graphing pains... (1 Viewer)

[complex]

Doing revision for my 4u test on friday... Wish I'd started earlier. God dammit.

Two questions, related to reciprocial function and division of ordinates in Chapter 1 of Cambridge 4u, exercise 1.5, question 3 and 4.

Question 3:

Graph y=x [easily done]
Graph y=e^x [easily done]
Now Graph y= x/e^x...

Ok, so first thing I did was sub in x=1, which gave me (1,1/e) as a coordinate.

After that I pretty much had no idea what to do, so I subbed values into my calculator to get the general shape of the function. I realize I could have just graphed x . 1/(e^x) which would have been easier - but I think there is an easier way to do it. Dunno how though. Something about analysis of what happens as it tends to 0 and infinity, but I dunno where to go from here;

as x ---> inf, e^x ---> inf
as x ---> -inf, e^x ---> 0

I mean shit, thats great and all but all I did was look at the graph of both functions and what they tend towards. Where do I go from there?

Question 4:

Graph y=ln(x) [easily done]
Graph y=x [easily done]
Graph y=ln(x)/x

Ok, first thing I did was put the asymtote of x=1 in.

Second, I subbed in e. Giving ln(e)/e; (e,1/e). Pretty useless so far.

Then I realized again, it would have been much easier to just graph ln(x) . 1/x, but i'm still pretty sure theres an easier way of going about this.

Help is greatly appreciated. <3

[edit: I have seen/know the solution - I just want some help on the method.]

 

[Slidey]

Is this graphing or sketching?

Sketch:
y=x
y=e^-x (or 1/e^x) - it is just the mirror image of e^x about the y axis

Now, y=xe^-x
y'=e^-x - xe^-x = e^-x(1-x)=0
x=1
y(1)=1/e

Now, do the extreme values tests, as well as test either side of the extreme point (1,1/e) and test x=0. You should note that the function tends to 0 as x-> inf and tends to negative infinity as x -> -inf, and the extreme value is a maximum.

So you've got an overall shape for the graph: it goes from 0 at x=0 to a peak of 1/e at x=1 and then settles down to 0 as it x approaches infinity. Easy.

What about the negative side of the x-axis though? Well, obviously the entire negative side of the x axis is negative, since there was only one root for the derivative and it was a max value. Now, we're dividing x by e^x. Ignoring that y=x is negative (x<0 means y<0) for the moment, what happens when you divide a number greater than one by a number between 0 and 1? E.g. 4/0.1 = 40. The number gets bigger. So basically, as x-> negative inf, y -> negative infinity at a very fast rate (essentially exponential).

If it is actually graphing, then it's probably easier. You can do some of the above stuff, but what I'd do is just plot every 1 points for each function then divide them (or multiply if you're going with xe^-x).

 

[Slidey]

Question 4:

Graph y=ln(x) [easily done]
Graph y=x [easily done]
Graph y=ln(x)/x

Ok, first thing I did was put the asymtote of x=1 in.

Second, I subbed in e. Giving ln(e)/e; (e,1/e). Pretty useless so far.

Then I realized again, it would have been much easier to just graph ln(x) . 1/x, but i'm still pretty sure theres an easier way of going about this.

Help is greatly appreciated. <3

[edit: I have seen/know the solution - I just want some help on the method.]

OK, so again, I'd be graphing y=1/x instead of y=x.

Asymptote of x=1, good.
y=lnx.x^-1
y'=1/x^2 -lnx.x^-2 = x^-2(1-lnx)=0
1=lnx
e=x
So stat pt or extreme pt at (e,1/e). Test either side to figure out max at x=e. Also find root of x=1.

Now, since we have ln(x), there will be nothing for values of x < 0. Asymptote at x=0, and since we've got a max at x=e, and a root at x=1, obviously for values of x<1, it's pretty similar to the graph of y=lnx.

Now, since y=1/x has an asymptote at y=0, so too will any composite functions. Given there's a max value at x=e and no roots after this, the function must approach zero as x-> infinity.