Algebra of real functions

\(\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2\), but (x\ne 2). Keep the original denominator restriction even after simplification.

A scalar multiple of a function is defined by \((3f)(x)=3f(x)\). Hence, \((3f)(x)=3(2x-3)=6x-9\). Option C changes only the coefficient of \(x\), but the constant term must also be multiplied by 3. Exam tip: distribute the scalar to every term of the function.

For the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Thus, \(f(3)=7-3=4\) and \(g(3)=3+1=4\), so \((f-g)(3)=4-4=0\). Exam tip: evaluate both functions at the same input before subtracting.

First, \(f(1)=1+3=4\) and \(g(1)=2\times1=2\). Since \((fg)(1)=f(1)g(1)\), we get \((fg)(1)=4\times2=8\), so option A is correct. The value 6 would result from adding the two function values, but the notation here requires multiplication. Exam tip: distinguish \((fg)(x)\), which denotes the product of functions, from \(f(g(x))\), which denotes composition.

\(\left(\frac{f}{g}\right)(2)=\frac{f(2)}{g(2)}=\frac{12}{6}=2\). The denominator value must not be (0).

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Thus, \(f(-1)=(-1)^2+1=2\) and \(g(-1)=-1\), so \((f+g)(-1)=2+(-1)=1\). Option C is only the value of \(f(-1)\), without adding \(g(-1)\). Exam tip: evaluate the two functions separately and then add their values.

For the product of two functions, \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x+5)\cdot 2=2x+10\). Option B represents addition rather than multiplication. Exam tip: distinguish \((fg)(x)\), the product, from \(f(g(x))\), the composition.

Adding the two functions gives \((f+g)(x)=f(x)+g(x)=(4x+1)+(x-2)=5x-1\). Therefore, the coefficient of \(x\) is 5. The value 4 is only the coefficient of \(x\) in \(f(x)\), not in their sum. Exam tip: add the coefficients of like-power terms when combining linear functions.

Here, \((fg)(x)\) denotes the product \(f(x)g(x)\). Thus, \((fg)(x)=(x-4)(x+4)\). Using the identity \((a-b)(a+b)=a^2-b^2\), we get \(x^2-4^2=x^2-16\). Option B is incorrect because the constant term should be negative, not positive. Exam tip: when two conjugate binomials are multiplied, apply the difference-of-squares identity directly.

The product of two functions is defined by \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=\frac{1}{x}\cdot x=1\), since \(x\ne 0\). The expression \(\frac{1}{x^2}\) would result from division, not multiplication. Exam tip: Rewrite \((fg)(x)\) as \(f(x)g(x)\) before substituting the given functions.

For (\sqrt{x}), we need (x\ge 0), while (x+1) is defined for all real (x). Hence the domain is ([0,\infty)).

The quotient is \(\left(\frac{f}{g}\right)(x)=\frac{x+2}{x-5}\), so its denominator must not be zero. Thus \(x-5\neq 0\), or \(x\neq 5\), giving the domain \(\mathbb{R}\setminus\{5\}\). In option B, \(-2\) is excluded even though it only makes the numerator zero, which is allowed. Exam tip: For a quotient, exclude only the values that make the denominator zero.

For a linear combination of functions, \((2f+g)(x)=2f(x)+g(x)\). Therefore, \(2(x^2+3)+(2x-1)=2x^2+6+2x-1=2x^2+2x+5\), so option A is correct. In option B, the constant term is calculated incorrectly; \(6-1=5\), not 2. Exam tip: multiply the entire function by its coefficient before combining like terms.

\((f+2g)(1)=f(1)+2g(1)\). Here, \(f(1)=3(1)+2=5\) and \(g(1)=1^2=1\), so \((f+2g)(1)=5+2(1)=7\). The value 5 results from forgetting the coefficient 2 of \(g(1)\). In an exam, write the rule for the combined function before substituting the value.

By the definition of the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Therefore, \((f+g)(x)=x+x^2\). The option \(x^3\) represents \(f(x)g(x)\), the product of the function values, not their sum. Exam tip: For \((f+g)(x)\), add the values of the two functions at the same input.

By the definition of the difference of two functions, \((f-g)(0)=f(0)-g(0)\). Since \(f(0)=0\) and \(g(0)=3\), we get \((f-g)(0)=0-3=-3\). Option B results from reversing the subtraction order. Exam tip: In \((f-g)(x)\), subtract the value of the second function from the value of the first function.

For the quotient function, \(\left(\frac{f}{g}\right)(0)=\frac{f(0)}{g(0)}\). Here, \(f(0)=0^2-1=-1\) and \(g(0)=0+1=1\), so the value is \(\frac{-1}{1}=-1\). Option B is incorrect because it ignores the negative sign in the numerator. Exam tip: For a quotient of functions, evaluate the numerator and denominator separately and verify that the denominator is not zero.

By definition of the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \((x+6)+(x-6)=2x\), since \(+6\) and \(-6\) cancel each other. Option B would arise from multiplying the functions, not adding them. In an exam, first write the definition of the function operation and then simplify.

For the product of two real functions, \((fg)(x)=f(x)g(x)\). Here, \(f(0)=1\) and \(g(0)=2\), so \((fg)(0)=1\times2=2\). Exam tip: interpret \(fg\) as the product of functions; composition is written as \(f\circ g\).

By definition, \((f-g)(x)=f(x)-g(x)\). Thus, \(f(2)=2^2=4\) and \(g(2)=2\times2=4\), so \((f-g)(2)=4-4=0\). Option B is the value of either function separately, not their difference. Exam tip: calculate both function values before subtracting.

For (\sqrt{x-1}), (x-1\ge 0), so (x\ge 1). The product domain is the intersection of both domains.

The sum is (f + g)(x) = x + 1 + 1/x. Since 1/x is undefined at x = 0, while x + 1 is defined for every real number, the sum is defined for all real numbers except 0. Thus its domain is ℝ \ {0}.

Here, \((fg)(x)=\frac{x}{x+2}(x-1)\). The denominator cannot be zero, so \(x+2=0\) gives \(x=-2\). Therefore, \(-2\) is excluded from the domain. Remember that the domain of a product is the intersection of the domains of the two functions; a zero in the numerator does not by itself exclude a value.

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(x)=2x^2+(x^2+3)=3x^2+3\). The like terms \(2x^2\) and \(x^2\) combine, while the constant 3 remains unchanged. In an exam, first write \(f(x)+g(x)\) when finding the sum of functions.

The difference of two functions is defined as \((f-g)(x)=f(x)-g(x)\). Hence, \((f-g)(x)=(x^2+5)-(x^2-2)=x^2+5-x^2+2=7\). The minus sign before the second bracket changes both signs inside it. Exam tip: Always write \(f(x)-g(x)\) with brackets before simplifying a function difference.