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Problem Points Score 1 16 2 16 3ab 18 3cd 18 4 18 5 14 ∑ 100 M161, Test 1, Fall 2007 Name: Section: Instructor: You may not use calculators on this exam cos2 θ = 1+cos 2θ 2 sin2 θ = 1−cos 2θ 2 Theorem (The Derivative Rule for Inverses) If f has an interval I as domain and f 0 (x) exists and is never zero on I, then f −1 is differentiable at every point in its domain. The value of ( f −1 )0 at a point b in the domain of f −1 is the reciprocal of the value of f 0 at the point a = f −1 (b): 1 . ( f −1 )0 (b) = 0 −1 f ( f (b)) 1) Let f (x) = xex for any nonnegative real number x ≥ 0. a) What is the range of f ? b) Explain why f has an inverse. (If you refer to a criterion or test, indicate why the test applies!) c) Find a value a such that f (a) = e. d) Let g(x) = f −1 (x) the inverse of f . Determine g0 (e) = ( f −1 )0 (e). 2) 0 a) Place ln(10) approximately on this line: (remember that e ≈ 2.71828) 1 2 3 4 b) Calculate the derivative 5 6 d ln(x4 · ex ). dx c) Evaluate sec(asin(5/7)). 7 8 3) Z Evaluate the following integrals. Show your work. ex a) dx 1 + e2x Z b) 2x dx Z sin(x) dx cos(x) Z 6 p dx 4 − (x + 1)2 c) d) 4) The table below gives three differential equations. For each equation, indicate which of the four possible general solutions I-IV (i.e. one is not used) is a general solution for this equation. (You do not need to solve the equations, just to verify which of the functions is a solution!) Furthermore, we list for each equation one initial value setting. Find the correct value for C in the respective general solution that will solve this given initial value problem. dy + 3y = ex dx dy + ex = x2 dx dy + 3y = e−x dx y(0) = 0 y(0) = 2 y(0) = 3/2 General solution Initial Value Problem Value of C to solve the IVP Possible General Solutions: I II III IV y(x) = 1/2 e2 x +C e−3 x y (x) = 1/4 ex +Ce−3 x y (x) = −1/2 x2 + ex +C y (x) = −ex + 1/3 x3 +C Hint: The correct values for C are to be chosen from the set {±1/4, 0, ±1, ±2, ±3}, no two of the initial value problems have the same C-value. 5) A radioactive substance has a half-life of 100 years (that is, after 100 years of decay, only half the amount of the substance remains). How long will it take until only 10% of the initial amount remains? (As you don’t have a calculator the result can be an expression and does not need to be an explicit number.)