Algebra of real functions

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(a)=f(a)+g(a)=a^2+2a\). Options B and C incorrectly mix \(x\) and \(a\), while option D does not represent the sum. Exam tip: when evaluating a function at \(a\), replace every occurrence of \(x\) with \(a\).

The direct answer is option A, 0. The notation \\((fg)(0)\\) means the product of the two function values at 0: \\(f(0)g(0)\\). First calculate \\(f(0)=0^2-2(0)=0\\). Next calculate \\(g(0)=0^2+2(0)=0\\). Therefore \\((fg)(0)=f(0)g(0)=0\cdot0=0\\). Option A is correct. Option B, 4, might arise from confusing the signs or substituting another value, but it is not the product at zero. Option C, -4, is also not obtained by direct substitution and may come from an incorrect sign calculation. Option D, 1, has no basis in either function value. The product rule is simple: evaluate each function at the given input, then multiply. Since both values are zero, the product must be zero. Remember that \\(fg\\) here means multiplication, not composition.

(f(x)) needs \(x \neq 1\), and \(f(x)=\frac{1}{x-1}\) is never zero. So the only restriction is \(x \neq 1\).

\(\left(\frac{f}{g}\right)(x)=\frac{(x-4)(x+4)}{x+4}=x-4\), so at \(x=-1\) the value is (-5). Check directly as \(\frac{-15}{3}=-5\).

\(\frac{x^2-16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4\), hence \(-1-4=-5\). Substitute carefully after simplification.

For the square root, (9-x^2 \geq 0), so (-3 \leq x \leq 3). The linear function gives no extra restriction.

Applying \(f\) first sends \(x\in A\) to \(f(x)\in B\). Applying \(g\) next gives \(g(f(x))\in C\), so the composite is \(g\circ f\). Exam tip: in a composition, the function on the right acts first.

\(\frac{x^2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2\), so at \(x=2\) the value is (4). Factorisation saves time.

(f(1)=1) and (g(1)=3), so ((f+g)(1)-(fg)(1)=4-3=1). In such questions, find individual values first.

((f-g)(x)=x^2-2x+2=(x-1)^2+1), which is never (0). Completing the square quickly shows real solutions.

\((g-f)(x)=g(x)-f(x)=(x^2+1)-(x^2-1)=2\) for every real \(x\). Thus, the difference is a constant function, whose range is the singleton set \(\{2\}\). \(\{0\}\) is incorrect because the constant terms give \(1-(-1)=2\). Exam tip: simplify \((g-f)(x)\) completely before determining its range.

The product function is (fg)(x) = f(x)g(x) = (x + 1)x² = x³ + x². The highest power of x is 3, so the degree is 3. Option B gives the degree of g(x) alone, not that of the product. In such questions, multiply the functions first and then identify the highest power of x.

The direct answer is option A, 0 and 2. The sum function is found by adding corresponding values: \\( (f+g)(x)=f(x)+g(x)=(x^2-5x)+3x=x^2-2x\\). A zero is an input for which the function value is zero, so set \\(x^2-2x=0\\). Factor: \\(x(x-2)=0\\). By the zero-product property, either \\(x=0\\) or \\(x-2=0\\), giving \\(x=2\\). Hence the zeroes are 0 and 2, option A. Option B, 0 and 5, incorrectly uses the zero of the original f term without combining it with g. Option C, 2 and 5, misses the zero x=0. Option D, only 2, misses one valid solution. Always form the requested sum first; do not find zeroes of f or g separately. Checking gives \\(0^2-2(0)=0\\) and \\(2^2-2(2)=0\\).

With the same denominator, subtract numerators: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1}). Because of denominator (x+1), (x \neq -1).

For a linear combination of functions, multiply each entire function by its coefficient: \((3f+2g)(x)=3f(x)+2g(x)=3(2x+1)+2(x-2)=6x+3+2x-4=8x-1\). Hence, option A is correct. Option B results from an incorrect sign in the constant term. In the exam, distribute each coefficient across every term of the corresponding function.

\(\frac{x^2+6x+9}{x+3}=\frac{(x+3)^2}{x+3}=x+3\), hence the value at \(x=0\) is (3). The perfect square identity is useful.

In \(f(x)\), \(\sqrt{x-1}\) is in the denominator. Thus, the radicand must satisfy \(x-1\geq 0\), and the denominator must not be zero, so \(x-1\neq 0\). Therefore, \(x>1\). Since \(g(x)=x\) is defined for all real numbers, the domain of \((f+g)\) is the intersection of the two domains, namely \((1,\infty)\). Exam tip: For a square root in the denominator, require the expression under the root to be strictly positive.

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

The difference of two functions is \((f-g)(x)=f(x)-g(x)\). Thus, \((x^2+x+1)-(x^2-x+1)=x^2+x+1-x^2+x-1=2x\). Option B results from failing to change the sign of the \(-x\) term when subtracting the entire function \(g(x)\). Exam tip: when a function is preceded by a minus sign, change the sign of every term inside its brackets.

The product of two real-valued functions is defined by \((fg)(x)=f(x)g(x)\). Thus, \((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2\), with the restriction \(x\neq 3\). Therefore, \((fg)(4)=4+2=6\). Option 1 is only the value of \(g(4)\), while option 4 incorrectly treats \(x=4\) as outside the domain. Exam tip: write \((fg)(x)=f(x)g(x)\) first and check the original domain restriction before cancelling factors.

((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), but both square roots require (x \geq 0). The simplified form does not change the original domain.

\(\frac{x^2-3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2\), so at \(x=3\) the value is (1). Factorisation makes division simple.

For the product of two functions, \((fg)(x)=f(x)g(x)\). Thus, \((fg)(x)=(x+4)(4-x)=4^2-x^2=16-x^2\), using the identity \((a+b)(a-b)=a^2-b^2\). Option B has the sign reversed. In an exam, first check whether \(fg\) denotes the product of functions or their composition.

By the definition of the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Therefore, \((4x-7)-(x+2)=4x-7-x-2=3x-9\), so option A is correct. Option C results from combining the constant terms incorrectly. Exam tip: when subtracting a bracket, change the sign of every term inside it.

The direct answer is A: \(x+3\), with \(x\ne3\). By definition, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-9}{x-3}\). Factor the numerator as a difference of squares: \(x^2-9=(x-3)(x+3)\). Therefore \(\frac{(x-3)(x+3)}{x-3}=x+3\), but cancellation is allowed only when \(x-3\ne0\), so \(x=3\) remains excluded. Option A is correct. Option B merely shows a cancelled factor and is not the simplified quotient. Option C incorrectly changes subtraction into addition. Option D, 1, would result only if numerator and denominator were identical, which they are not. Also, the simplified rule does not restore the missing value at \(x=3\). The key cue is: factor first, cancel common factors, and keep denominator restrictions.