Algebra of real functions

For the quotient function, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\). Here, \(f(2)=2^2-3(2)+2=0\) and \(g(2)=2-1=1\), so \(\left(\frac{f}{g}\right)(2)=\frac{0}{1}=0\). Option D is incorrect because the denominator, \(g(2)=1\), is not zero. Exam tip: evaluate the numerator and denominator separately and check that the denominator is non-zero.

For a linear combination of functions, combine their values at the same input: \((f-2g+h)(x)=f(x)-2g(x)+h(x)=x^2-2x+1=(x-1)^2\). Option B has the wrong sign for the middle term, while the other options do not match the expansion. Exam tip: substitute the function values first, then simplify or complete the square.

With a common denominator, (f-g=\frac{x^2+1-(x^2-1)}{x}=\frac{2}{x}), where (x\ne 0). In subtraction, both signs of the second numerator change.

For the product function, \((fg)(x)=f(x)g(x)\). Here, \(f(0)=0^2+2(0)+2=2\) and \(g(0)=0^2-2(0)+2=2\). Therefore, \((fg)(0)=f(0)g(0)=2\times2=4\). The value 2 is the value of each individual function, not their product. Exam tip: For \((fg)(a)\), evaluate both functions at \(a\) and then multiply the results.

Substituting \(x=2\), we get \(f(2)=2^2-4=0\) and \(g(2)=|2|-2=0\). Therefore, \((f-g)(2)=f(2)-g(2)=0-0=0\), so option A is correct. Remember that \((f-g)(x)\) means the difference of the two functions, not their sum.

In \(\frac{f}{g}\), both (f) and (g) must be defined and (g(x)\ne 0); this excludes \(x=\pm2\). In a quotient, zero of the second function is also forbidden.

(f(x)) simplifies to (x+4), but (f) is undefined at (x=4). For equal functions, check both rule and domain.

Both square roots require \(2x-1\ge 0\) and \(7-3x\ge 0\). Hence the domain is \(\left[\frac{1}{2},\frac{7}{3}\right]\).

\(f(-2)=0\) and \(g(-2)=-1\), so the quotient is (0). At a point, first check whether the denominator is zero.

((f-g)(x)=x^2+2x+4=(x+1)^2+3), so the minimum value is (3). Completing the square is a fast method for minimum value.

The denominator of (g(x)) is (x-3), so at (x=3), (g) and (f+g) are undefined. In a sum, both functions must be defined.

\((fg)(x)=\left(x+\frac{2}{x}\right)\left(x-\frac{2}{x}\right)=x^2-\frac{4}{x^2}\). Apply \((a+b)(a-b)=a^2-b^2\).

(|x-1|=|x+1|) means (x) is equally distant from (1) and (-1), so (x=0). The distance idea is useful in modulus questions.

For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.

For a real square root, the expression inside the root must be non-negative. For \(f(x)\), \(x-2\ge 0\), so \(x\ge 2\). For \(g(x)\), \(10-2x\ge 0\), so \(x\le 5\). The sum \((f+g)(x)\) is defined only where both conditions hold; hence its domain is \([2,5]\). The closest distractor, \((2,5)\), is incorrect because square roots may equal zero at \(x=2\) and \(x=5\), so both endpoints are included. Exam tip: the domain of a sum is the intersection of the domains of the given functions.

(f(x)) simplifies to (x+5), but (f) is not defined at (x=5). For equal functions, check both rule and domain.

((f-g)(x)=x^2+4x+12=(x+2)^2+8), so the minimum value is (8). Completing the square is fast for such questions.

((f-g)(x)=\frac{x+4-(x-2)}{(x-2)(x+4)}=\frac{6}{(x-2)(x+4)}). Keep numerator signs carefully while using a common denominator.

The equation \((f-g)(x)=0\) means \(f(x)=g(x)\), so \(|x-3|=|x+1|\). Thus, \(x\) must be equally distant from 3 and −1. Their midpoint is \(x=1\), so option C is correct. At \(x=3\) or \(x=-1\), only one of the two function values is zero; they are not equal. Exam tip: For \(|x-a|=|x-b|\), the solution is generally the midpoint of \(a\) and \(b\).

\((fg)(x)=\left(x+\frac{3}{x}\right)\left(x-\frac{3}{x}\right)=x^2-\frac{9}{x^2}\). Use \((a+b)(a-b)=a^2-b^2\) here.

For (f), \(x^2-16\ne 0\), so (x\ne -4,4), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.

The correct answer is option A: \((f+g)(x)=3x^2-2x-4\). The symbol \(f+g\) means add the two function values: \((f+g)(x)=f(x)+g(x)\). Substitute the expressions: \((2x^2-3x+1)+(x^2+x-5)\). Combine like terms by degree: \(2x^2+x^2=3x^2\), \(-3x+x=-2x\), and \(1-5=-4\). Thus the result is \(3x^2-2x-4\). Option A matches every combined term. Option B changes the coefficient of \(x^2\) and the other terms, so it is not the sum. Option C incorrectly makes the linear coefficient positive, as if \(-3x+x=2x\). Option D leaves out one \(x^2\) contribution and has the wrong linear coefficient. Always add corresponding powers separately; do not multiply the functions or combine unlike terms. Memory cue: align the \(x^2\), x, and constant columns before adding.

The square root needs (x+5\ge 0), and the denominator needs (x\ne 1). So the domain is ([-5,\infty)\setminus{1}).

The direct answer is option A, 9. Add the functions pointwise: \((f+g)(x)=f(x)+g(x)=x^2-8x+20+5=x^2-8x+25\). Complete the square: \(x^2-8x+25=(x-4)^2+9\). For every real number \(x\), a square such as \((x-4)^2\) is at least zero. Its smallest value is zero when \(x=4\). Thus the smallest possible value of the whole expression is \(0+9=9\). Option A is correct. Option B, 5, forgets the quadratic part and is not the minimum of the complete expression. Option C, 25, is the constant term before completing the square and is not the minimum; for example, at \(x=4\), the value is 9. Option D, 4, is the value of \(x\) where the minimum occurs, not the minimum value itself. The minimum is attained at \(x=4\), but the requested answer is the output value, 9. Memory cue: rewrite a quadratic as \((x-h)^2+k\); its minimum is \(k\) when the squared term is zero.

The denominator of (g(x)) is (x-2), so at (x=2), (g) and (f+g) are undefined. In a sum, both functions must be defined.