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)
Question
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)
Solution
(a)
Para encontrar este límite, observamos el comportamiento de la función cuando se aproxima a 0 desde la izquierda (valores negativos de ). Según la descripción, la curva entra en la ventana en el segundo cuadrante y se dirige hacia el punto cerrado . Por lo tanto, el valor del límite es:
(b)
Para este límite, observamos el comportamiento de la función cuando se aproxima a 0 desde la derecha (valores positivos de ). Según la descripción, la curva comienza de nuevo en el punto abierto y se dirige hacia arriba y a la derecha. Por lo tanto, el valor del límite es:
(c)
El límite existe si y solo si los límites laterales son iguales. En este caso, , por lo tanto:
(d)
Para encontrar este límite, observamos el comportamiento de la función cuando se aproxima a 2 desde la izquierda. Según la descripción, la curva se dirige hacia el punto abierto . Por lo tanto, el valor del límite es:
(e)
Para este límite, observamos el comportamiento de la función cuando se aproxima a 2 desde la derecha. Según la descripción, la curva comienza de nuevo en el punto abierto y se dirige hacia arriba y a la derecha. Por lo tanto, el valor del límite es:
(f)
El límite existe si y solo si los límites laterales son iguales. En este caso, , por lo tanto:
(g)
Para encontrar el valor de , observamos el valor de la función en . Según la descripción, hay un punto cerrado en . Por lo tanto:
(h)
Para encontrar este límite, observamos el comportamiento de la función cuando se aproxima a 4. Según la descripción, la curva pasa por el punto y sale de la ventana en el primer cuadrante. Por lo tanto, el valor del límite es:
Similar Questions
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 graphs of f and g are given. Use them to evaluate each limit, if it exists. (If an answer does not exist, enter DNE.)The x y coordinate plane is given. There is one curve and one point on the graph. The point occurs at (−1, 3). The curve enters the window at (−2, −1), goes up and right, passes through the open point (−1, 1), changes direction at (0, 2), goes down and right, and passes through the open point (2, −1). At this point, the curve turns sharply forming a cusp, goes up and right, and exits the window at (4, 2). The x y coordinate plane is given. There are two curves on the graph. The first curve enters the window at (−2, 0), goes up and right, and ends at the closed point (0, 3). The second curve begins at the open point (0, 1), goes up and right, and passes through (2, 2). At this point, the curve turns sharply forming a cusp, goes down and right traveling in a straight line, and exits the window at (4, −2).(a)lim x→2 [f(x) + g(x)](b)lim x→0 [f(x) − g(x)](c)lim x→−1 [f(x)g(x)](d)lim x→3 f(x)g(x)(e)lim x→2 [x2f(x)](f)f(−1) + lim x→−1 g(x)
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 graph of g consists of two straight lines and a semicircle as shown in the figure.There are two line segments and a semicircle connected on the x y-coordinate plane. The graph is labeled y = g(x). The first line segment starts at the point (0, 20), goes down and right, and ends on the x-axis at the point (10, 0). The semicircle starts at the point (10, 0), passes through (20, −10), and ends at the point (30, 0). The second line segment starts at the point (30, 0), goes up and right, and ends at the point (35, 5).Evaluate each integral by interpreting it in terms of areas.(a)10g(x) dx0100−50π (b)30g(x) dx1050π−50 (c)35g(x) dx0
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