The function \( f(x)=x^{2}-7, x \geq 0 \) is one-to-one. (a) Find the inverse of \( f \) and check the answer. (b) Find the domain and the range of \( f \) and \( f^{-1} \) (c) Graph \( f, f^{-1} \), and \( y=x \) on the same coordinate axes. (a) \( f^{-1}(x)=\sqrt{x+7} \) (Simplify your answer Use integers or fractions for any numbers in the expression.) (b) The domain of \( f \) is \( (-\infty, \infty) \). (Type your answer in interval notation.)

We are given the function   f(x) = x² – 7  with  x ≥ 0. Because the domain is restricted to x ≥ 0, f is one-to-one and an inverse exists. We will find the inverse, determine the domains and ranges, and describe how to graph them. ────────────────────────────── Step (a): Find the Inverse of f and Check the Answer 1. Write y = f(x):   y = x² – 7  with x ≥ 0. 2. Solve for x:   y + 7 = x²   x = ±√(y + 7) Since x is restricted to x ≥ 0, we take the nonnegative square root:   x = √(y + 7). 3. Swap the roles of x and y to obtain the inverse:   f⁻¹(x) = √(x + 7). So the inverse function is   f⁻¹(x) = √(x + 7). 4. Checking the Inverse: a. Check f(f⁻¹(x)):   f(f⁻¹(x)) = f(√(x + 7))     = (√(x + 7))² – 7     = (x + 7) – 7     = x. b. Check f⁻¹(f(x)):   f⁻¹(f(x)) = f⁻¹(x² – 7)     = √((x² – 7) + 7)     = √(x²)     = x     (since x ≥ 0). Both compositions yield the identity on the appropriate domains, so f⁻¹ is indeed the inverse of f. ────────────────────────────── Step (b): Find the Domain and the Range of f and f⁻¹ For f(x) = x² – 7 with x ≥ 0: • Domain of f:   Since x is given as x ≥ 0,   Domain: [0, ∞). • Range of f:   The smallest value occurs at x = 0:     f(0) = 0² – 7 = –7.   Since as x increases f(x) increases without bound,   Range: [–7, ∞). Now for the inverse function f⁻¹(x) = √(x + 7): • Domain of f⁻¹:   Inside the square root, we require x + 7 ≥ 0,   so x ≥ –7.   Domain: [–7, ∞). • Range of f⁻¹:   Since square root outputs are nonnegative,   Range: [0, ∞). ────────────────────────────── Step (c): Graph f, f⁻¹, and y = x on the Same Coordinate Axes 1. Graph of f(x) = x² – 7:   • It is a parabola opening upward.   • Its vertex is at (0, –7).   • Since the domain is x ≥ 0, only the right branch of the parabola is drawn (starting at x = 0). 2. Graph of f⁻¹(x) = √(x + 7):   • This is the reflection of f across the line y = x.   • The function is defined for x ≥ –7.   • Its starting point is when x = –7: f⁻¹(–7) = √(0) = 0, so it starts at (–7, 0).   • It increases to the right. 3. Graph of y = x:   • This is the line with slope 1 passing through the origin.   • It is the mirror line for any function and its inverse. To sketch these: • First draw the line y = x. • Then plot the graph of f(x) = x² – 7 for x ≥ 0. Mark the vertex at (0, –7) and plot a few points (for example, at x = 2, f(2) = 4 – 7 = –3; at x = 3, f(3) = 9 – 7 = 2). • Next, plot f⁻¹(x) = √(x + 7) which is the mirror image of the f graph. Start at (–7, 0) and plot points accordingly (for example, if x = –3, then f⁻¹(–3) = √(4) = 2, so the point (–3, 2) is on f⁻¹). Because these graphs are reflections, every point (a, b) on f corresponds to a point (b, a) on f⁻¹. ────────────────────────────── Final Answers (a) Inverse Function:   f⁻¹(x) = √(x + 7). (b) Domains and Ranges:   For f(x) = x² – 7, Domain: [0, ∞) and Range: [–7, ∞).   For f⁻¹(x) = √(x + 7), Domain: [–7, ∞) and Range: [0, ∞). (c) Graphs:   • f(x) = x² – 7 is an upward-opening parabola starting at (0, –7) for x ≥ 0.   • f⁻¹(x) = √(x + 7) is the reflection of f across the line y = x, starting at (–7, 0).   • The line y = x is drawn for reference. These complete the answers for this problem.