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BONUS Week 2 Homework
Due Jan 30 at 11:59pm Points 8 Questions 8 Available Jan 23 at 8am - Feb 2 at 11:59pm Time Limit None
Instructions
This quiz was locked Feb 2 at 11:59pm.
Attempt History
Attempt Time Score LATEST Attempt 1 78 minutes 6 out of 8
Score for this quiz: 6 out of 8 Submitted Jan 30 at 11:51pm This attempt took 78 minutes. Correct answer
Question 1 1 / 1 pts
a. 2 x b.
1 2 ℓn (^2 x^ −^3 ) c. 2 / ( 2 x − 3 ) This follows by the chain rule,
f ′^ ( x ) = [ ℓn ( 2 x − 3 )]′^ =
( 2 x − 3 )′ 2 x − 3 =^
2 2 x − 2 d. x / 2
(c). This follows by the chain rule,
f ′^ ( x ) = [ ℓn ( 2 x − 3 )]′^ =
( 2 x − 3 )′ 2 x − 3 =^
2 2 x − 3
Correct answer
Please answer all the questions below.
(Lesson 2.1: Derivatives.) BONUS: If f ( x ) = ℓn ( 2 x − 3 ), find the derivative f ′^ ( x ).
Question 2 1 / 1 pts
a. cos( 1 / x^2 ) b. sin( 1 / x^2 ) c. −^
1 x^2 sin(^1 /^ x ) d.
1 x^2 sin(^1 /^ x ) By the chain rule,
[cos( 1 / x )]′^ = − sin( 1 / x )[ 1 / x ]′^ =
1 x^2 sin(^1 /^ x )
(d). By the chain rule,
[cos( 1 / x )]′^ = − sin( 1 / x )[ 1 / x ]′^ =
1 x^2 sin(^1 /^ x ) Correct answer
Question 3 1 / 1 pts
a. x = 0 b. x = 1 c. x = ℓn ( 2 ) = 0. d. x = 12 ℓn ( 2 ) = 0.
This doesn't take too much work. Namely, set 0 = f ( x ) = e^4 x^ − 4 e^2 x^ + 4 = ( e^2 x^ − 2 )^2.
This is the same as e^2 x^ = 2 , or x =
1 2 ℓn (^2 )^ =^ 0.
(d). This doesn't take too much work. Namely, set 0 = f ( x ) = e^4 x^ − 4 e^2 x^ + 4 = ( e^2 x^ − 2 )^2.
This is the same as e^2 x^ = 2 , or x =
1 2 ℓn (^2 )^ =^ 0.
Correct answer
Question 4 1 / 1 pts
(Lesson 2.1: Derivatives.) BONUS: If f ( x )^ =^ cos(^1 /^ x ), find the derivative f ′( x ).
(Lesson 2.2: Finding Zeroes.) BONUS: Suppose that f ( x ) = e^4 x^ − 4 e^2 x^ + 4. Use any method you want to find a zero of f ( x ), i.e., x such that f ( x ) = 0.
If we let f ( x ) = sin( x ) − x and g ( x ) = x , then lim x → 0 f ( x ) = 0 and lim x → 0 g ( x ) = 0 , so that lim x → 0
f ( x ) g ( x )
seems to get us into a 0/0 issue. Thus, we'll need to employ L'Hôspital's Rule:
lim x → 0
f ( x ) g ( x ) =^ x lim → 0
f ′^ ( x ) g ′^ ( x ) =^ x lim → 0
cos ( x ) − 1 1 =^
0 1 =^ 0. c. ∞ d. undetermined
(b). If we let f ( x ) = sin( x ) − x and g ( x ) = x , then lim x → 0 f ( x ) = 0 and lim x → 0 g ( x ) = 0 , so that lim x → 0
f ( x ) g ( x )
seems to get us into a 0/0 issue. Thus, we'll need to employ L'Hôspital's Rule:
lim x → 0
f ( x ) g ( x ) =^ x lim → 0
f ′^ ( x ) g ′^ ( x ) =^ x lim → 0
cos ( x ) − 1 1 =^
0 1 =^ 0.
Wrong answer
Question 7 0 / 1 pts
a. - b. 1/ c. 3/ d. 3
(c). We have
∫^20 ( x − 1 )^2 dx ≈
b − a n
n ∑ i = 1
f ( a +
( b − a ) i n )^ =^
2 4
4 ∑ i = 1
2 i 4 −^1 )
Well, this is sort of close to the true integral of 2/3. Of course, we could've done even better if n had been bigger or if we had used the midpoint of each interval instead of the right endpoint.
(Lesson 2.4: Numerical Integration.) BONUS: Find the approximate value of the integral
∫^20 ( x − 1 )^2 dx (^) using the lesson's form of the Riemann sum with f ( x ) = ( x − 1 )^2 , a = 0 , b = 2 , and n = 4.
Unanswered
Question 8 0 / 1 pts
a. Normal b. Unif(0,1) c. Exponential d. Weibull
(a). This is the Box-Muller method to generate normal random variables. We'll learn much more about this later on. Here's some example Matlab code that works well...
%Matlab code clear all;close all;clc; z_vec=zeros(1000,1); for i = 1: u = rand; v = rand; z_vec(i)= sqrt(-2log(u))cos(2piv); end nbins = 30; bins = linspace(-5,5,nbins); histogram(z_vec,bins)
Of course, you can do this easily in R, Excel, Python etc.
Quiz Score: 6 out of 8
(Lesson 2.6: Simulating Random Variables.) BONUS: Suppose U and V are independent Uniform(0,1) random variables. (You can simulate these using the RAND() function in Excel, for instance.) Consider the nasty-looking random variable
Z = (^) √− 2 ℓn ( U )cos( 2 πV ),
where the cosine calculation is carried out in radians (not degrees). Go ahead and calculate Z... don't be afraid. Now, repeat this task 1000 times (easy to do in Excel) and make a histogram of the 1000 Z 's. What distribution does this look like?
