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c1-t2m-a

# c1-t2m-a - NAME Brief Answers TEST2M/MAC2311 Page 1 of 5...

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NAME: Brief Answers TEST2M/MAC2311 Page 1 of 5 Read Me First: Show all essential work very neatly. Use correct notation when presenting your computations and arguments. Write using complete sentences. Be careful. Remember this: "=" denotes "equals" , " " denotes "implies" , and " " denotes "is equivalent to". Do not "box" your answers. Communicate. Show me all the magic on the page. 1. (20 pts.) Compute the derivatives of the following functions. You may use any of the rules of differentiation that are at your disposal. Do not attempt to simplify the algebra in your answers. (a) f(x) = 3x 5 - 7x -6 + 4tan(x) f (x) = 15x 4 + 42x -7 + 4sec 2 (x) (b) g(x) = (4x -3 - x)cos(x) g (x) = (-12x -4 - 1)cos(x) - (4x -3 - x)sin(x) 2sec(t) (c) h(t) = 3t 5 - cot(t) 2sec(t)tan(t)(3t 5 - cot(t)) - (2sec(t)(15t 4 + csc 2 (t)) h (t) = (3t 5 - cot(t)) 2 (d) L(x) = 2 tan 2 (3x 3 ) L (x) = 4 tan(3x 3 ) sec 2 (3x 3 ) 9x 2 (e) y = sin(3x) + 5cos(10x) + sec( π /4) dy = 3cos(3x) - 50sin(10x) dx Note: sec( π /4) is a constant.

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TEST2M/MAC2311 Page 2 of 5 2. (10 pts.) (a) Using complete sentences and appropriate notation, state the Intermediate Value Theorem. If f is continuous on a closed interval [a,b], and k is any number between f(a) and f(b), inclusive, then there is a number x 0 in the interval [a,b] with f(x 0 ) = k. (b) Can the Intermediate Value Theorem be used to show the equation (x - 5)/(x 2 - 4x - 5) = 0 has a solution in the interval [-2,6]? Explain completely. [Hint: Let f(x) = (x - 5)/(x 2 - 4x - 5) on [-2,6]. Discuss appropriate properties of f and then draw a suitable conclusion.] Let f(x) = (x - 5)/(x 2 - 4x - 5) on [-2,6]. Observe that f(-2) = -1, f(6) = 1/7, and k = 0 is a number between f(-2) and f(6). Unfortunately, f is not continuous on the interval [-2,6].
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