I just got back my test, I am probably going to fail this class. I got a 20/60. Anyways it might be worth my time to continue and to review the test to see what I did wrong.

  1. In terms of definitions, what does it mean for f not to be differnetiable at a number in its domain, a? I put "It means that there is no derivative at f(a) either because it is not defined or it approaches infinity.

That was, like most of the test, wrong. What about that is incorrect? I am pretty stupid so it is entirely possible I don't know what a lot of those words actually mean.

3) Show that the graph of $f(x) = x^3 + 5x^2 + 9x + 17$ has no horizontal tangent line, by working with the derivative.

I have never seen a problem like this before and I have no idea how to solve it, I stated that the derivative of 17 is zero so that would be a horizontal tangent line, of course that is wrong. I don't need the answer to this, just how to solve it. I know the answer, just not the why.

6) $\displaystyle\lim\limits_{t\to0} \frac{\sin{2t}}{\tan{6t}}$

This was actually a homework problem I did a couple minuted before the test and somehow couldn't figure out on the test. Not sure why.

I guess that is all the questions I have, I got pretty much rest of the test wrong but I think I know what to do on those problems.

  • 5
    $\begingroup$ I don't mean to sound condescending, but all the self-deprecation in your posts (not just this one) is unnecessary at best and harmful at worst. Deciding that you are "pretty stupid" is no way to succeed at anything. $\endgroup$ – Austin Mohr Oct 5 '11 at 22:58
  • $\begingroup$ Well smart people don't study for a test and then get the fifth lowest score out of 40 people do they? $\endgroup$ – toby yeats Oct 5 '11 at 22:59
  • 1
    $\begingroup$ Dosn't mean your stupid. The fact that you can study this type of mathematics proves so. I am in my 4th year now and the precise definition of a derivative is confusing to me. I get it a little bit but no way can apply it in a problem solving situation. I am talking about the episilon delta proof. However I don't consider my self stupid just because I don't understand a concept. $\endgroup$ – Tyler Hilton Oct 5 '11 at 23:01
  • 4
    $\begingroup$ Scoring well on an exam is not equivalent to being intelligent. Perhaps your study habits need revision (as mine do) or you have difficulty with memorization (as I do). In any case, it does you no good to doubt yourself. Ask yourself this: If two people are identical in their intelligence and study habits, but one thinks he will succeed and one thinks he will fail, who will perform better? $\endgroup$ – Austin Mohr Oct 5 '11 at 23:03
  • $\begingroup$ @ Tyler Thats fine, but you probably would have done better on the test than me and I put 20 hours of studying into it +homework +all the in class time. @Austin Probably the person who thinks he will succeed. If I think I will succeed at anything at this point it is just me lying to myself. The vast history of evidence saying I will not do well it hard to ignore. $\endgroup$ – toby yeats Oct 5 '11 at 23:04

For #3 you should recall that a function $f(x)$ has a horizontal tangent line if there is a real solution to $f'(x) = 0$. For your specific example, if $f(x) = x^{3} + 5x^{2} + 9x + 17$ then $f'(x) = 3x^{2} + 10x + 9$. Solving the quadratic equation $0 = 3x^{2} + 10x + 9$ gives us solutions that are complex: $\frac{-5 - i2^{\frac{1}{2}}}{3}$ and $\frac{-5 + i2^{\frac{1}{2}}}{3}$. Since $f'(x) = 0$ has no solution for real $x$ then the graph of $f(x)$ does not have a horizontal tangent line.

  • $\begingroup$ I don't get it, isn't the horizontal line $\frac{-5 - i2^{\frac{1}{2}}}{3}$ $\endgroup$ – toby yeats Oct 5 '11 at 23:00
  • $\begingroup$ as @Unreasonable Sin explained, this root (solution) is not a real number since it involves an complex number (i=sqrt(-1)), so the tangent can't be drawin in a graph where axis are real numbers. $\endgroup$ – NoChance Oct 5 '11 at 23:04
  • $\begingroup$ It's important to distinguish between the function and its derivative. The derivative of a function $f(x)$ at some point $x = a$ gives you the slope of the tangent line at that point. A horizontal tangent line has slope $0$, so we are looking for solutions to $f'(x) = 0$. The two solutions I gave in the answer are complex numbers. (The $i$ makes them complex.) If they didn't have that $i$ they'd be real and so the function $f(x)$ would have at least one horizontal tangent line, or even two. $\endgroup$ – Unreasonable Sin Oct 5 '11 at 23:06

I am sure you will be able to score better if you try harder. I suggest that you study lesson by lesson and not to wait till just before the exam because math takes time to digest in your brain. Facts need to be built gradually. Also, many students find Calculus hard because of Algebra. Make sure the basics are clear in your head. Devote sometime to build you understanding gradually. There are 100s of books on these topics, browse some in your library and use it as a reference when you get stuck.

As per your question "what does it mean for f not to be differnetiable at a number in its domain", your answer is OK but it does not reveal the real reason. The real reason is that f would (not be continuous function) at that point (see an example at: example)

As for number (6): The problem here is that if you stick a zero in Tan(x) you will get very large value (infinity), this tells you that this is not working. So, we need to do something about Tan(x). The only thing that can be done is to say Tan(x)=Sin(x)/Cos(x). However, for this to be useful, we have to shape Sin(2x) to a function containing (x) not (2x). Trigonometric identities come to help here. We know that

sin ( a + β) = sin a cos β + cos a sin β

so Sin (2x) = Sin(x) Cos(x)+ Cos(x)Sin(x) = 2 Sin(x) Cos(x)

now we can write:

sin (2x)/Tan(x) = 2 Sin(x) Cos(x) / (sin (x) / Cos (x) ) so sin (2x)/Tan(x) = 2 Cos(x) Sin(x) Cos(x) / Sin (x) = 2 Cos(x) Cos(x)

The limit of the L.H.S as x --> 0 = limit of the R.H.S as x --> 0 = 2 Cos(0) Cos(0) = 2

Check this on the following graph of Sin (2x)/Tan(x)enter image description here

  • $\begingroup$ Thanks, I completely forgot about that identity. I have a very bad memory and this will likely be a problem again. I tried very hard on this, I can't try much harder in this class without sacrificing my grades in other classes. I doubt I will do much better on any tests, historically my test scores decline the longer a class goes on because the material and tests just get more difficult. I started with a D, got an F and that trend will continue most likely. $\endgroup$ – toby yeats Oct 5 '11 at 23:38
  • $\begingroup$ You just need to take some rest. Look at your previous exams and figure out what went wrong. Identities like the one above are easy to forget because there are many of those. Why not memorize 2 each day? By the end of your course, you will know them all. $\endgroup$ – NoChance Oct 5 '11 at 23:48
  • $\begingroup$ I have to memorize all the hyperbolic sin functions and the derivatives and inverse of all the sin functions as well. No time really, I have a test on monday as well which I plan on doing about as well as I did on this test. $\endgroup$ – toby yeats Oct 5 '11 at 23:53
  • $\begingroup$ Try to workout solved examples and problems and have some confidence and some good sleep before the test. You will do well. I am sure. $\endgroup$ – NoChance Oct 5 '11 at 23:57
  • $\begingroup$ Now you are lying :P thanks anyways though. If I get a D on this test I will be surprised. $\endgroup$ – toby yeats Oct 5 '11 at 23:59

On question 3, you take the derivative $f'(x)=3x^2+10x+9$ and show this is never zero for real $x$.

For example you could try to solve the quadratic and find two complex roots, or complete the square with $f'(x)=3\left(x+\frac{5}{3}\right)^2 + \frac{2}{3}$, which is clearly always positive.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.