Destiny of Pi Posted March 4, 2018 Report Share Posted March 4, 2018 Hey guys, I was sick and missed some math classes. Here are two questions that I can partially solve them but not for the second part. It would be great if some of you could help me with the steps and process. For this question, I got that when x=0.4, y=1.573, but I don't know how to justify whether the approximation if greater or smaller than the actual value. For this question, I found that the Maclaurin series is ln(2) +1/2x+1/8x^2. I wonder how could we use this to solve part b. Reply Link to post Share on other sites More sharing options...
kw0573 Posted March 4, 2018 Report Share Posted March 4, 2018 1b. Because slope increases with x and y, and you are always underestimating the derivative, you would be underestimating the value at x = 0.4 2b Limits with indeterminate form (0/0, infinity/infinity etc) can be evaluated by substituting expansions at the desired x value for both numerator and denominator. For example, limit of sin x / x as x approaches 0 can be evaluated as i) sin x expand at x = 0, is x - x^3 / 6 + ..., x is just x ii) (x - x^3 /6 + ...) / x = x / x - x^3/6 / x + ... = 1 - x^3/6 /x + ..., which when evaluated at x = 0, is just 1. iii) hence for sin x / x at x -->0 the limit is 1. You can do a similar analysis for the desired expression. Reply Link to post Share on other sites More sharing options...
Destiny of Pi Posted March 4, 2018 Author Report Share Posted March 4, 2018 Thanks, I have another question. For this question part b, I can use a graphing calculator. But I don't know how to find the horizontal asymptote to the function. The function is f(x)= x^(1/x) Reply Link to post Share on other sites More sharing options...
kw0573 Posted March 5, 2018 Report Share Posted March 5, 2018 (edited) Horizontal asymptotes only at x --> ± infinity. Because f(x) is not real in x <0, x --> + infinity gives asymptote at f(x) = 1. You should get f(x) = 1 as horizontal asymptote from graphing calculator. An alternative (more rigorous) way is to transform the function. define g(x) = ln (f(x)) = 1/x * ln x We know that the limit as g(x) of x approaches infinity is 0 f(x) = e^(g(x)), same limit would evaluate to e^0 = 1 Edited March 5, 2018 by kw0573 Reply Link to post Share on other sites More sharing options...
Destiny of Pi Posted March 6, 2018 Author Report Share Posted March 6, 2018 Thanks! I just have one more question For this question, I got the slopes are all zero for part a and b. For part c, isn't the slope just y=3? since it passes through (0,3)? How is it 8 marks? Reply Link to post Share on other sites More sharing options...
kw0573 Posted March 6, 2018 Report Share Posted March 6, 2018 x and y with 5 values each create 25 (x,y) pairs. So for example the slope at (0,3) is -3. Reply Link to post Share on other sites More sharing options...
Destiny of Pi Posted March 6, 2018 Author Report Share Posted March 6, 2018 3 minutes ago, kw0573 said: x and y with 5 values each create 25 (x,y) pairs. So for example the slope at (0,3) is -3. sorry, my bad Then how do you do part c? Reply Link to post Share on other sites More sharing options...
kw0573 Posted March 6, 2018 Report Share Posted March 6, 2018 (edited) You would use the method of integrating factor (which should be in your textbook). You can also find it online. It's just an application of both e^x derivatives and product rule. Edited March 6, 2018 by kw0573 not chain rule Reply Link to post Share on other sites More sharing options...
Destiny of Pi Posted March 6, 2018 Author Report Share Posted March 6, 2018 ok thanks!!! Reply Link to post Share on other sites More sharing options...
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