Algebra of real functions

Using the algebra of functions, \((f+2g)(x)=f(x)+2g(x)\). Therefore, \(3x^2+2(2x+1)=3x^2+4x+2\), so option A is correct. In option C, the factor 2 has not been properly distributed to \(g(x)\). In the exam, multiply the entire function \(g(x)\) by 2 before combining like terms.

For both roots, (x-2\ge 0) and (5-x\ge 0), so (2\le x\le 5). For a sum, take the intersection of domains.

The direct answer is option A: \(\mathbb{R}\setminus\{-2,1\}\). A rational function is defined only when its denominator is nonzero. For \(f(x)=1/(x-1)\), we need \(x-1\ne0\), so \(x\ne1\). For \(g(x)=1/(x+2)\), we need \(x+2\ne0\), so \(x\ne-2\). The sum \((f+g)(x)\) exists only where both functions exist, so we take the common domain: all real numbers except -2 and 1. Option A states exactly this. Option B removes only 1 and wrongly allows -2, where \(g\) has division by zero. Option C removes only -2 and wrongly allows 1, where \(f\) is undefined. Option D lists only the excluded numbers, but a domain contains the allowed inputs, not merely the forbidden ones. Even if the fractions are combined algebraically, the original functions are not defined at these points. Memory cue: for addition, intersect the domains; for denominators, never allow zero.

The product of two functions is defined as (fg)(x)=f(x)g(x). Thus, f(3)=3^2+1=10 and g(3)=3-2=1, so (fg)(3)=10×1=10. The value 12 would result from adding the two function values, not multiplying them. Exam tip: For (fg)(a), evaluate both f(a) and g(a) first, then multiply.

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \(f(-2)=2(-2)-1=-5\) and \(g(-2)=(-2)^2=4\). Therefore, \((f+g)(-2)=-5+4=-1\). Option C results from incorrectly taking the sign of \(f(-2)\), while option D treats the square of \(-2\) as negative. Exam tip: evaluate each function separately before adding their values.

For algebraic operations on functions, \((f-2g)(x)=f(x)-2g(x)\). Therefore, \((f-2g)(x)=(5x+4)-2(2x-3)=5x+4-4x+6=x+10\), so option A is correct. Option B results from combining the coefficients incorrectly. In an exam, remember that distributing \(-2\) changes the signs of both terms inside the bracket.

\((f-g)(x)=f(x)-g(x)=x^2-3x=x(x-3)\). By the zero-product rule, \(x(x-3)=0\) gives \(x=0\) or \(x=3\). Option B is incorrect because \((f-g)(1)=1-3=-2\), not zero. Exam tip: first form the difference of the functions, then factorise it.

First, \((f+g)(x)=f(x)+g(x)=(x+4)+(2x-1)=3x+3\). Also, \((fg)(x)=f(x)g(x)=(x+4)(2x-1)=2x^2+7x-4\). Therefore, the required equation is \(2x^2+7x-4=3x+3\). Remember that \(fg\) denotes the product of the functions here, not their composition; while expanding, the middle term is \(8x-x=7x\).

(g(x)) is defined for all real (x), but (f(x)) needs (x-2\ne 0). For product also, take the common domain.

For the difference of two functions, (f − g)(x) = f(x) − g(x). Hence, (f − g)(x) = (2x² − 3) − (x² + 5) = 2x² − x² − 3 − 5 = x² − 8. Option B results from effectively adding the functions, so it is incorrect. In exams, remember to subtract the entire g(x), including the constant term.

The product of two functions is defined by \((fg)(x)=f(x)g(x)\). Therefore, \((fg)(x)=3(x^2-2x)=3x^2-6x\), so option A is correct. Option B results from adding the two functions instead of multiplying them. Exam tip: do not confuse \((fg)(x)\), the product, with \(f(g(x))\), the composite function.

The sum of functions is defined by (f + g)(x) = f(x) + g(x). At x = -4, f(-4) = |-4| = 4 and g(-4) = -4, so (f + g)(-4) = 4 + (-4) = 0.

((fg)(x)=\frac{1}{x}\cdot x^2=x), but the original domain keeps (x\ne 0). Restrictions do not disappear after simplification.

