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Using the Intermediate Value Theorem In Exercises 99–102, verify that the Intermediate Value Theorem applies to the indicated interval and find the value of c guaranteed by the theorem.
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- In Exercises 83–86, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false. If the graph of a function has three x-intercepts, then it musthave at least two points at which its tangent line is horizontalarrow_forwardIn Exercises 11–18, use the function f defined and graphed below toanswer the questions. (a) Does f (-1) exist?arrow_forwardSuppose f and g are the piecewise-defined functions defined here. For each combination of functions in Exercises 51–56, (a) find its values at x = -1, x = 0, x = 1, x = 2, and x = 3, (b) sketch its graph, and (c) write the combination as a piecewise-defined function. f(x) = { (2x + 1, ifx 0 g(x) = { -x, if x 2 8(4): 51. (f+g)(x) 52. 3f(x) 53. (gof)(x) 56. g(3x) 54. f(x) – 1 55. f(x – 1)arrow_forward
- a) Find the domain of f, g, f + g, f – & fg, ff, f/ g b) Find (f + g)(x), (f – g)(x), (fg)(x), (ff)(x), For each pair of functions in Exercises 17–32: 15. (8 and g/f. Find f+ g)(x), (f – g)(x), (fg)(x), (ff)(x), (f/8)(x), and (g/f)(x). 17. f(x) = 2x + 3, g(x) = 3 – 5x %3D 18. f(x) = -x + 1, g(x) = 4x – 2 19. f(x) = x – 3, g(x) = Vx + 4 20. f(x) = x + 2, g(x) = Vx – 1 21. f(x) = 2x – 1, g(x) = – 2x² 22. f(x) = x² – 1, g(x) = 2x + 5 23. f(x) = Vx – 3, g(x) : = Vx + 3arrow_forwardIn Exercises 13-14, find the domain of each function. 13. f(x) 3 (х +2)(х — 2) 14. g(x) (х + 2)(х — 2) In Exercises 15–22, let f(x) = x? – 3x + 8 and g(x) = -2x – 5.arrow_forwardIn Exercises 83–85, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Per-form the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where ƒ′ = 0. (In some exercises, you may have to use the numerical equation solver to ap-proximate a solution.) You may want to plot ƒ′ as well. c. Find the interior points where ƒ′ does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function’s absolute extreme values on the interval and identify where they occur. 83. ƒ(x) = x4 - 8x2 + 4x + 2, [-20/25, 64/25] 84. ƒ(x) = -x4 + 4x3 - 4x + 1, [-3/4, 3] 85. ƒ(x) = x^(2/3)(3 - x), [-2, 2]arrow_forward
- In Exercises 17–20, the linear function. use the limit definition to calculate the derivative ofarrow_forwardIn Exercises 139–142, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. x2 – 25 = x - 5 5 139. X - x? + 7 140. = x? + 1 7 7 domain of f(x) = is x(x – 3) + 5(x - 3) 141. The (-0, 3) U (3, 0). 142. The restrictions on the values of x when performing the division f(x) h(x) g(x) k (x) are g(x) + 0, k(x) # 0, and h(x) + 0.arrow_forwardIn Exercises 39–44, each function f(x) changes value when x changes from x, to xo + dx. Find a. the change Af = f(xo + dx) – f(xo); b. the value of the estimate df = f'(xo) dx; and c. the approximation error |Af – df|. y = f(x)/ Af = f(xo + dx) – f(x) df = f'(xo) dx (xo, F(xo)) dx Tangent 0| xo + dx 39. f(x) 3D х? + 2x, хо —D 1, 40. f(x) = 2x² + 4x – 3, xo = -1, dx = 0.1 41. f(x) = x³ - x, xo = 1, dx = 0.1 dx = 0.1 %3D 42. f(x) 3 х, Хо —D 1, dx %3D 0.1 43. f(x) — х 1, Хо —D 0.5, dx %3D0.1 44. f(x) 3D х3 — 2х + 3, Хо — 2, dx 3D 0.1arrow_forward
- In Exercises 51–54, graph the function ƒ to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at x = 0. If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function’s value(s) should be?arrow_forwardIn Exercises 104–105, express the given function h as a composition of two functions f and g so that h(x) = (f• g)(x). 104. h(x) = (x² + 2x – 1)* 105. h(x) = V7x + 4 %3! %3!arrow_forward
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