GCSE2016Troop
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f(x) = e^-2x + 3/x + 3 , x>0
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RDKGames
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(Original post by GCSE2016Troop)
f(x) = e^-2x + 3/x + 3 , x>0
Go ahead.
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the bear
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you can put in different x values to see what happens. look out for asymptotes etc.
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GCSE2016Troop
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(Original post by RDKGames)
Go ahead.
(Original post by the bear)
you can put in different x values to see what happens. look out for asymptotes etc.
Well I can see now that e^-2x tends to 0 and so does 3/x, so the answer should be f(x) > 3, as this is where the asymptote is? if this is right how would I demonstrate this in working out? Also, for future questions similar to this do you have any tips to try find the range quicker as I seem to struggle and take a very long time on the difficult ones which will cost me in the exam
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(Original post by GCSE2016Troop)
Well I can see now that e^-2x tends to 0 and so does 3/x, so the answer should be f(x) > 3, as this is where the asymptote is? if this is right how would I demonstrate this in working out? Also, for future questions similar to this do you have any tips to try find the range quicker as I seem to struggle and take a very long time on the difficult ones which will cost me in the exam
Easy as that. Though a lot of the time you may need to determine the max/min value of the function in order to get the right bound(s). In this example it's obvious that there is no upper bound since we can make x as small as we want and the second term blows up to infinity. So there is no finite bound. But there is a lower bound as you have shown.

Just practice the hard ones, take your time, and then you get used to them and can do them quicker.

Have a go at f(x) = e^{-x}\sin x for x > 0 as 'hard' example. Find its range. HINT: This would involve finding max/min of the function, it is not sufficient to say "oh this function tends to 0 as x goes to infinity, so 0 must be a bound"
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begbie68
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e^(-2x) does not "blow up to infinity" as x->0 ... ?
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GCSE2016Troop
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(Original post by RDKGames)
Easy as that. Though a lot of the time you may need to determine the max/min value of the function in order to get the right bound(s). In this example it's obvious that there is no upper bound since we can make x as small as we want and the first two terms blow up to infinity. So there is no finite bound. But there is a lower bound as you have shown.

Just practice the hard ones, take your time, and then you get used to them and can do them quicker.

Have a go at f(x) = e^{-x}\sin x for x > 0 as 'hard' example. Find its range. HINT: This would involve finding max/min of the function, it is not sufficient to say "oh this function tends to 0 as x goes to infinity, so 0 must be a bound"
so I differentiated, made equal to 0 and my solving gets 0 so I am struggling with the max part? From looking at the function I would assume 0 < f(x) < 1 but dont think im correct
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RDKGames
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(Original post by begbie68)
e^(-2x) does not "blow up to infinity" as x->0 ... ?
Whoops, must've been sleeping.
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RDKGames
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(Original post by GCSE2016Troop)
so I differentiated, made equal to 0 and my solving gets 0 so I am struggling with the max part? From looking at the function I would assume 0 < f(x) < 1 but dont think im correct
What did you differentiate to?

And no, 0&lt;f(x)&lt;1 is not correct.
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GCSE2016Troop
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(Original post by RDKGames)
What did you differentiate to?

And no, 0&lt;f(x)&lt;1 is not correct.
cosx e^-x - e^-x sinx
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RDKGames
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(Original post by GCSE2016Troop)
cosx e^-x - e^-x sinx
Right, so make it =0.

Notice that e^{-x} \neq 0 so you can cancel it out. Proceed to solve for the first solution from there.
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begbie68
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or factorise and e^-x CANNOT =0, so we must solve for the other factor = 0 .....
(Original post by RDKGames)
Right, so make it =0.

Notice that e^{-x} \neq 0 so you can cancel it out. Proceed to solve for the first solution from there.
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RDKGames
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(Original post by begbie68)
or factorise and e^-x CANNOT =0, so we must solve for the other factor = 0 .....
I see no advantage of doing it in that order TBH.
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GCSE2016Troop
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(Original post by RDKGames)
Right, so make it =0.

