Question The graph of a function f is shown.The x y-coordinate plane is given. The curve begins at y = 1 on the positive y-axis, appears to go horizontally right, passes through the approximate point (3, 1), goes up and right becoming more steep, and ends at the point (5, 3) nearly vertical.Does f satisfy the hypotheses of the Mean Value Theorem on the interval [0, 5]?Yes, because f is continuous on the open interval (0, 5) and differentiable on the closed interval [0, 5].Yes, because f is increasing on closed interval [0, 5].    Yes, because f is continuous on the closed interval [0, 5] and differentiable on the open interval (0, 5).No, because f is not differentiable on the open interval (0, 5).No, because f does not have a minimum nor a maximum on the closed interval [0, 5].No, because f is not continuous on the open interval (0, 5).If so, find a value c that satisfies the conclusion of the Mean Value Theorem on that interval. (If an answer does not exist, enter DNE.)c = 3    0.1

ZFKBVS The Asker · Calculus

The graph of a function f is shown.

The x y-coordinate plane is given. The curve begins at y = 1 on the positive y-axis, appears to go horizontally right, passes through the approximate point (3, 1), goes up and right becoming more steep, and ends at the point (5, 3) nearly vertical.

Does f satisfy the hypotheses of the Mean Value Theorem on the interval [0, 5]?

Yes, because f is continuous on the open interval (0, 5) and differentiable on the closed interval [0, 5].Yes, because f is increasing on closed interval [0, 5].    Yes, because f is continuous on the closed interval [0, 5] and differentiable on the open interval (0, 5).No, because f is not differentiable on the open interval (0, 5).No, because f does not have a minimum nor a maximum on the closed interval [0, 5].No, because f is not continuous on the open interval (0, 5).

If so, find a value c that satisfies the conclusion of the Mean Value Theorem on that interval. (If an answer does not exist, enter DNE.)

c = 

3