Algebra of real functions

\(\frac{f}{g}=\frac{(x-2)^2}{x-2}=x-2\) where (x\ne 2), so at \(x=3\) the value is (1). First note the restriction then substitute.

(g(x)=(f+g)(x)-f(x)=x^2+3x+5-(3x+2)=x^2+3). To find the unknown function subtract the known function from the given sum.

(f(x)=\frac{x^2-1}{x-1}=x+1), but (x\ne 1) because (g(x)\ne 0). While finding an unknown factor check the division restriction.

First, f(1)=1²=1 and g(1)=2(1)+1=3. Therefore, (f+g)(1)=f(1)+g(1)=1+3=4, while (fg)(1)=f(1)g(1)=1×3=3. Hence, the required value is 4+3=7. Exam tip: (f+g)(x) denotes the sum of the function values, whereas (fg)(x) denotes their pointwise product; do not confuse the two operations.

The first function needs (x\ne -2) and the second needs (x\ne 2). The domain of the sum is obtained by excluding both restrictions.

The direct answer is option A, \(-4\). A product function is evaluated by multiplying the separate function values: \((fg)(x)=f(x)g(x)\). Put \(x=0\) into the first function: \(f(0)=0^2+2=2\). Put \(x=0\) into the second function: \(g(0)=0^2-2=-2\). Now multiply: \((fg)(0)=2\times(-2)=-4\). Therefore option A is correct. Option B, 0, would be obtained only if one factor were zero, but the factors are 2 and \(-2\), not zero. Option C, 4, ignores the negative sign in \(g(0)\), because \(2\times2=4\) but the second value is actually \(-2\). Option D, \(-2\), uses only one value and does not perform the required multiplication. The important distinction is that \((fg)(0)\) means product, not composition; composition would be written differently. Memory cue: evaluate each function first, then multiply.

\(\frac{g}{f}=\frac{x^2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2\), but (x\ne -2). Exclude the zero of the cancelled factor too.

The direct answer is option A: \(x=\pm1\). To find where a function is zero, set its expression equal to zero. Here \((f-g)(x)=f(x)-g(x)=2x^2-(x^2+1)=x^2-1\). We need \(x^2-1=0\). Rearranging gives \(x^2=1\), so taking both square roots gives \(x=1\) or \(x=-1\), written \(x=\pm1\). Option A is correct. Option B, \(x=0\), is not a zero because \((f-g)(0)=0-1=-1\). Option C, \(x=\pm2\), is not correct because substituting either value gives \(x^2-1=4-1=3\), not zero. Option D is wrong because real solutions do exist, namely 1 and \(-1\). The symbols \(\pm\) are important because both positive and negative numbers have square 1. Memory cue: for zeros, write the function equal to zero before solving.

\((f+g)(x)=f(x)+g(x)=(x^2+x)+(x^2-x)=2x^2\). Since \(2x^2\geq 0\) for every real \(x\), and the minimum value \(0\) is attained at \(x=0\), the range is \([0,\infty)\). Option D is incorrect because the minimum is 0, not 1. Exam tip: a positive multiple of \(x^2\) has a range beginning at its minimum value, 0.

((f\circ g)(x)=\frac{1}{x-1}) needs (x\ne 1) and ((g\circ f)(x)=\frac{1}{x}-1) needs (x\ne 0). In the sum both restrictions combine.

(f(g(x))=(x+1)^2) and (f(h(x))=(x-1)^2), so the difference is (4x). The difference of squares formula is fast here.

(f(x)+1=\frac{x+3}{x-3}+1=\frac{2x}{x-3}), where (x\ne 3). Make a common denominator when adding a constant.

By the definition of the sum of two functions, (f+g)(x)=f(x)+g(x)=(2x−5)+(5−2x)=0. Therefore, the result is the zero function, whose value is 0 for every x. Option B equals 2f(x), not f(x)+g(x). Exam tip: combine the variable terms and constant terms separately when adding functions.

