Dekoyl's question thread

[dekoyl]

Hopefully this thread won't get much activity. Otherwise it spells trouble.
I occasionally run across a few questions which I hope can be cleared up.
Most sincere thanks to those who help out.

1. has only one real solution if: b < 0. Why?
I let Then I use the discriminant to let .
I see the fault in my working out as I seemingly change the whole question by working with a different equation. How else can I tackle this?

[Pandemonium]

if you look at the equation

2^2x is a positive 'exponentially increasing' curve
2^x is also a positive 'exponentially increasing curve'

both of them do not touch the x axis, but they have an asymptote of y=0
also, they are one-to-one increasing functions. so when you add them together you will get another one-to-one increasing function. also, adding the asymptotes together (y=0 and y=0) will get you y=0 as a result. without any vertical translation [EDIT: in the negative y direction], there will be no x-intercept.

if we shift this asymptote below the x axis (ie. y=b), then the graph will ONLY intersect the x axis once.

therefore, when b<0 the graph only has one x-intercept.

[dekoyl]

Ah thank you very much, Pandemonium. Your explanation of the question makes sense but I wasn't sure how to get the answer algebraically. However, thanks again for the explanation.

[shinny]

You were on the right track actually. The discriminant is useless however, and your logic was incorrect anyway. Despite having only one solution, in actual fact, it has two, but one is undefined for . The actual discriminant would be 1-4b, and since b is less than 0, then -4b is always positive and hence . And yes, that does mean there are two solutions, but continue by using the quadratic formula and you'll get , and substitute your back in. Using similar logic to before, and considering that the range of , then reject the negative of the because , and hence the overall fraction will be less than 0 for any value of b, and hence is undefined for . Hence;

Therefore, only one solution. Q.E.D. etcetc. blargh.

EDIT: Whoops forgot to consider 0<b<1/4 as in humphdogg's post, so ignore the

[humph]

Hopefully this thread won't get much activity. Otherwise it spells trouble.
I occasionally run across a few questions which I hope can be cleared up.
Most sincere thanks to those who help out.

1. has only one real solution if: b < 0. Why?
I let Then I use the discriminant to let .
I see the fault in my working out as I seemingly change the whole question by working with a different equation. How else can I tackle this?

Let . Then we can rewrite the original equation as , which, by the quadratic formula, yields .
If , then , which is impossible, as has a strictly positive range (that is, ).
If , then the discriminant is negative and so there are no real solutions for .
If , then the discrimant is positive and so both and are real numbers. , however, is clearly negative and so cannot be a solution.
Similarly, if , then , and so , which is again impossible.
Finally, if , then , and so , and so is positive, and hence is a real solution.

That's obviously in a fair bit more detail than you'd ever need to cover, but that's the formal algebraic way of proving it.

[Pandemonium]

oh well, you can still use either and get full marks in the methods exam whether you write it algebraically or not.

methods isn't strictly formal maths anyway.

[Glockmeister]

Heh. the algebraic way is the only way I've been taught

[dekoyl]

Are the answers wrong? Or am I doing more than the question requires?

If ∞ is such that and . Find the rule for
Well I found the rule which is the answer.
But since the domain is ∞, shouldn't the answer be ?

I just want to make sure as I often misinterpret questions. Thanks.

[Mao]

they are equivalent.



you are not wrong in your answer, nor is their answer. It's just different ways of expressing the same thing. both answers are defined over the specified domain and are algebraically equivalent. Outside this domain though, they have very different implications.

[dekoyl]

they are equivalent.

But if it were wouldn't the domain be \{3} instead of the ∞ required? Or am I making a fool of myself?

[Mao]

they are equivalent.

But if it were wouldn't the domain be \{3} instead of the ∞ required? Or am I making a fool of myself?

that's the maximal or implied domain.

however, in this case the domain is restricted. so long as the restricted domain does not conflict with the implied domain [in which case we take the intersection of the two sets], we only work with the restricted domain.

the maximal domain only applies when there are no domain restrictions.

[dekoyl]

maximal[/b] or implied domain.

however, in this case the domain is restricted. so long as the restricted domain does not conflict with the implied domain [in which case we take the intersection of the two sets], we only work with the restricted domain.

the maximal domain only applies when there are no domain restrictions.

Ah right of course. Thank you Mao.

[dekoyl]

Another one that doesn't require working out I think
1.

Is the inverse or just (without the minus) The answer only has the addition sign so I'm just checking.

2. Use Euler's linear approximation method to estimate to 2 d.p., given
I can't seem to get the answer required. Note that this is non-calculator.

Much thanks.

[shinny]

1. There can't be a in a function, so definitely not with the . As to whether it's plus or minus, was there a domain restriction with the original function? If so, graph that and reflect it and it should be obvious as to whether it's the positive or negative branch for the inverse.
2. Wow, you sure that's non-calc? I'll give it a go later, but despite linear approximation being an approximation, it still often requires calculators, and a log2(19) doesn't look pretty right now, and definitely not to 2 d.p. Which paper was this in?

[dekoyl]

1. There can't be a in a function, so definitely not with the . As to whether it's plus or minus, was there a domain restriction with the original function? If so, graph that and reflect it and it should be obvious as to whether it's the positive or negative branch for the inverse.

Nope there were no domain restrictions.

2. Wow, you sure that's non-calc? I'll give it a go later, but despite linear approximation being an approximation, it still often requires calculators, and a log2(19) doesn't look pretty right now, and definitely not to 2 d.p. Which paper was this in?

Yep this is exam 1, iTute 2007. It's alright this isn't important.