# User talk:AiliShao

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## Contents

Thread title | Replies | Last modified |
---|---|---|

Solution | 4 | 17:28, 7 March 2018 |

MATH110/April 2017/Question 11 (a) | 2 | 00:46, 5 March 2018 |

MATH110/April 2017/Question 11 (b) | 3 | 00:18, 5 March 2018 |

MATH110/April 2017/Question 11 (c) | 0 | 22:19, 4 March 2018 |

MATH110/April 2017/Question 04 (b) (ii) | 0 | 21:46, 4 March 2018 |

MATH110/April 2017/Question 04 (a) (iv) | 0 | 21:40, 4 March 2018 |

Math 152 2015 Questions A13 - A15 | 1 | 07:03, 3 March 2018 |

MATH105/April 2017/Question 01 (j) | 3 | 22:14, 13 February 2018 |

Math 152 2017 Question A13 | 1 | 22:05, 20 January 2018 |

Dear Aili,

For the last argument in this question, I believe the counterexample you wrote is not correct, your example is still symmetric. Would you mind correcting it?

Best,

Han

Hi Aili,

I think what you used is not the mean value theorem, but the intermediate value theorem. When you use it, it would be great if you mention that the function f' satisfies the assumption of the theorem. i.e., the continuity of the function. Also, in the definition of the critical value c, we also have one more condition that f' changes its sign around c. So, please add explanation on this part, too. Thanks!

Umm, I have made some changes to my solutions. I am not sure about where I should put "in the definition of the critical value c, we also have one more condition that f' changes its sign around c"... but I did verify the assumptions of IVT.

Sorry, Aili. I don't think we need to add that part. I was confused about the definition. What you did is enough.

Thanks for revising the solution!

Again, since MATH 110 students has slightly week background on pre-calculus, it would be helpful for students if you explain the range of on (0,\frac{\pi}{2}).

Also, when you put fraction inside of the bracket, please use \left( and \right).

Thanks!

p.s., please change the flags to R after you revised!

It would be great, if you recall from part (a) that c lies on (0,pi/2) :)

I made some changes... but not sure about whether I need to further explain why \sin x is positive on this interval.

Thanks, Aili. No you don't need to explain further, because they learn about trigonometric function in the class.

Hi Aili,

Can you add the explanation on the relation between extreme values and f'(x)=0?

Also, I guess some students have difficulty in plotting and accurately in the exam. So, they might not recognize that it is only intersect at once.

How about adding alternative way to show that there's only one x satisfying ? (I guess we can use everywhere. )

Hi Aili,

It would be great if you explain why the limit goes to infinity. (you can explain the denominator goes... and the numerator goes.... , so ....) As I mentioned before, we should provide more explanatory solutions for MATH110 students.

Also, please explain why you consider only. (In other words, why the other points are not the candidate for vertical asymptotes. )

It would be helpful if you recall the definition again.

Thanks,

Hi Aili,

As far as I know, we should consider both, to find horizontal asymptotes. If you agree, please revise your solution. Also, it would be great if you recall the definition of the horizontal asymptotes.

Thanks,

Hi Aili,

Thanks for your work so far. Please see my corrections to your solutions to the problems above, and make sure you understand them.

In particular, note that there is a voltage drop across the current source. I also suggest reading this page on Kirchoff's laws.

Hi Aili,

If I understand your solution correctly, your solution says that -\infty+ \infty =0, which is not true. If you agree with this, can you correct the solution? Thanks!

Best, Hyunju

I think that "reduced row echelon form" means that the first "1" in each row has to come after the first "1" in the previous row, so the matrix appearing in the hint of that problem is not in reduced row echelon form.

In fact, any invertible matrix has the identity as its reduced row echelon form: a standard algorithm to invert an invertible matrix is to augment the matrix by the identity and write the resulting matrix in reduced row echelon form; the left half of the matrix will then be the identity matrix.

If you think I've made a mistake, please let me know.