The graph of f is given.The x y coordinate plane is given. The curve starts at the closed point (−2, −1), goes up and right becoming more steep, passes through the point (0, 0), passes through the point (1, 1), goes up and right becoming less steep, passes through the point (2, 2), and ends at the closed point (4, 3).(a)Why is f is one-to-one?f is one-to-one because it passes the .(b)What are the domain and range of f −1? (Enter your answers in interval notation.)domain     range     (c)What is the value of f −1(1)?(d)Estimate the value of f −1(0) to the nearest tenth.

The graph of a function f is given in the figure.A curve is shown on the x y coordinate plane. It begins at the point (−2, −1), goes up and to the right, passes through the approximate point (−1, −0.2), and passes through the negative x-axis at the approximate point (−0.8, 0). It then continues up and right, passes through the positive y-axis at the point (0, 1), and reaches a high point at (1, 3). It then goes down and right, passes through the points (2, 2) and (3, 1), and ends at the approximate point (4, 0.5).(a)Find the value of f(1).(b)Estimate the value of f(−1).(c)For what values of x is f(x) = 1? (Enter your answers as a comma-separated list.) (d)Estimate the value of x such that f(x) = 0.x = (e)State the domain and range of f. (Enter your answers in interval notation.)domain range (f)On what interval is f increasing? (Enter your answer using interval notation.)

nterpret the following graph in detail: (i) Identify the domain and range.(ii) Does this graph represent a function and a one-one function. Why/Why not? Provide a detailed explanation/justification.

QUESTION 26Choice D is correct. The definition of the graph of a function f in thexy-plane is the set of all points (x, f(x)). Thus, for −4 ≤ a ≤ 4, the valueof f(a) is 1 if and only if the unique point on the graph of f with x-coordinate a has y-coordinate equal to 1. The points on the graph of fwith x-coordinates −4, , and 3 are, respectively, (−4, 1), , and(3, 1). Therefore, all of the values of f given in I, II, and III are equalto 1.Choices A, B, and C are incorrect because they each omit at least onevalue of x for which f(x) = 1

The graph of a function g is shown. The x y-coordinate plane is given. The curve begins at the point (−2, 0), goes up and right, passes through the point (−1.5, 1), goes up and right, changes direction at the point (−1, 1.5), goes down and right, passes through the point (−0.5, 1), goes down and right, passes through the origin, goes down and right, passes through the point (0.5, −1), goes down and right, changes direction at the point (1, −1.5), goes up and right, passes through the point (1.5, −0.5), goes up and right, changes direction at the point (2, 0.5), goes down and right, crosses the x-axis at x = 2.5, goes down and right, changes direction at the point (3, −1), goes up and right, passes through the point (3.5, −0.5), goes up and right, and ends at the point (4, 0.5). Estimate 4 −2 g(x) dx with six subintervals using the following. (a) right endpoints 0 Correct: Your answer is correct. (b) left endpoints -0.5 Correct: Your answer is correct. (c) midpoints

The graph of a function g is shown.The x y-coordinate plane is given. A function labeled y = g(x) with 4 parts is graphed.The first part is a curve, enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 2.5, and ends at the open point (2, 3).The second part is a curve begins again at the open point (2, 1), goes up and right becoming less steep, and ends at the open point (5, 2).The third part is the closed approximate point (5, 1.2).The fourth part is a curve, begins at the open point (5, 2) goes down and right becoming more steep, and exits the window in the first quadrant.Use it to state the values (if they exist) of the following:(a)  lim x → 2− g(x)(b)  lim x → 2+ g(x)(c)  lim x → 2 g(x)(d)  lim x → 5− g(x)(e)  lim x → 5+ g(x)(f)  lim x → 5 g(x)SolutionLooking at the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right.Therefore (a) lim x → 2− g(x) =     and    (b) lim x → 2+ g(x) = .Since the left and right limits are different, we conclude that (c) the limit as x approaches 2 of g(x) does not exist.The graph also shows that (d) lim x → 5− g(x) =     and    (e) lim x → 5+ g(x) = .This time, the left and right limits are the same and so, by this theorem, we have (f) lim x → 5 g(x) = Despite this fact, notice that g(5) ≠ 2.