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 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.)

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.

Graphs of f and g are shown.There are two curves on the x y coordinate plane.A curve labeled f enters the top of second quadrant, goes down and right, crosses the negative x-axis and reaches a minimum in third quadrant, then goes up and right and intersects the origin, and reaches a maximum in first quadrant the same distance away from the x and y axes as its minimum was away from them. It then goes down and right, intersects the positive x-axis the same distance away from the origin as it intersected the negative x-axis, and exits the bottom of fourth quadrant, the same distance away from the x and y axes as was its entry point.A curve labeled g is always above the curve labeled f. g enters the top of second quadrant, goes down and right, reaches a minimum in second quadrant, goes up and right, then reaches a local maximum as it crosses the positive y-axis, and goes down and right. Then it reaches a minimum in first quadrant the same distance away from the x and y axes as was its first minimum. It then goes back up and exits the top of first quadrant the same distance away from the x and y axes as was its entry point.Is f even, odd, or neither?evenodd    neitherExplain your reasoning.It is symmetric about the origin.It is symmetric with respect to the y-axis.    It is symmetric with respect to the x-axis.It is not symmetric about the origin or the y-axis.Is g even, odd, or neither?evenodd    neitherExplain your reasoning.It is symmetric about the origin.It is symmetric with respect to the y-axis.    It is symmetric with respect to the x-axis.It is not symmetric about the origin or the y-axis.

The x y-coordinate plane is given. A function composed of several line segments is on the graph. The function begins at y = −1 on the negative y-axis, goes up and right in a linear fashion, crosses the x-axis at x = 2, sharply changes direction at (4, 1), goes down and right in a linear fashion, sharply changes direction at (6, 0), goes up and right in a linear fashion, sharply changes direction at (8, 2), goes horizontally right, sharply changes direction at (12, 2), goes down and right in a linear fashion, sharply changes direction at (14, 1), goes up and right in a linear fashion, and ends at (16, 3).

Use the graph of g to find the value of each expression. (If an answer does not exist, enter DNE.)The x y coordinate plane is given. The curve enters the window in the second quadrant, goes down and right to the closed point (0, −1). The curve starts again at the open point (0, −2), goes up and right to the open point (2, 2). There is a closed point at (2, 1). The curve starts again at the open point (2, 0), goes up and right, passes through the point (4, 3), and exits the window in the first quadrant.(a)lim x → 0− g(x)(b)lim x → 0+ g(x)(c)lim x → 0 g(x)(d)lim x → 2− g(x)(e)lim x → 2+ g(x)(f)lim x → 2 g(x)(g)g(2)(h)lim x → 4 g(x)