For algebraic operations on functions, \(g^2(x)\) means \((g(x))^2\), not the composition \(g(g(x))\). Thus, \(g^2(x)=(x+1)^2=x^2+2x+1\). Therefore, \((f-g^2)(x)=(x^2+2x+1)-(x^2+2x+1)=0\). Option B results from an incorrect subtraction of only some terms. Exam tip: distinguish the pointwise square \(g^2\) from the composition \(g\circ g\).

The product of two real functions is defined by \((fg)(x)=f(x)g(x)\). Therefore, \((fg)(x)=(x^2-3x)(x^2+3x)\). This is of the form \((a-b)(a+b)=a^2-b^2\), so \((x^2)^2-(3x)^2=x^4-9x^2\). Option B has the wrong sign, while C and D do not result from the correct expansion. Exam tip: recognize conjugate binomials and apply the difference-of-squares identity directly.

((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x^2-4}), where (x\ne \pm 2). Keep restrictions while multiplying denominators.

The direct answer is option A, 9. A composition means applying the function on the right first. Thus \\(f\\circ g)(1)=f(g(1))\\). First, \\(g(1)=2\\). Then \\(f(2)=2^2+1=5\\). In the other order, \\(g\\circ f)(1)=g(f(1))\\). Now \\(f(1)=1^2+1=2\\), so \\(g(2)=2(2)=4\\). Therefore the required sum is \\(5+4=9\\). Option A works because it gives this complete calculation. Option B, 7, is not the sum of the two correct composition values. Option C, 6, is also incorrect and may result from adding or evaluating only partly. Option D, 5, is just one of the two values, not their total. Remember: in \\(f\\circ g\\), do g first, then f.

The direct answer is option A: \(\mathbb R\setminus\{2\}\). A quotient \((f/g)(x)\) equals \(f(x)/g(x)\), so the denominator must not be zero. Here the original denominator is \(g(x)=x-2\). Set it equal to zero: \(x-2=0\), giving \(x=2\). Therefore every real number is allowed except 2. Although \(f(x)=x^2-5x+6=(x-2)(x-3)\), cancellation may simplify the formula to \(x-3\\) only for values where the original quotient was defined; it cannot restore the missing value \(x=2\). Option A is correct. Option B wrongly excludes 3; the denominator at 3 is 1, so the quotient is defined. Option C excludes both 2 and 3, but 3 is allowed. Option D includes 2, where division by zero occurs. Memory cue: for a quotient, inspect the original denominator before cancelling factors.

Here (f^2(x)+g^2(x)=\frac{(x^2-1)^2+(2x)^2}{(x^2+1)^2}=\frac{(x^2+1)^2}{(x^2+1)^2}=1). Recognise the identity while adding squares.

In a quotient, the denominator (g(x)) must not be zero, so (x=2) is excluded. In exams, check the domain before simplifying.

For subtraction, both functions must be defined, so (x=-3) and (x=4) are excluded. Always remove values making any denominator zero.

Here, \((f-g)(x)=f(x)-g(x)=|x|-x\). For \(x\ge 0\), \(|x|=x\), so \(|x|-x=0\). For \(x<0\), \(|x|=-x\), giving \(|x|-x=-2x\), which is not zero. Hence option A is correct; option C is incomplete because every non-negative value, not just zero, is a solution. Exam tip: For modulus expressions, always check the cases \(x\ge 0\) and \(x<0\).

For a linear combination of functions, the coefficients multiply the entire functions: \((3f-2g)(x)=3f(x)-2g(x)=3(x^2+1)-2(2x-3)\). Simplifying gives \(3x^2+3-4x+6=3x^2-4x+9\), so option A is correct. In option C, the constant term has the wrong sign. Exam tip: distribute a negative coefficient across every term inside the parentheses.

Even if the product simplifies to (1), the original functions forbid (x=1) and (x=-1). The domain should not be changed after simplification.

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \(f(-2)=(-2)^2=4\) and \(g(-2)=2(-2)+1=-3\), so \((f+g)(-2)=4+(-3)=1\). The value 3 may result from an incorrect operation such as subtracting \(g(-2)\) from \(f(-2)\). Exam tip: evaluate the two functions separately before adding their values.