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Assignment 3 Solutions.pdf

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Department
Calculus
Course
Calculus 1000A/B
Professor
Chris Brandl
Semester
Fall

Description
MATH 127 Fall 2012 Assignment 3 Solutions 1. Solve the following equations for x. q ▯x 2x 5 x (a) e 3e ▯ 4 = e r 3e2x▯ 5 = e2x 4 5 3e2x▯ = e4x 4 5 e4x▯ 3e2x+ = 0 4 2x 2 2x 5 (e ) ▯ 3e + = 0 4 Notice that this is like a quadratic with respect tope . Using the quadratic 3 ▯ 9 ▯ 5 3 ▯ 2 5 1 formula we get that the roots are given by ex = = = ; . 2 2 2 2 2x 5 First consider e = 2. 5 e2x= 2 ▯ ▯ 5 2x = ln 2 ▯5▯ x = ln 2 2 ▯1▯ Now consider e2x = 1 . Solving for x we get x =n 2 . 2 2 ▯5▯ ▯ 1 Thus, the two solutions to the equation are x =n 2 and x = ln 2 . 2 2 (b) ln(2ln(x)) = 3 ln(2ln(x)) = 3 3 2ln(x) = e 3 ln(x) = e 2 e x = e 2 1 (c) ln(x ▯ 2e) + ln(x ▯ 3e) = 2 We ▯rst simplify the left hand side using properties of logarithms. ln(x ▯ 2e) + ln(x ▯ 3e) = ln((x ▯ 2e)(x ▯ 3e)) 2 2 = ln(x ▯ 5ex + 6e ) Thus, we have that ln(x ▯ 5ex + 6e ) = 2. From this we get x ▯5ex+6e = e 2 2 2 2 2 which in turp gives x ▯ 5ex + 5e = 0. Tpe quadratic formula gives that 2 2 5e ▯ 25e ▯ 20e (5 ▯ 5)e x = = . Now since we need x > 3e for 2 2 p (5 + 5)e ln(x ▯ 2e) + ln(x ▯ 3e) to be de▯ned we have the single solution x = . 2 2. Let f(x) = ▯e ▯x + 2. (a) Sketch the graph of f, labelling any asymptotes and x and y-intercepts. 3 2.5 2 y=2 1.5 1 (0,1) 0.5 (-ln(2),0) -4 -3.5-3-2.5-2-1.5-1-0.50 0.5 1 1.5 2 2.5 3 3.54 -0.5 -1 -1.5 -2 -2.5 -3 ▯1 (b) Explain why the inverse exists based on the graph. Find f . We can see that the graph is strictly increasing and so, passes the horizontal line test. That is, we see that the function is one-to-one and from this we get that the 2 inverse exists. Solving for the inverse we get y = ▯e ▯x + 2 ▯x y ▯ 2 = ▯e ▯x 2 ▯ y = e ln(2 ▯ y) = ▯x x = ▯ln(2 ▯ y) ▯1 f (x) = ▯ln(2 ▯ x) (c) Sketch f ▯1 and label the x and y-intercepts on the same axes used in part a. 3 2.5 y=2 2 1.5 1 (0,1) 0.5 (-ln(2),0) -4 -3.5-3 -2.5-2-1.5-1 -0.50 0.5 1 1.5 2 2.5 3 3.5 4 -0.5 (1,0) x=2 (0,-ln(2)) -1 -1.5 -2 -2.5 -3 3. For each of the following functions, state the domain of the function, ▯nd its inverse, and state the domain of the inverse. p x (a) f(x) = 2 ▯ e
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