Find the range of f

f(x) = e^-2x + 3/x + 3 , x>0

Go ahead.

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

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"

e^(-2x) does not "blow up to infinity" as x->0 ... ?

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, $0<f(x)<1$ is not correct.

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.

or factorise and e^-x CANNOT =0, so we must solve for the other factor = 0 .....

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?

Yep, but its with $\leq$ sign and not just $<$. 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.

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.

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?