Algebra of real functions

(f(x)=x-2), but (x=-2) is excluded, so the difference is (0) for all (x\ne -2). Domain restriction is part of the answer.

(\frac{x}{x-1}-\frac{1}{x-1}=\frac{x-1}{x-1}=1), but (x\ne 1). The original restriction remains after simplification.

\(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.

(f(0)=2) and (g(0)=2), so the difference is (0). In modulus, substitute first and then take positive distance.

Here, \((fg)(x)=f(x)g(x)=\frac{x+1}{x-1}\), with the domain restriction \(x\ne1\). Solving \(\frac{x+1}{x-1}=1\) gives \(x+1=x-1\), which is impossible. Hence, there is no real solution. The choice \(x=1\) is also invalid because \(f(x)\) is undefined there. Exam tip: For an equation containing a fraction, first state that its denominator cannot be zero.

The numerator root needs (2x-3\ge 0), and the denominator root needs (7-x>0). Therefore the domain is ( \left[\frac{3}{2},7\right) ).

((f+g)(x)=x^2-2x-1=(x-1)^2-2), so the minimum value is (-2). Completing the square is fast for such questions.

The governing concept is composition of functions, where (f∘g)(x)=f(g(x)) and (g∘f)(x)=g(f(x)). Function composition is generally not commutative, so the order must be followed carefully. First, (f∘g)(x)=f(x+1)=(x+1)^2=x^2+2x+1. Next, (g∘f)(x)=g(x^2)=x^2+1. Subtracting the second result from the first gives (x^2+2x+1)-(x^2+1)=2x. Therefore option B is correct. Option A misses the coefficient 2, option C gives a different expression rather than the requested difference, and option D incorrectly assumes that both compositions are equal.

The governing concept is the domain of a quotient of functions. Both f(x) and g(x) must be defined, and the divisor g(x) must not equal zero. For f(x)=(x+1)/(x−3), the denominator is zero at x=3, so f is undefined there. Also, g(3)=3−3=0, so the quotient is independently forbidden at the same value. Hence every real number except 3 belongs to the domain. The numerator x+1 may be zero at x=−1; a zero numerator is allowed and does not remove −1 from the domain. Therefore the domain is ℝ\{3}, making option A correct. Option B incorrectly includes x=3. Option C unnecessarily excludes −1, and option D excludes the wrong point. Even if the quotient is written in a different algebraic form, the original restrictions must be preserved.

The governing concept is equality of functions together with preservation of the original domain. Factor the numerator of f: x²−1=(x−1)(x+1). Whenever x≠−1, cancellation is valid and f(x)=[(x−1)(x+1)]/(x+1)=x−1, which equals g(x). At x=−1, the original formula for f has denominator zero, so f is undefined. The cancelled expression x−1 may have a numerical value there, but cancellation does not repair a missing point in the original domain. Thus the functions agree on all real x except −1, represented by x≠−1 or ℝ\{−1}. Option A is correct. x=1 is permitted, so option B is wrong. All real numbers incorrectly includes −1, and option D is far too restrictive because equality holds at infinitely many other inputs.