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Warm up Problems Assume for a continuous and differentiable function, f (-1) = -2 and f (3) = 6. Answer the following true or false, then state the theorem that was used: 1) f (c) = 1 for some c on the interval 2) f (c) = 0 for some c on the interval 3) -2 ≤ f (c) ≤ 6 for all c on the interval 4) f (c) = 2 for some c on the interval
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Review of Chapter 5
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Strategy for Related Rates “PGWEDA” P) Draw a picture G) Identify given information, including rates (derivatives) W) Identify what you want to find E) Find an equation to relate the variables D) Take the derivative with respect to time A) Plug in values to get your answer “Changes at a rate of …” Involves time
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Ex. Gravel is dumped from a conveyor belt at a rate of 30 ft 3 /min, and the pile formed is a cone whose base diameter and height are always equal. How fast is the height increasing when the pile is 10 ft high?
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Ex.
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Ex. Let for x ≠ 0. Find and classify all critical points. Find all inflection points. Find the global max/min values on [1,100].
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Strategy for Optimization 1) Draw a picture, if appropriate 2) Write down given information, including an equation 3) Find the function to be optimized 4) Substitute to get one variable 5) Take the derivative 6) Set equal to zero and solve Find the largest, smallest, closest Maximize, minimize
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Calculators Allowed 1.B2. B3. B4. B5. A 6. B7. 375,000 ft. 2 8. a. b. 32.725 cm 3 c. -5.890 cm 3 /hr. d. constant = -0.3 No Calculators 1. D2. C3. E4. A5. E 6. a. a(t) = ln tb. t > ec. -1
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7. a. x = -2 because f goes from pos. to neg. b. x = 4 because f goes from neg. to pos. c. -1 < x < 1 and 3 < x < 5 because f is incr. d. 8. a. horiz. tan at and, both local min. because f goes from neg. to pos. b. concave up for all x 0 because f > 0 c. below, because graph is concave up
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9. a. x = 0 and x = 3 because slope of f changes signs b. Abs. min at x = 4 because f decreases a lot until x = 4, then there is only increase. Abs. max at x = -1 because the only increase after x = 4 is not enough to cancel out all the decrease c. y = 4x + 4
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