Let's get this straight.

[golden]

For the graph y = |x|:
The graph is strictly decreasing across (-infinity, 0] and strictly increasing across [0, infinity).
Is that correct?

[acinod]

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.

[AleksIlia]

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

[luffy]

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

[paulsterio]

Luffy, you love the discussions on Strictly Increasing/Decreasing don't you?

[luffy]

Luffy, you love the discussions on Strictly Increasing/Decreasing don't you?

No, not at all. That is why I literally just copied and pasted from a previous post I made.

[paulsterio]

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!

[golden]

MAV 2008 Exam 2 Multiple Choice - Q10.

e^2x + be^x + 1 = 0.

Solve for b for which there are no real solutions.

[b^3]

MAV 2008 Exam 2 Multiple Choice - Q10.

e^2x + be^x + 1 = 0.

Solve for b for which there are no real solutions.

Let a=e^2x
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.

[luffy]

MAV 2008 Exam 2 Multiple Choice - Q10.

e^2x + be^x + 1 = 0.

Solve for b for which there are no real solutions.

Let a=e^2x
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 . In this particular question, it will make no difference because the answer you get is which has no impact on b^3's answer above.

EDIT 2: My bad - it actually did make a difference to b^3's answer as outlined by jane1234 below

[jane1234]

MAV 2008 Exam 2 Multiple Choice - Q10.

e^2x + be^x + 1 = 0.

Solve for b for which there are no real solutions.

Let a=e^2x
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]

[b^3]

Yeh sorry guys my bad, e^x and I always forget to sub back in for logs and e^x's to check wether it works or not.