Derivative test review 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.


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*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.
 A: 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. 
A: 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)
A: 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.
