Post by: golden on October 15, 2011, 11:12:51 am
The graph is strictly decreasing across (-infinity, 0] and strictly increasing across [0, infinity).
Is that correct?
Post by: acinod on October 15, 2011, 11:19:41 am
The cusp is actually strictly decreasing and strictly increasing. VCAA talked about this in of their assessor's report I think.
Post by: AleksIlia on October 15, 2011, 06:26:01 pm
Post by: luffy on October 15, 2011, 06:44:44 pm
Yep, I actually got this wrong in my school's SAC :(
The cusp is actually strictly decreasing and strictly increasing. VCAA talked about this in of their assessor's report I think.
Its mentioned in the sample 2010 questions for methods.
Strictly increasing is pretty much the same as increasing just include the endpoints and dy/dx = 0 where relevant? At least that's what I was told
In most cases, it is (i can't think of a case where it isn't.) However, make sure you go by the definition rather than just remembering it as "include endpoints + increasing." The definition of strictly increasing goes along the lines of "A relation is strictly increasing over an interval [a,b] if b>a implies f(b) > f( a)." Something along those lines.
EDIT: Have a read of this: http://www.vcaa.vic.edu.au/correspondence/bulletins/2011/April/2011AprilSup2.pdf
Post by: paulsterio on October 15, 2011, 07:00:09 pm
Post by: luffy on October 15, 2011, 07:01:18 pm
Luffy, you love the discussions on Strictly Increasing/Decreasing don't you? :P
No, not at all. That is why I literally just copied and pasted from a previous post I made.
Post by: paulsterio on October 15, 2011, 07:21:57 pm
No, not at all. That is why I literally just copied and pasted from a previous post I made.
It's just that you're always around whenever this topic is brought up! :P
Post by: golden on October 16, 2011, 10:36:31 am
e2x + bex + 1 = 0.
Solve for b for which there are no real solutions.
Post by: b^3 on October 16, 2011, 10:47:03 am
MAV 2008 Exam 2 Multiple Choice - Q10.Let a=e^2x
e2x + bex + 1 = 0.
Solve for b for which there are no real solutions.
then a^2+ba+1=0
for no solutions, discriminant <0
b^2-4(1)()<0
draw y=x^2-4 and find when it is less than zero
b^2<4, i.e. intercepts at +2, -2
So there are no solutions for -2<b<2
NOTE: probably should post this in the methods help thread next time, instead of here.
Post by: luffy on October 16, 2011, 11:18:28 am
MAV 2008 Exam 2 Multiple Choice - Q10.Let a=e^2x
e2x + bex + 1 = 0.
Solve for b for which there are no real solutions.
then a^2+ba+1=0
for no solutions, discriminant <0
b^2-4(1)()<0
draw y=x^2-4 and find when it is less than zero
b^2<4, i.e. intercepts at +2, -2
So there are no solutions for -2<b<2
NOTE: probably should post this in the methods help thread next time, instead of here.
I think you meant "let a = e^x"
EDIT: Also, to add to b^3's post, in the exam, just to show that e^x > 0 for all real x values, you might want to show that
EDIT 2: My bad - it actually did make a difference to b^3's answer as outlined by jane1234 below :D
Post by: jane1234 on October 16, 2011, 11:23:17 am
MAV 2008 Exam 2 Multiple Choice - Q10.Let a=e^2x
e2x + bex + 1 = 0.
Solve for b for which there are no real solutions.
then a^2+ba+1=0
for no solutions, discriminant <0
b^2-4(1)()<0
draw y=x^2-4 and find when it is less than zero
b^2<4, i.e. intercepts at +2, -2
So there are no solutions for -2<b<2
NOTE: probably should post this in the methods help thread next time, instead of here.
I think you meant "let a = e^x"
There are also no solutions at b=2 as you would get e^x = -1 which produces no real solutions.
So the answer is b E (-2,2]
Post by: b^3 on October 16, 2011, 11:28:56 am