Notice that e^{-x} \neq 0 so you can cancel it out. Proceed to solve for the first solution from there.
Noticed my mistake now I have looked back. I had cancelled it out liked you said, on the other part when I had sinx=cosx and then divided both sides I made a silly mistake and put it equal to 0 not 1. So I get x = arctan 1. When I sub that in for y I get 0.322 (3s.f) so would that be the max? max and therefore f(x) is less than this?
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(Original post by GCSE2016Troop)
Noticed my mistake now I have looked back. I had cancelled it out liked you said, on the other part when I had sinx=cosx and then divided both sides I made a silly mistake and put it equal to 0 not 1. So I get x = arctan 1. When I sub that in for y I get 0.322 (3s.f) so would that be the max? max and therefore f(x) is less than this?
Yep, but its with \leq sign and not just &lt;. Strictly speaking, you should use the 2nd derivative to show it is a max point and also reason that this is the most +ve max point along the curve, but I hope it's obvious from the eqn. that it must be. x begins at 0 increases slightly and so both e^{-x} and \sin x are +ve, so their product is +ve hence the first stationary point will be when the curve starts decreasing... i.e. it will be a max point.

Now onto determining the lower bound.


P.S. Just say that x = \dfrac{\pi}{4} hence f(x) \leq \dfrac{\sqrt{2}}{2}e^{-\frac{\pi}{4}} and prevent any rounding. Precision answer like that should be taken to be as your automatic answer at this level, unless stated otherwise.
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GCSE2016Troop
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(Original post by RDKGames)
Yep, but its with \leq sign and not just &lt;. Strictly speaking, you should use the 2nd derivative to show it is a max point and also reason that this is the most +ve max point along the curve, but I hope it's obvious from the eqn. that it must be. x begins at 0 increases slightly and so both e^{-x} and \sin x are +ve, so their product is +ve hence the first stationary point will be when the curve starts decreasing... i.e. it will be a max point.

Now onto determining the lower bound.


P.S. Just say that x = \dfrac{\pi}{4} hence f(x) \leq \dfrac{\sqrt{2}}{2}e^{-\frac{\pi}{4}} and prevent any rounding. Precision answer like that should be taken to be as your automatic answer at this level, unless stated otherwise.
Yeah I realised with my answer that you never use it for a question like the range, I just couldn't seem to get an exact answer like you have shown. Lazy on my part.

Have you ever seen an a level question where it wants you to give reason why this is the most +ve max point along the curve. If so, would you working out be explaining like you have or would you perhaps have to sub in to show the +ve product or draw a graph to show the slight increase after 0?

I will finish with the question tomorrow as I really need to be getting to sleep, do you have any more questions like this I could try tomorrow also?
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(Original post by GCSE2016Troop)
Have you ever seen an a level question where it wants you to give reason why this is the most +ve max point along the curve.
Is this any different to the very very standard question of : "given y=f(x), find the max/min and determine (justify) the nature" ?
?!
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(Original post by GCSE2016Troop)
Yeah I realised with my answer that you never use it for a question like the range, I just couldn't seem to get an exact answer like you have shown. Lazy on my part.

Have you ever seen an a level question where it wants you to give reason why this is the most +ve max point along the curve. If so, would you working out be explaining like you have or would you perhaps have to sub in to show the +ve product or draw a graph to show the slight increase after 0?
I think it helps to know what the function vaguely looks like. So you know that e^{-x} is a function that starts at 1 and decays to 0 (as x goes from 0 to infinity) and sin(x) just oscillates between -1 and 1, so combine the two, and you will get decaying oscillations. If you can't imagine what it looks like still, then just look at https://www.desmos.com/calculator/ryff2i0gjm
So providing a small sketch of this situation would be sufficient justification for your analysis.


Otherwise, with a little more tweaking on your analytical work to the solutions of \tan x = 1, you may notice that the local maximums keep occurring at x = \dfrac{\pi}{4}, \qquad \dfrac{\pi}{4} + 2 \pi, \qquad\dfrac{\pi}{4} + 4 \pi, \quad \ldots.

At these values, \sin x = \dfrac{\sqrt{2}}{2}... so treat this as a constant in the function as far as these points of interest go, hence the value of the function really hinges on what happens to e^{-x} as you go down these values ..... obviously e^{-x} decreases, hence f(x) decreases as you go down these local maxima, so the first solution will be the global maximum and the rest will be smaller than this.

Similar approach can be adapted to the minimum of the function.


But clearly, there are different approaches to the justification and I have shown you two.

I will finish with the question tomorrow as I really need to be getting to sleep, do you have any more questions like this I could try tomorrow also?
I don't have any but you can probably make a thread asking for some similar style problems from others, or look up those resources online.
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