\\) equal to zero would give a wrong result.

The product function is defined by (fg)(x)=f(x)g(x). At x=2, f(2)=2²−4(2)+4=4−8+4=0, while g(2)=2−2=0. Therefore (fg)(2)=0×0=0, so option A is correct. This also follows from the zero-product property: if at least one factor is zero, their product is zero. Option B and option D confuse the product with a value from one factor or involve an arithmetic sign error. Option C may result from evaluating only 2² and ignoring the remaining terms. Both functions are defined at x=2 because they are polynomials, so no domain restriction changes the calculation.

((f+g)(x)=x^2+2x=(x+1)^2-1), so the minimum value is (-1). Completing the square helps find the minimum of a quadratic.

((f-g)(x)=(x^2+1)-(x^2-3)=4), so at (x=5) the value is also (4). Simplifying first reduces calculation.

The direct answer is option A, \\([4,\\infty)\\). For a square root to be real, its radicand must be at least zero. For \\(f(x)=\\sqrt{x-4}\\), require \\(x-4\\ge0\\), so \\(x\\ge4\\). For \\(g(x)=\\sqrt{x+1}\\), require \\(x+1\\ge0\\), so \\(x\\ge-1\\). The product \\(fg\\) is defined only where both functions are defined, so take the intersection of these two sets: \\([4,\\infty)\\cap[-1,\\infty)=[4,\\infty)\\). Option B includes values from -1 to 4 that make \\(x-4\\) negative, so it is too large. Option C contains only a bounded interval and also includes invalid values. Option D includes negative values that make square roots non-real. Endpoint 4 is included because \\(\\sqrt0\\) exists.

\((f+g)(x)=f(x)+g(x)=(x-1)+(x+1)=2x\). Also, \((f-g)(x)=f(x)-g(x)=(x-1)-(x+1)=-2\). Therefore, the correct pair is \((2x,-2)\). Option B results from an incorrect sign while subtracting, and option D corresponds to the product \(f(x)g(x)\), not the sum or difference. Exam tip: when subtracting \(g(x)\), change the signs of both terms inside the parentheses.

(x^2-1=(x-1)(x+1)), so the denominator is zero at (x=1) and (x=-1). Values making the denominator zero are excluded from the domain.

Using the rule for a linear combination of functions, \((5f-2g)(x)=5f(x)-2g(x)\). Therefore, \(5(2x^2)-2(3x^2)=10x^2-6x^2=4x^2\), so option A is correct. Option B represents only \(5f(x)\) and ignores the subtraction of \(2g(x)\). In an exam, apply the coefficients first and then combine like terms.

\(f(2)=0\) and \(g(2)=-1\), so the quotient is (0). A zero numerator is allowed when the denominator is non-zero.

((f+g)(x)=x^2+x+2), whose highest power is (2). The degree of a polynomial is given by its highest exponent.

\((f-g)(x)=f(x)-g(x)=(x^2-1)-(x-1)=x^2-1-x+1=x^2-x=x(x-1)\), so option A is correct. Option B results from mishandling the signs in the subtraction. Exam tip: when subtracting a polynomial, change the sign of every term in the polynomial being subtracted before simplifying.

((fg)(x)=\frac{x+2}{x-1}(x-1)=x+2), but because of the original (f(x)), (x\neq 1). Write the domain restriction with the simplification.

\(g(-1)=0\), so the quotient is not defined at (-1). Check the original denominator before cancellation.

Function addition is performed pointwise: \((f+2g+h)(x)=f(x)+2g(x)+h(x)=x^2+2x+1=(x+1)^2\). Option B omits the coefficient 2 of \(g(x)\). Exam tip: In a linear combination, multiply each function by its stated coefficient before adding.