Algebra of real functions

The sum of two functions is defined pointwise: \((f+g)(x)=f(x)+g(x)\). Thus, \((f+g)(x)=(x^2-1)+(2x+3)=x^2+2x+2\). In option B, the constant terms \(-1\) and \(3\) have been combined incorrectly to obtain \(4\); option C corresponds to \(f-g\), while option D is the result of \(g(f(x))\). Exam tip: distinguish carefully between function addition, subtraction, and composition.

By definition, (g − f)(x) = g(x) − f(x). Thus, x² + 4 − (3x − 5) = x² + 4 − 3x + 5 = x² − 3x + 9, so option A is correct.

For the product of two functions, (fg)(x)=f(x)g(x). Thus, (fg)(x)=(x+2)(x-3)=x^2-3x+2x-6=x^2-x-6, so option A is correct. In option B, the sign of the middle term is incorrect. Exam tip: Do not confuse (fg)(x), which denotes multiplication here, with composition f∘g(x).

\(\left(\frac{f}{g}\right)(x)=\frac{x^2-4}{x-2}=x+2\) but (x\ne 2). Even after simplifying do not forget the original denominator restriction.

Both denominators cannot be zero so (x\ne 1) and (x\ne -1). For addition the common domain is the intersection of both domains.

For square roots we need (x-2\ge 0) and (5-x\ge 0) so (2\le x\le 5). In such questions apply both conditions together.

The composition (f ∘ g)(x) means f(g(x)). Substituting g(x) = x − 4 into f gives (x − 4)² + 1 = x² − 8x + 16 + 1 = x² − 8x + 17, so option A is correct.

The composition (g ∘ f)(x) means g(f(x)). Since g(u) = u², substituting f(x) = 2x + 1 gives (2x + 1)² = 4x² + 4x + 1. Therefore, option A is correct.

Multiplication gives ((fg)(x)=1) but original denominators require (x\ne 1,-1). Simplification does not remove restrictions.

If (x\ge 0) then (|x|-x=0) and if (x<0) then (|x|-x=-x-x=-2x). For modulus use cases according to sign.

((f\circ g)(x)=|x|^2=x^2) and ((g\circ f)(x)=|x^2|=x^2). Sometimes changing composition order still gives the same result.

The direct answer is option A, 28. The notation \((f+g)h\) means the product of the sum function \(f+g\) and the function h. First evaluate at 2: \(f(2)=2+1=3\), \(g(2)=2(2)=4\), and \(h(2)=2^2=4\). Then \((f+g)(2)=3+4=7\), and \(((f+g)h)(2)=7\times4=28\). Option A is correct. Option B, 20, may come from an incorrect multiplication or addition. Option C, 16, is only \(h(2)^2\) or another unrelated operation, not the required product. Option D, 12, is \(f(2)h(2)\) and forgets g. The safest method is evaluate each function first, add f and g, then multiply by h.

In \(\frac{f}{g}\) the denominator is \(g(-3)=0\) so the value is undefined. Check restrictions before simplifying \(\frac{x^2-9}{x+3}\).

The direct answer is option A: \([-4,\infty)\setminus\{1\}\). First, \(f(x)=\sqrt{x+4}\) is real only when the radicand is non-negative: \(x+4\ge0\), so \(x\ge-4\). Next, \(g(x)=1/(x-1)\) is defined only when its denominator is not zero: \(x-1\ne0\), so \(x\ne1\). For the product \((fg)(x)=f(x)g(x)\), both functions must be defined, so we take the intersection of these conditions. Thus all numbers from \(-4\) onward are allowed except \(1\). Option A is correct. Option B incorrectly removes \(-4\), although \(\sqrt{0}\) is defined. Option C removes every number below \(1\), which is unnecessary. Option D wrongly allows values below \(-4\) and excludes \(-4\), even though the square root is valid at \(-4\). Memory cue: for a square root use radicand \(\ge0\), and for a denominator use \(\ne0\).

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Therefore, \((f+g)(-2)=(-2)^3+3(-2)=-8-6=-14\), so option A is correct. As an exam tip, remember that an odd power preserves the negative sign of a negative number.

By definition, (f − g)(x) = f(x) − g(x). Thus, (x² + 2x) − (x² − 2x) = x² + 2x − x² + 2x = 4x. The x² terms cancel, and subtracting the second function changes the sign of each of its terms.

((f-g)(x)=\frac{2x+1}{3}-\frac{x-4}{2}=\frac{4x+2-3x+12}{6}=\frac{x+14}{6}). Use the least common denominator for fractions.

\((f\circ g)(x)=f(g(x))=f(x^2)=\frac{1}{x^2}\). This is defined only when the denominator is not zero, so \(x^2\ne 0\Rightarrow x\ne 0\). Hence the domain is \(\mathbb{R}\setminus\{0\}\).

\((f\circ g)(x)=2\left(\frac{x-3}{2}\right)+3=x\). This shows the two functions behave like inverses of each other.

((f\circ g)(x)=\frac{3(5x+2)-2}{5}=\frac{15x+4}{5}=3x+\frac{4}{5}). Substitute the whole (g(x)) in place of (x).

(g − f)(x) = (x² + 1) − (x² − 1) = 2. Its value remains 2 for every real value of x, so it is a constant function. The x² terms cancel because they are identical in both functions; therefore, the result is not quadratic. In exams, take care to change the signs of every term when subtracting the second function.

\(\frac{x^2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2\), but (x\ne -1). It is necessary to exclude the zero of the denominator.

For both square roots (x+1\ge 0) and (x-3\ge 0), hence (x\ge 3). Even in subtraction both functions must be defined.

The sum of two functions is defined pointwise: \((f+g)(x)=f(x)+g(x)\). Therefore, \((f+g)(x)=(x^2-2x)+(2x-x^2)=x^2-x^2-2x+2x=0\). Hence, option A is correct. Option B results from not handling the opposite signs correctly. Exam tip: combine like powers only after preserving the signs of all terms.

Since (x^2+1>0) for all real (x), the denominator is never zero. Therefore the domain is (\mathbb{R}).