(f − g)(x) = f(x) − g(x) = (x² + 2x) − (x² − 2x). On removing the brackets, x² − x² + 2x + 2x = 4x. Therefore, the correct answer is 4x. Exam tip: when subtracting a function, change the signs of every term in the subtracted function; hence −2x becomes +2x.

The sum of the two functions is \((f+g)(x)=f(x)+g(x)=(2x+1)+(x^2-1)=x^2+2x\). Hence, \((f+g)(2)=2^2+2(2)=4+4=8\), so option A is correct. Remember that \((f+g)(a)\) is found by adding the values of both functions at \(a\).

The product function is (fg)(x) = √x/(x − 4). Its domain must satisfy every restriction inherited from both factors. The square root requires x ≥ 0, because a real square root is defined only for a non-negative radicand. The denominator requires x − 4 ≠ 0, so x ≠ 4. Intersecting these conditions gives [0, ∞) with the single point 4 removed. Therefore option A is correct. Option B wrongly excludes zero, even though √0 is defined. Option C wrongly excludes all values greater than 4, and option D includes negative numbers where √x is not real. Simplifying the expression does not remove the original denominator restriction.

\(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2+4x+4}{x+2}\). Factoring the numerator as \((x+2)^2\) gives \(\frac{(x+2)^2}{x+2}=x+2\), provided \(x\ne -2\). Option C is not correct because it equals the quotient only for a particular value, not for all allowed values of \(x\). Exam tip: factor the numerator first and exclude any value that makes the original denominator zero.

Here, \((fg)(x)\) denotes the product \(f(x)g(x)\), not the composition of the functions. Thus, \(f(0)=-3\) and \(g(0)=5\), so \((fg)(0)=f(0)g(0)=(-3)(5)=-15\). Exam tip: For \((fg)(a)\), find the values of both functions at \(a\) and then multiply them.

The simplified form is (\frac{x}{x+1}), but original denominators keep (x\ne 1,-1). Remember restrictions from cancelled factors too.

The algebraic combination of functions is evaluated pointwise: \((f+g-h)(x)=f(x)+g(x)-h(x)=2x+(x+3)-(x-1)\). On removing the brackets, \(2x+x+3-x+1=2x+4\), so option A is correct. Option B results from combining the \(x\)-terms incorrectly. Exam tip: change the sign of every term inside a function that is being subtracted.

Use the definitions of addition and multiplication of functions. Since f(x) = x² and g(x) = 2, we have (f + g)(x) = x² + 2 and (fg)(x) = 2x². Set them equal: x² + 2 = 2x². Subtracting x² from both sides gives x² = 2, so the real solutions are x = √2 and x = −√2. Therefore option A is correct. Option B comes from confusing x² = 2 with x² = 4, option C does not satisfy the equation because 0 + 2 is not 0, and option D is false because two real solutions exist. Both functions are defined for every real x, so no additional domain restriction changes the answer.

\(\frac{x^2-1}{x+1}=x-1\), so at \(x=2\) the value is (1). Directly, \(f(2)=3\) and \(g(2)=3\), so the value is (1).

The governing concept is that the domain of a sum or difference of functions is the intersection of the domains of both component functions. For f(x) = √(x + 1), the radicand must be nonnegative: x + 1 ≥ 0, so x ≥ −1. For g(x) = √(x − 3), we need x − 3 ≥ 0, so x ≥ 3. Both conditions must hold simultaneously; therefore the common domain is [−1, ∞) ∩ [3, ∞) = [3, ∞). The endpoint 3 is included because √0 is defined, so excluding it would be incorrect. Hence option A is correct. Option B ignores g’s restriction, option C incorrectly gives an upper endpoint, and option D excludes a valid endpoint.

Factor the numerator as \(x^2+3x+2=(x+1)(x+2)\). Thus, \(\left(\frac{f}{g}\right)(x)=\frac{(x+1)(x+2)}{x+1}=x+2\), provided \(x\ne -1\). Option B is only \(g(x)\), not the quotient. Exam tip: always exclude values that make the denominator zero, even after cancellation.

For a linear combination of functions, \((2f-g)(x)=2f(x)-g(x)\). Thus, \(2f(x)=4x^2+2x\), and \((2f-g)(x)=(4x^2+2x)-(x^2-x)=3x^2+3x\). Therefore, option A is correct. Remember to distribute the minus sign to both terms of \(g(x)\): \(-(x^2-x)=-x^2+x\).