Functions

Step 1: (f^{-1}(5)) means the (x) for which (f(x)=5). Step 2: From (x+2=5), we get (x=3). Step 3: Therefore (f^{-1}(5)=3).

Step 1: For (x^3), different real inputs give different outputs, so it is one-one. Step 2: For every real (y), (x=\sqrt[3]{y}) is real. Step 3: Hence every (y) has a preimage, so the function is onto as well.

Step 1: For one-one, different inputs must have different images. Step 2: (2\ne -2), but (|2|=|-2|=2). Step 3: Therefore this function is not one-one.

Step 1: To be onto, every element of (R) must be an image. Step 2: (-2\in R), but (|x|) is never negative. Step 3: Hence the range is not (R), and the function is not onto.

Step 1: The denominator of a fraction cannot be zero. Step 2: At (x=0), (\frac{1}{0}) is undefined. Step 3: So the domain is all real numbers except (0).

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

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

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

Step 1: For (f(a)), replace (x) by (a). Step 2: Replacing (x) by (a) in (2x-3) gives (2a-3). Step 3: In symbolic evaluation, change only the given variable.

Step 1: For (f(a+h)), replace (x) by the whole expression (a+h). Step 2: Putting (x=a+h) in (x^2) gives ((a+h)^2). Step 3: Keep the whole expression in brackets.

Step 1: For (f(x+2)), replace the original (x) by (x+2). Step 2: (f(x+2)=(x+2)+1=x+3). Step 3: Treat the entire inside expression as the input.

Step 1: ((f+g)(x)=f(x)+g(x)). Step 2: (2x+(x-1)=3x-1). Step 3: In the sum of functions, add their values.

Step 1: ((fg)(x)=f(x)g(x)). Step 2: (f(2)=4) and (g(2)=2). Step 3: Therefore ((fg)(2)=4\cdot2=8).

Step 1: This is a constant function, so its value is (5) for every input. Step 2: (f(3)=5) and (f(1)=5). Step 3: Hence (f(3)-f(1)=5-5=0).

Step 1: (f:A\to B) means the function maps from (A) into (B). Step 2: The image of any element of (A) lies in (B). Step 3: Therefore (f(a)\in B).

Step 1: First (f) sends an element of (A) to (B). Step 2: Then (g) sends that element of (B) to (C). Step 3: Hence the composite function (g\circ f) maps from (A) to (C).

Step 1: ((g\circ f)(1)=g(f(1))). Step 2: (f(1)=2\cdot1+1=3). Step 3: (g(3)=3^2=9), so the value is (9).

The direct answer is A, \\(x\\). The composition \\(f\\circ g\\) means apply \\(g\\) first and then apply \\(f\\): \\( (f\\circ g)(x)=f(g(x)) \\). Here \\(g(x)=x+2\\), so replace the input of \\(f\\) by \\(x+2\\). Since \\(f(t)=t-2\\), we get \\(f(x+2)=(x+2)-2=x\\). Thus the two opposite changes cancel. Option A is correct because the final output is \\(x\\). Option B, \\(x+4\\), would result from adding both changes instead of applying the minus sign correctly. Option C, \\(x-4\\), uses the wrong direction or repeats subtraction. Option D, \\(2x\\), incorrectly multiplies the variable; no multiplication by 2 occurs. A common mistake is reversing the order: in a composition, the right-hand function acts first. Memory cue: read \\(f\\circ g\\) from right to left—first \\(g\\), then \\(f\\).

Step 1: (f(g(x))=f(x-3)=(x-3)+3=x). Step 2: (g(f(x))=g(x+3)=(x+3)-3=x). Step 3: Both composites give the identity function, so they are inverses.

Step 1: In (x+1), (x) can take all real values. Step 2: For any real (y), choosing (x=y-1) gives (f(x)=y). Step 3: Therefore the range is all of (R).

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

Step 1: In a many-one function, different inputs may have the same image. Step 2: For (f(x)=x^2), (f(1)=f(-1)=1). Step 3: Hence (x^2) is many-one, while the other given simple functions are one-one.

Step 1: The images of (1) and (2) are different, (3) and (4), so the function is one-one. Step 2: The element (5) of (B) is not an image of any element. Step 3: Hence it is not onto.

Step 1: Both elements (a) and (b) of (B) appear as images, so the function is onto. Step 2: Both (1) and (2) map to (a), so it is not one-one. Step 3: Check onto and one-one separately.

Step 1: Put (x=3) in the given function. Step 2: (f(3)=4\cdot3-5=12-5=7). Step 3: In a linear function, multiply first and then add or subtract.