Functions

The governing concept is substitution followed by algebraic simplification of a function expression. Since f(x) = x + 2, replace f(x) in the requested expression with x + 2. This gives f(x) − 2 = (x + 2) − 2. The constants +2 and −2 cancel, leaving x. Therefore option A is correct. Option B adds 2 instead of subtracting it, option C incorrectly doubles the variable, and option D subtracts 2 from x even though the function already contains +2 that must also be included. The result is an identity: for every value of x in the domain, evaluating f(x) and then subtracting 2 returns the original input x.

Step 1: Replace (x) by (x+1). Step 2: (f(x+1)=2(x+1)-1=2x+2-1). Step 3: Simplifying gives (2x+1).

Step 1: (f(x+1)=(x+1)^2) and (f(x)=x^2). Step 2: The difference is ((x+1)^2-x^2). Step 3: Simplifying gives (x^2+2x+1-x^2=2x+1).

The governing concept is addition of functions, defined pointwise by (f + g)(x) = f(x) + g(x). Evaluate each function at x = 2: f(2) = 2 + 3 = 5 and g(2) = 2(2) = 4. Add the results: (f + g)(2) = f(2) + g(2) = 5 + 4 = 9. Hence option A is correct. Option B is f(2) plus an incorrect value, option C may result from using only 2 + 2, and option D is only f(2), omitting g(2). It is important not to confuse function addition with composition; the notation f + g means adding corresponding outputs, not placing one function inside the other.

The governing concept is subtraction of functions, which is defined pointwise as (f − g)(x) = f(x) − g(x). Substitute the given expressions: (f − g)(x) = 3x − (x + 1). The parentheses are essential because the minus sign applies to both terms inside g(x). Distribute the subtraction: 3x − x − 1 = 2x − 1. Thus option A is correct. Option C results from failing to change the sign of the constant, option B adds the functions instead of subtracting them, and option D incorrectly multiplies expressions and introduces a square. The final expression is linear, as expected from subtracting two linear functions.

Here (fg)(3) denotes the product of the two function values, so the governing rule is (fg)(x) = f(x)g(x). Evaluate both functions at 3: f(3) = 3 + 2 = 5 and g(3) = 3 − 2 = 1. Multiply them: (fg)(3) = 5 × 1 = 5. Therefore option A is correct. Option B is only g(3), while option C and option D arise from other incorrect combinations of the input or outputs. This should not be confused with composition, written f(g(3)); composition would give f(1) = 3, which is not what the product notation (fg)(3) asks for in this context. The direct product calculation uniquely confirms 5.

The governing concept is composition of functions: (f ∘ g)(x) means that g is applied first and f is applied to the result. Therefore, (f ∘ g)(2) = f(g(2)). First calculate g(2) = 2 + 1 = 3. Next apply f to this output: f(3) = 3² = 9. Hence the required value is 9, so option A is correct. Option B, 5, could result from adding the two function values incorrectly; option C is only 2² and ignores g; option D does not follow the composition rule. The order of composition is essential because f ∘ g generally differs from g ∘ f.

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=x-1) into (f). Step 3: (f(x-1)=2(x-1)+3=2x+1).

The governing concept is finding an inverse function by interchanging the input and output and solving for the original input. Let y = f(x) = x − 5. Add 5 to both sides: y + 5 = x, or x = y + 5. To express the inverse in the usual variable x, replace y by x. Thus f⁻¹(x) = x + 5, which is option A. Option B repeats the original function rather than reversing it. Option C changes the sign incorrectly and would represent 5 − x, while option D multiplies by 5 instead of undoing the subtraction. A quick check confirms the result: f(f⁻¹(x)) = (x + 5) − 5 = x.

Step 1: Let (y=\frac{x}{3}). Step 2: Multiplying both sides by (3), we get (x=3y). Step 3: Hence the inverse is (f^{-1}(x)=3x).

Step 1: (f^{-1}(4)) means the (x) for which (f(x)=4). Step 2: From (2x-6=4), we get (2x=10). Step 3: Hence (x=5), so (f^{-1}(4)=5).

Step 1: Different values of (x) give different values of (x-7). Step 2: For any real (y), choosing (x=y+7) gives (f(x)=y). Step 3: Hence it is both one-one and onto.

Step 1: Since (x^2\ge 0), (x^2+4\ge 4). Step 2: The codomain (R) contains (2), but no real (x) can map to it. Step 3: Therefore the range is not all of (R), so the function is not onto.

Step 1: For every real (x), (x^2\ge 0). Step 2: Therefore (x^2+4\ge 4). Step 3: The minimum value is (4), so the range is ([4,\infty)).

Step 1: For every (x), the value of the function remains (2). Step 2: Such a function is called a constant function. Step 3: In a constant function, all inputs give the same image.

The governing concept is the range of a constant function. A range consists of all output values that the function actually produces as the input runs through its domain. Here the domain is R, but f(x) is always equal to 2, regardless of whether x is positive, negative, or zero. Consequently, every real input has the same image, 2, and no other output can occur. Therefore Range(f) = {2}, so option A is correct. Option B confuses the codomain R with the actual range. Option C would be correct only for the constant function f(x) = 0, while option D incorrectly suggests that every positive real number is attained. The singleton set notation is essential because the range contains exactly one value.

Step 1: The expression inside the square root, (x+2), must be non-negative. Step 2: (x+2\ge 0) gives (x\ge -2). Step 3: Hence the real domain is ([-2,\infty)).

Step 1: The denominator (x+3) must not be zero. Step 2: (x+3=0) gives (x=-3). Step 3: So all real numbers except (-3) form the domain.

Step 1: The expression inside the square root must satisfy (5-x\ge 0). Step 2: This gives (x\le 5). Step 3: Therefore the domain is ((-\infty,5]).

Step 1: The denominator of the function is (x+1). Step 2: Setting (x+1=0) gives (x=-1). Step 3: A fraction is undefined when the denominator is zero.