Algebra of real functions

Here, \((fg)(x)\) denotes the product of the two functions: \((fg)(x)=f(x)g(x)=(x-2)(x+2)=x^2-4\). Therefore, \((fg)(x)+4=x^2-4+4=x^2\), so option A is correct. Option B results from adding 4 to the product without accounting for the \(-4\) term. Exam tip: apply the identity \((a-b)(a+b)=a^2-b^2\).

The sum of two functions is found by adding their values pointwise: \((f+g)(x)=f(x)+g(x)=(x^2+1)+(2x-3)=x^2+2x-2\). Option B has the wrong sign for the linear term and an incorrect constant term. In exams, combine like powers separately and check the signs carefully.

Here, \((fg)(x)\) denotes the product of the two functions: \((fg)(x)=f(x)g(x)=(x+2)(x-5)=x^2-5x+2x-10=x^2-3x-10\). Option D, \(2x-3\), is \(f(x)+g(x)\), the sum rather than the product. Exam tip: distinguish \((fg)(x)\), which means a product, from \(f(g(x))\), which means composition.

\(\left(\frac{f}{g}\right)(x)=\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3\), where \(x \neq 3\). Always check that the denominator is not zero.

The denominators (x-2) and (x+4) must not be zero, so (x \neq 2) and (x \neq -4). In rational functions, check denominators first.

For the difference of two functions, \((f-g)(x)=f(x)-g(x)=2x^2-(x+1)\). At \(x=-2\), \(f(-2)=2(-2)^2=8\) and \(g(-2)=-2+1=-1\). Therefore, \((f-g)(-2)=8-(-1)=9\). The distractor 7 results from mishandling the negative value of \(g(-2)\). Exam tip: evaluate both functions separately before subtracting.

First, \(f(2)=2^2-1=3\) and \(g(2)=2(2)+1=5\). Therefore, \((fg)(2)=f(2)g(2)=3\times5=15\). Here, \((fg)\) denotes the product of the two functions, not their composition. In an exam, carefully distinguish \((fg)(x)\) from \((f\circ g)(x)\).

The direct answer is A: \((g+f)(x)=(f+g)(x)\). Function addition is defined pointwise: \((f+g)(x)=f(x)+g(x)\). Thus \((g+f)(x)=g(x)+f(x)\). Ordinary addition is commutative, so changing the order does not change the sum. Substituting the given functions gives \((f+g)(x)=(4x-1)+(x^2+2)=x^2+4x+1\), and \((g+f)(x)=(x^2+2)+(4x-1)=x^2+4x+1\). Option A is correct. Option B is wrong because the expressions are equal, not opposites. Option C is wrong because both functions are defined for every real number, so their domains are the same. Option D is wrong because equality holds for every real \(x\), not only at zero. Memory cue: in addition, order can be exchanged.

The product of two functions is defined by \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x+6)(x-6)=x^2-6^2=x^2-36\). Option B incorrectly suggests a sum-of-squares pattern, whereas this is a product of conjugate binomials. Exam tip: recognize the identity \((a+b)(a-b)=a^2-b^2\).

In the original (f(x)), (x-1 \neq 0), so (x \neq 1), even if simplification occurs after multiplication. Decide the domain from original expressions.

((fg)(x)=\frac{2}{x}\cdot x^2=2x), but (x \neq 0) because of the original denominator. Keep original restrictions after simplification.

\(\frac{x^2+4x+4}{x+2}=\frac{(x+2)^2}{x+2}=x+2\), where \(x \neq -2\). Write the zero of the denominator separately.

By the definition of subtraction of functions, \((f-g)(x)=f(x)-g(x)\). Hence, \((f-g)(-5)=|-5|-(-5)=5+5=10\). Option B is incorrect because the two function values are not equal. Exam tip: \(|-5|=5\), while \(g(-5)=-5\); do not replace the latter by its absolute value.

For the algebraic combination of functions, \((2f-g)(x)=2f(x)-g(x)\). Therefore, \(2x^3-(2x^3-1)=2x^3-2x^3+1=1\). Option B results from handling the negative sign incorrectly. In an exam, remember that a minus sign before brackets changes the signs of all terms inside.

(x^2) is defined for all real (x), but (\sqrt{x}) needs (x \geq 0). The domain of the sum is the intersection.

(\sqrt{x-2}) requires (x \geq 2) and (\frac{1}{x}) requires (x \neq 0); the intersection is (x \geq 2). Identify the strongest restriction.

\(\left(\frac{f}{g}\right)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1\). In division, first check whether the denominator value is zero.

By definition, (g − f)(x) = g(x) − f(x). Substituting the given functions gives (x² + 4) − (x² − 4) = x² + 4 − x² + 4 = 8. The x² terms cancel, so the difference function is the constant 8.

By the definition of addition and subtraction of functions, \((f+g-h)(x)=f(x)+g(x)-h(x)\). Therefore, \((2x+3)+(x-4)-x^2=-x^2+3x-1\), so option A is correct. In option B, the sign of the \(x^2\) term is incorrect because \(h(x)\) is being subtracted. Exam tip: write the complete expressions in brackets before combining like terms.

\(\frac{x^2+x}{x}=\frac{x(x+1)}{x}=x+1\), where \(x \neq 0\). Do not forget the restriction while cancelling a common factor.

((f+g)(x)=\frac{x+2}{x+2}=1), but (x+2 \neq 0). It is important to write the domain restriction with the simplified answer.

The product function is defined by (fg)(x) = f(x)g(x). Substituting the given functions gives 3(x² − 7x) = 3x² − 21x, so option A is correct.

By the definition of the sum of functions, (f+f)(x)=f(x)+f(x)=2f(x). Therefore, (f+f)(x)=2(x²+1)=2x²+2, so option A is correct. Option B incorrectly doubles only the constant term instead of the entire function. Exam tip: the sum of a function with itself is twice the function.

By the definition of the product of functions, \((ff)(x)=f(x)\cdot f(x)\). Therefore, \((ff)(x)=(x-2)(x-2)=(x-2)^2=x^2-4x+4\), so option A is correct. Option B represents \(2f(x)\), not the product \(f(x)f(x)\). In an exam, distinguish \((ff)(x)\) from \(f(f(x))\), which denotes composition.

For pointwise function operations, \((f^2+g^2)(x)=[f(x)]^2+[g(x)]^2=\sin^2 x+\cos^2 x\). Using the fundamental trigonometric identity \(\sin^2 x+\cos^2 x=1\), the value is \(1\). Option C uses subtraction, while option D is the double-angle expression \(\sin 2x\). Exam tip: immediately recognize \(\sin^2 x+\cos^2 x=1\).