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- Determine whether each of the following statements is true or false, and explain why. The chain rule is used to take the derivative of a product of functions.3- Show that this function is differentiable for all and find its derivative f2) = f(x+ jy) = x* – 15x*y + 15x*y* - y6 + j(6x®y- 20x*y + 6xy")Let f be the continuous function defined on [−1,8] whose graph, consisting of two line segments, is shown above. Let g and h be the functions defined by g(x)=sqrt(x^2−x+3) and h(x)=5e^x−9sinx. (a) The function k is defined by k(x)=f(x)g(x). Find k′(0). (b) The function m is defined by m(x)=f(x)/2g(x). Find m′(5). (c) Find the value of x for −1<x<2 such that f′(x)=h′(x).
- Let y = f (x) be a function with domain all real numbers. It is known thatf ′(x) = e^x(x −2)^3(x + 3)^7Find and classify the critial numbers of y = f (x). (i.e determine if they are relative max, relative min, orneither.let f(x), g(x) be two contiously differentiable functions satistfying the relationships f'(x)=g(x) and f''(x)=-f(x). Let h(x)=f^2(x)+g^2(x). If h(0)=5, find h(10).Consider the following function and closed interval. f(x) = √6 - Xx, [-19, 6] Is f continuous on the closed interval [-19, 6]? Yes No If f is differentiable on the open interval (-19, 6), find f'(x). (If it is not differentiable on the open interval, enter DNE.) f'(x) = Find f(-19) and f(6). f(-19) = f(6) Find = f(b) f(a) b-a f(b) f(a) b - - a C = = for [a, b] = [-19, 6]. Determine whether the Mean Value Theorem can be applied to f on the closed interval [a, b]. (Select all that apply.) 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 interval (a, b) such that f'(c) list. If the Mean Value Theorem cannot be applied, enter NA.) f(b) – f(a). If the Mean Value Theorem cannot be applied, explain why not. (Enter your answers as a comma-separated b-a
- Prob. 6 (a) (10 point) Let f(x) = 2x² – 3. Find ƒ'(−2) using only the limit definition of derivatives. (b) (10 p.) If ƒ(x) = √√x + 6, find the derivative f'(c) at an arbitrary point c using only the limit definition of derivatives.Find the derivative of the function f(x) = 6x² – 18z by the definition of derivatives f(x+h) – f(z) h f'(z) = lim O 12x + 18 O 6z O 12z 18 - O 12z O 6x + 18Let f(x) be a function that is differentiable for all x. Let g(x) be defined by g(x) = f(x) + f(3 – x). Show that g'(x) has a root in the interval (0,3). (a) (b) Let f(x) be continuous and differentiable on the interval [-8,0]. Suppose f(-8) = -2 and f'(x) < 3 for all x. What is the largest possible value for f (0)? Justify your answer.
- f:[0,1] -> [0,1] such that f(x)=x^2 has two fixed points. Is this true?Let g(x) = x/x^2+75 ..g'(x) = Ax2 + B (x2 + 75)C , where A = B = C = g''(x) = Dx(x2 + E) (x2 + F)G , whereD = E = F = G = Find the interval(s) on which the graph of the function is concave upward and those on which it is concave downward. Enter the values for the endpoint(s) in the appropriate blanks and DNE in any empty blanks.The graph of the function is concave upward on the interval(s): (−∞, ∞) (−∞, a) (−∞, a] (a, ∞) [a, ∞) (−∞, a) ∪ (b, ∞) (−∞, a] ∪ [b, ∞) (−∞, a) ∪ (b, c) (a, b) ∪ (c, ∞) (a, b) [a, b] None of the above. a = b = c = The graph of the function is concave downward on the interval(s): (−∞, ∞) (−∞, d) (−∞, d] (d, ∞) [d, ∞) (−∞, d) ∪ (f, ∞) (−∞, d] ∪ [f, ∞) (−∞, d) ∪ (f, h) (d, f) ∪ (h, ∞) (d, f) [d, f] None of the above. d = f = h =