Algebra of real functions

Adding the functions gives \((f+g)(x)=(x^3-2x)+(x^3+2x)=2x^3\). Therefore, \(\frac{f+g}{2}=\frac{2x^3}{2}=x^3\), so option A is correct. Option C is only \(g(x)\), not the average of the two functions. In an exam, combine like terms first and then divide the entire result by 2.

(\frac{x^2+2}{x^2+5}=1-\frac{3}{x^2+5}), so it approaches (1) but never equals (1). Hence no actual maximum is attained.

Since (f + g)(2) = f(2) + g(2), we get 18 = [3(2) − 2] + [2k + 4] = 2k + 8. Thus 2k = 10, so k = 5.

Using (f^2-g^2=(f-g)(f+g)), we get (2\cos x\cdot 2\sin x=4\sin x\cos x). Algebraic identities also apply to trigonometric functions.

\((f-g)(x)=[x^2+px+4]-[x^2-px+4]=2px\). Hence, \((f-g)(2)=2p\cdot2=4p\). Given \(4p=12\), we get \(p=3\). Option 4 is incorrect because it would give \(4p=16\). Exam tip: simplify the difference of the two functions before substituting the given value of \(x\).

\(\lfloor\frac{7}{3}\rfloor=2\) and \(g\left(\frac{7}{3}\right)=\frac{1}{3}\), so the value is \(2+\frac{2}{3}=\frac{8}{3}\). Find integer and fractional parts separately.

((f-g)(x)=x^2-2x+1=(x-1)^2), so the only solution is (x=1). A square is zero only when its inside term is (0).

In (f), the denominator is zero at (x=-2), so (f) and (f-g) are undefined. Check the original domain before simplification.

The square root needs (16-x^2\ge 0), meaning (-4\le x\le 4), and the denominator needs (x\ne 1). The domain is the intersection of these conditions.

(f=(x+2)^2), so (\frac{f}{g}=x+2), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.

((fg)(x)=3x^3), whose range is all real numbers. An odd-degree polynomial extends in both directions.

First, \((f+g)(x)=f(x)+g(x)=x^2+2x+5+2x+1=x^2+4x+6\). Completing the square gives \(x^2+4x+6=(x^2+4x+4)+2=(x+2)^2+2\), so option A is correct. Option B would produce the middle term \(-4x\), not \(+4x\). Exam tip: for \(x^2+bx+c\), take half of the coefficient of \(x\), square it, and add and subtract that value.

With a common denominator, (f-g=\frac{x^2+4-(x^2-4)}{x}=\frac{8}{x}). While subtracting the second numerator, both signs change.

This item should be checked by solving ((f-g)(x)=x^2-2x-1). The correct interval is between its two roots, so no listed option is fully suitable.

(\frac{f}{g}=\frac{(x+3)^2+1}{(x+3)^2+5}), which gives the minimum (\frac{1}{5}) at ((x+3)^2=0). It is useful to set the common square term as (t\ge 0).

First multiply g(x) and h(x): g(x)h(x)=(x−2)·1=x−2. Hence, (f−2gh)(x)=(x²−4x+4)−2(x−2)=x²−4x+4−2x+4=x²−6x+8. Therefore, option A is correct. Option D results from failing to subtract the complete term 2gh correctly. Exam tip: In algebraic combinations of functions, evaluate the product and its numerical coefficient before performing subtraction.

Since \(g(-x)=-g(x)\) and \(f\) is even, \(h(-x)=f(-g(x))=f(g(x))=h(x)\). Hence \(h\) is even, not necessarily constant. Exam tip: apply the parity of the inner function first in a composition.

For the sum of two functions, \((f+g)(x)=f(x)+g(x)=(x^2-5x)+x=x^2-4x=x(x-4)\). Thus, the expression is zero when \(x=0\) or \(x=4\), so the correct set is \(\{0,4\}\). Option B gives the zeroes of \(f(x)\) alone, not those of \(f(x)+g(x)\). Exam tip: first form the sum of the functions and then factorise it to find its zeroes.

For product functions, \((fg)(x)=f(x)g(x)\). Thus, \(f(1)=1^2+2(1)=3\) and \(g(1)=1^2-2(1)=-1\). Therefore, \((fg)(1)=3\times(-1)=-3\), so option B is correct. Option C would result from applying an incorrect operation instead of multiplication. Exam tip: To evaluate \((fg)(a)\), first find \(f(a)\) and \(g(a)\), then multiply them.

((f+g)(x)=2|x|), and for (x<0) it is also (-2x). Among the options, (2|x|) is a general correct form.

\(f(2)=\frac{4-9}{2-3}=5\) and \(g(2)=\frac{4-1}{2-1}=3\). Therefore, \((f-g)(2)=f(2)-g(2)=5-3=2\), so option A is correct. The functions are undefined at \(x=3\) and \(x=1\), but \(x=2\) belongs to the domain of both functions. Exam tip: For a difference function, evaluate the two functions separately and then subtract.

(f − g)(x) = [x² + ax + 1] − [x² + bx + 1] = (a − b)x. Since this equals 5x, comparing the coefficients of x gives a − b = 5. Option B results from a sign error. Exam tip: subtract like-powered terms first, then compare their coefficients.

The product may simplify to (1), but the original functions forbid (x=3) and (x=-3). The domain is decided by original restrictions.

Function addition and subtraction are performed pointwise: (f−g+h)(x)=(2x²+1)−(x²−3)+(x²+4). Subtracting g(x) changes the signs of both of its terms, giving 2x²+1−x²+3+x²+4=2x²+8. Option B results from combining the terms incorrectly. Exam tip: when subtracting a function, place it in parentheses and reverse the sign of every term inside.

First evaluate f(0): f(0)=|0+2|+|0-2|=|2|+|-2|=2+2=4. Since g(x)=4, g(0)=4. Therefore, (f-g)(0)=f(0)-g(0)=4-4=0. Remember that (f-g)(a)=f(a)-g(a); evaluating only f(a) gives the distractor 4, not the required difference.