(a) Write down the inverse of \( f(x) \) in the form \( y=\ldots \) (b) Sketch the graph of the function and its inverse on the same set of axer. (c) Is the inverse a function? Justify your answer. (d) Restriet the domain of the given function in two dilterent ways to form une-to-one function (e) Sketch the graphs of each new function and inverse on the same set of axe. (f) Rewrite the equation of each inverse function in tho form \( f^{-1}(x) \) (g) Write down the domain and the ranye of ench manh dram

In this exercise we are meant to start with the function   f(x) = x² and investigate its inverse. (Note that if no function is given, it is common in such problems that f is taken to be x².) Let’s go through each part. ────────────────────────────── (a) Write down the inverse of f(x) in the form y = … Step 1. Write f in “y=” form:   y = x² Step 2. To “find the inverse” we interchange the roles of x and y:   x = y² Step 3. Solve for y. Taking square roots we have   y = √x or y = –√x. Thus, the inverse relation is   y = ±√x. Note: Since the original f(x) = x² is not one‑to‑one (because, for example, f(2) = f(–2) = 4), its inverse “relation” is given by these two branches. ────────────────────────────── (b) Sketch the graph of the function and its inverse on the same set of axes. To sketch: • The graph of f(x) = x² is a parabola opening upward with vertex at (0, 0). • The “graph” of the inverse consists of the two halves of the sideways parabola:   y = √x (the upper half) and y = –√x (the lower half). • Notice that the graph of the inverse relation is the mirror image of f(x) about the line y = x. (When you sketch, draw the parabola along with the line y = x and then reflect the curve across this line to get the inverse relation.) ────────────────────────────── (c) Is the inverse a function? Justify your answer. No, the inverse is not a function. For a relation to be a function it must assign exactly one output value to each input. Here, for a given x (with x > 0) the inverse relation gives two outputs: one positive and one negative (√x and –√x). Equivalently, the graph of y = ±√x fails the horizontal line test. Therefore, the inverse of f(x) = x² (with its natural domain) is not a function. ────────────────────────────── (d) Restrict the domain of the given function in two different ways to form a one‐to‐one function. Since f(x) = x² is not one‐to‐one on ℝ, we can restrict its domain to either of the following: 1. Restriction 1: Let f be defined on [0, ∞). That is, define   f₁(x) = x² for x ≥ 0. 2. Restriction 2: Alternatively, let f be defined on (–∞, 0]. That is, define   f₂(x) = x² for x ≤ 0. Both f₁ and f₂ are one‐to‐one. ────────────────────────────── (e) Sketch the graphs of each new function and its inverse on the same set of axes. For the first restricted function, f₁(x) = x², x ≥ 0: • f₁(x) is the right half of the parabola. • Its inverse is obtained by solving x = y² with y ≥ 0. This gives   f₁⁻¹(x) = √x. • Sketch f₁(x) on the x ≥ 0 portion and the graph of y = √x; note that these graphs are reflections of each other in the line y = x. For the second restricted function, f₂(x) = x², x ≤ 0: • f₂(x) is the left half of the parabola. • Its inverse is obtained by solving x = y² with y ≤ 0. This gives   f₂⁻¹(x) = –√x. • Similarly, when you graph f₂ and y = –√x, the two graphs are reflections of each other in the line y = x. ────────────────────────────── (f) Rewrite the equation of each inverse function in the form f⁻¹(x). Thus we have: 1. For f₁(x) = x² with x ≥ 0, the inverse function is   f₁⁻¹(x) = √x     (with x ≥ 0) 2. For f₂(x) = x² with x ≤ 0, the inverse function is   f₂⁻¹(x) = –√x    (with x ≥ 0) ────────────────────────────── (g) Write down the domain and range of each main graph. For the original f(x) = x² (without restriction): • Domain: all real numbers, ℝ. • Range: [0, ∞). • Its inverse “relation” (y = ±√x) has domain [0, ∞) and range (–∞, ∞), but note that it is not a function. For the restricted functions: 1. For f₁:   • Domain of f₁: [0, ∞)   • Range of f₁: [0, ∞)   • Domain of f₁⁻¹ (i.e. f₁⁻¹(x) = √x): [0, ∞)   • Range of f₁⁻¹: [0, ∞) 2. For f₂:   • Domain of f₂: (–∞, 0]   • Range of f₂: [0, ∞)   (since squaring any number is nonnegative)   • Domain of f₂⁻¹ (i.e. f₂⁻¹(x) = –√x): [0, ∞)   • Range of f₂⁻¹: (–∞, 0] (because –√x is never positive) ────────────────────────────── Summary (a) The inverse relation of f(x)=x² is: y = ±√x. (b) The graph of y = x² (a parabola) and the graph of its inverse (the two branches y = √x and y = –√x) are reflections of each other in the line y = x. (c) The inverse relation is not a function (since, for most x, there are two y‐values). (d) By restricting the domain to [0, ∞) or to (–∞, 0], f becomes one-to-one. (e) The graphs of   f₁(x)=x² (x ≥ 0) with inverse f₁⁻¹(x)=√x, and   f₂(x)=x² (x ≤ 0) with inverse f₂⁻¹(x)= –√x are each symmetric about y = x. (f) Inverse functions in proper notation:   For x ≥ 0: f₁⁻¹(x) = √x.   For x ≤ 0: f₂⁻¹(x) = –√x. (g) Their domains and ranges are as stated above. Any answer equivalent to the one given is correct.