2006 EXAM 1 Q4. b) NEED HELP!

[tyrone]

Hi I'm having a lot of difficulty in 2006 EXAM 1 Q4. b), I'm not understanding how to calculate the x and y intercepts for the graph.
I have correctly worked out  in the previous question that the amplitude=5 and period=pi

tp://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006mmcas1-w.pdf  (link to the question)

For the function f:[−π,π]→R,f(x) =5cos(2(x+pi/3))
sketch the graph of the function f on the set of axes below. Label axes intercepts with their coordinates.
Label endpoints of the graph with their coordinates.


thanks in advance.

[achre]

I really think that horizontal translations of trig functions are the hardest thing on the methods course, worse than yr. 11 prob 
So to get the axis intercepts, like any other function, you solve f(x)=0, so you get the general solution to cos(2x + pi/3):
=> 2x+pi/3 = 1 + 2*pi*n
=> x= (6*pi*n + pi + 3)/6
then proceed to find the solutions inside the domain
y-int is somewhat easier, f(0)=cos(2pi/3)=cos(pi/2 + pi/6)=-1/2

[DefyingGravity]

y-int is somewhat easier, f(0)=cos(2pi/3)=cos(pi/2 + pi/6)=-1/2

Achre, I don't know if that is quite right...

Here are my worked solutions, or at least what I did when I completed this practice exam.

[achre]

Achre, I don't know if that is quite right...

Here are my worked solutions, or at least what I did when I completed this practice exam.

Yep that's right, I forgot to multiply by the amplitude -_-

[DefyingGravity]

Yep that's right, I forgot to multiply by the amplitude -_-

I swear, this just proves the point that these kinds of questions are so difficult for so many students.
I reckon this kind of question takes me way too long for my liking, then I end up stressing that I'm not running on time.
Let's just hope something like this doesn't pop up tomorrow

[Lejn]

I'm fairly sure that this was the most recent calc free to have a trig function to be graphed, so it wouldn't surprise me if I see it come tomorrow. I'm so ridiculously comfortable with trig, but it always feel like these lag on.

[tyrone]

Thankyou very much Achre and DefyingGravity

In the same exam, i was stuck on Q.3 b)
Let y=xtan (x). Evaluate dy/dx when x=pi/6

I used the product rule and got to this stage--> xsec^2(x)+tan(x)  and then i think i have to sub in pi/6
and i get:

pi/6 sec^2 (pi/6) +tan (pi/6)

I know tan (pi/6) is square root of 3. From here im a little confused.
 I have looked at the worked answers on VCAA, however i feel like they just like jumped steps and bang got an answer if someone could please show me the full working out or help direct me to the correct answer, much appreciated. (p.s sorry I'm not sure how do use latex)

[achre]

Confused about what sec^2 (pi/6) is?
Don't write d/dx (tan(x)) as sec^2(x), just do it as 1/cos^2(x), it's much more straightforward.

[DefyingGravity]

Yep, I completely agree.

You've derived it fine, that's no worries. But if you use sec^2(x)=1/cos^2(x), then you have very little problems.

So, of course, step by step:
f'(x)=xsec^2(x)+tan(x)
f'(x)=1/cos^2(x)+tan(x)
f'(pi/6)=1/((cos(pi/6))^2)+tan(pi/6)
 

[DefyingGravity]

Should you have any more problems with that, just let us know.
Honestly, my advice is to just take your time with trigonometric functions. It's way too easy to miss, say, the amplitude or get something in the wrong quadrant.

Good luck for tomorrow - Let us know how you go!