Let f be the function given by f () = on the closed interval [-7,7]. Of...
Determine whether the Mean Value Theorem can be applied to f on the closed interval (a, b). (Select all that apply.) f(x) = 16 - xl, [3, 7] Yes, the Mean Value Theorem can be applied. No, because f is not continuous on the closed interval [a, b]. No, because f is not differentiable in the open interval (a, b). None of the above. If the Mean Value Theorem can be applied, find all values of c in the open...
23. Let be a function defined and continuous on the closed interval (a,b). If f has a relative maximum at cand a<c<b, which of the following statements must be true? 1. f'(c) exists. II. If f'(c) exists, then f'(c)= 0. III. If f'(c) exists, then f"(c)<0. (A) II only (B) III only (C) I and II only (D) I and III only (E) II and III only
Determine whether the Mean Value Theorem can be applied to fon the closed interval (a, b). (Select all that apply.) RX) - 17 - xl. 14,8) Yes, the Mean Value Theorem can be applied. No, because fis not continuous on the closed interval [a, b]. No, because is not differentiable in the open interval (a, b). None of the above. (Ь) - Ka) ba If the Mean Value Theorem can be applied, find all values of c in the open...
Determine whether the Mean Value theorem can be applied to fon the closed interval [a, b]. (Select all that apply.) F(x) - 2 - X. [-7,2) Yes, the Mean Value Theorem can be applied. No, because fis not continuous on the dosed interval (a, b). No, because is not differentiable in the open interval (a, b). None of the above. of the Mean Value Theorem can be applied, find all values of e in the open interval () such that...
Let us verify the Mean Value Theorem with the function f(x) = VE on the interval (2,8). Solution. We have f is continuous on (2,8) f is differentiable on (2,8). f'(o) – f(8) – f(2) 8 - 2 We have f'(x) = The only value that satisfies the Mean Value Theorem is
Graph of A continuous function fis defined on the closed interval - 4sxs6. The graph of consists of a line segment and a curve that is tangent to the x-axis at x-3, as shown in the figure above. On th interval Dexc6, the function fis twice differentiable, with f(x)>0. Is there a value of a -4sach, for which the Mean Value Theorem applied to the interval (a 6), guarantees a value ca cx6, at which f'(c) = ? Justify your...
7. [-17 Points) DETAILS LARCALCET7 4.2.043.EP. Emal. Campbell, Outlook MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Consider the following function and closed interval FX)-3X, [1, 2] Is i continuous on the closed interval [1, 2]? Yes NO Iff is differentiable on the open interval (1, 2), find ). (If it is not differentiable on the open interval, enter ONE.) Find (1) and R2). (1) - R2) - Find for (a, b) - (1,21 b) - Determine whether the Mean Value...
Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = In(x), (1,91 Yes, it does not matter if is continuous or differentiable, every function satisfies the Mean Value Theorem. Yes, f is continuous on [1, 9] and differentiable on (1,9). No, f is not continuous on 1, 9). No, f is continuous on [1, 9] but not differentiable on (1,9). There is not enough information to verify if this function satisfies the Mean...
please explain in detail 4 -11 23 4 Graph of f Let f be a continuous function defined on the closed interval -1Sxs4. The graph of f, consisting of three line segments, is shown above. Let g be the function defined by g(x) = 5 +1.f(t) dt for-1 $154. (A) Find g(4). (B) On what intervals is gincreasing? Justify your answer. (C) On the closed interval 1 s xs 4, find the absolute minimum value of g and find the...
9. (-/7 Points] DETAILS LARCALCET7 4.2.048.EP. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Consider the following function and closed interval. MX) -10% (1,9) Isf continuous on the closed interval (1,9]? Yes No Iff is differentiable on the open interval (1,9), find f'(x). (if it is not differentiable on the open interval, enter DNE.) ) - Find (1) and (9) R1) - R9) - Find b) - Ra) for (a, b) - (1,9). b) - RO) Determine whether the Mean Value...