Find the value or values of c that satisfy the equation the function and interval. 18) f(x) = x² + 2x + 4, [2, 3] f(b) f(a) b-a = f'(c) in the conclusion of the Mean Value Theorem for 18)
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- What is the purpose of the Intermediate Value Theorem?Find the absolute extremas as well as all values of x where they occurson the specified domain. 4)f(x) =1/3x^3 - 2x^2 + 3x - 4; [-2, 5] Determine all values on the interval that hold for the mean value theorem 5) f(x) = x^3 + 12x + 36 on [-2,5] 1) A small frictionless cart, attached to the wall by a spring, is pulled 10 cm back from its rest position and releasedat time t = 0 to roll back and forth for 4 sec. Its position at time t iss = 1 - 10 cos pi t. What is the cart's maximum speed? When is the cart moving that fast? What is the magnitudeof of the acceleration then?a. Locate the critical points of ƒ.b. Use the First Derivative Test to locate the local maximum and minimumvalues.c. Identify the absolute maximum and minimum values of the functionon the given interval (when they exist). ƒ(x) = x - 2 tan-1 x on [-√3, √3]
- The function f(x) is continuous and differentiable on the interval [−2,6]. It is known that f(−2)=4 and the derivative on the interval satisfies the condition f′(x)≤3 for all x∈(−2,6). Determine an upper bound of the function at the right endpoint x=6.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 = 3a. Locate the critical points of ƒ.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the functionon the given interval (when they exist). ƒ(x) = -x2 - x + 2 on ⌊-4, 4⌋
- Find the critical numbers of f, if any, (a) find the open intervals on which the function is increasing or decreasing, (b) apply the First Derivative Test to identify all relative extrema, and (c) use a graphing utility to confirm your results. f(x) = cos x − sin x, (0, 2 )a. Locate the critical points of ƒ.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the functionon the given interval (when they exist). ƒ(x) = x2 + 3 on ⌈-3, 2⌉Consider a differentiable function f with domain R and derivativesf'(x)=-aebx(1+bx) and f"(x)=-abebx(2+bx) , with a and b nonzero real numbers.The function has only one critical point x=-1/b and a local maximum at x=-1/bUse the Second Derivative test to find the value(s) of a and b
- The function f(x) is continuous and differentiable on the interval [2,10]. It is known that f(2)=8 and the derivative on the given interval satisfies the condition f′(x)≤4 for all x∈(2,10). Determine the maximum possible value of the function at x=10.Deduction of the second derivative criterion for optimization of functions <br> of two variables using and as related to Taylora. Locate the critical points of ƒ.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the functionon the given interval (when they exist).