Sketching functions of the form y=xf(x)

An example of a question would be this:

For part (a), my thinking process goes something like this:

We know there are vertical asymptotes at

x= 0, 2

.

Considering the branch of the graph at

x<0

, we can see the graph of

y = \frac{1}{f(x)}

would have a horizontal asymptote at

y=\frac{1}{2}

and would continually increase with a vertical asymptote at

x=0

.

Considering the branch of the graph for

0 < x < 2

we see that there will a negative maximum point, since the original function

f(x)

has a minimum negative point, and hence the reciprocal will produce a maximum negative point. The graph will then be concave down.

Considering the branch of the graph for

x > 2

, we see the orignal function is increasing for those values of

x

, starting out small and positive before becoming big and positive as x tends to infinity This means that the reciprocal will start out big and positive before gradually moving onto becoming small and positive as x tends to infinity.

Leading to this graph:

I was hoping if someone could share their thinking process behind sketching the function

y=xf(x)

, in as much date as possible, thank you! :smile:

(edited 9 years ago)

Untitled.png14.7KB

10255351_536003439842331_11429864260898805_n.jpg48.3KB

Reply 1

wmavfc

2

for part (b) you're stretching the function parallel to the y axis by a factor of x. I think this would mean the horizontal asymptote would then be y=2x? Not sure what other points you'd have to include

Reply 2

Zacken

OP

22

Update: I've spent some more time thinking on the question, is the following correct?

Considering the branch of the graph for values of

x<0

, we can see that this is basically starting out as multiplying a very large negative

x

value by a value close to 2 and the negative x values decreasing but remaining positive, leading to an increasing curve in the third quadrant.

Considering the branch of the graph for values of

x>2

, we can see that this is starting out as positive increasing values of x multiplied by increasing positive values of f(x), meaning that this will lead to another increasing curve that remains positive for values of

x>2

.

I get this shape:

I remain unable to come up with how to sketch the graph for values of

0<x<2

.

10799609_10152537831430773_160306132_n.jpg35.8KB

Reply 3

Zacken

OP

22

Original post by wmavfc

for part (b) you're stretching the function parallel to the y axis by a factor of x. I think this would mean the horizontal asymptote would then be y=2x? Not sure what other points you'd have to include

Wouldn't you only be able to stretch functions with constant factors, that is, how would you stretch a function with a factor that is constantly varying?

Reply 4

TeeEm

19

Original post by Zacken

An example of a question would be this:

For part (a), my thinking process goes something like this:

We know there are vertical asymptotes at

x= 0, 2

.

Considering the branch of the graph at

x<0

, we can see the graph of

y = \frac{1}{f(x)}

would have a horizontal asymptote at

y=\frac{1}{2}

and would continually increase with a vertical asymptote at

x=0

.

Considering the branch of the graph for

0 < x < 2

we see that there will a negative maximum point, since the original function

f(x)

has a minimum negative point, and hence the reciprocal will produce a maximum negative point. The graph will then be concave down.

Considering the branch of the graph for

x > 2

, we see the orignal function is increasing for those values of

x

, starting out small and positive before becoming big and positive as x tends to infinity This means that the reciprocal will start out big and positive before gradually moving onto becoming small and positive as x tends to infinity.

Leading to this graph:

I was hoping if someone could share their thinking process behind sketching the function

y=xf(x)

, in as much date as possible, thank you! :smile:

which exam is this question out of please?

Reply 5

wmavfc

2

Original post by Zacken

Wouldn't you only be able to stretch functions with constant factors, that is, how would you stretch a function with a factor that is constantly varying?

I think all you'd be able to do is show the transformation on the points you already have. Not sure apart from that though mate. Tried finding a mark scheme?

Reply 6

Zacken

OP

22

Original post by wmavfc

I think all you'd be able to do is show the transformation on the points you already have. Not sure apart from that though mate. Tried finding a mark scheme?

Yeah, I've looked at the markscheme and all. I understand it to a certain degree, but I'd rather fully understand the thought process. :smile:

Thanks for your help, though!

Reply 7

Zacken

OP

22

Original post by TeeEm

which exam is this question out of please?

It's from the IB exams, May 11, TZ2, paper 1, non-calc.

You'll find quite a few interesting questions to add to your collection if you peruse through their past papers! :wink:

Reply 8

TeeEm

19

Original post by Zacken

It's from the IB exams, May 11, TZ2, paper 1, non-calc.

You'll find quite a few interesting questions to add to your collection if you peruse through their past papers! :wink:

I have all their papers up to 2011 but I have no time (I think in order to complete all my maths tasks in my to do lists I need to live to around 700)

This question is already stolen (adapted of course...)

thanks

Reply 9

Hasufel

16

I believe your logic on question b) is sound (except on 0<x<2). The function is decreasing all the way to the turning point, but another key is the change in concavity at just before x=1

This is a suspicious-looking function to me - almost certainly likely to be piece-wise.

when f is multiplied by x, the zeros remain unchanged.

The left half of your original (for x<0) is exactly:

\displaystyle \frac{-4}{\pi} \arctan(x)

(the limit of this function as x approaches -infinity is 2)

and the right half of your original function (for x>=0) is exactly

\displaystyle \frac{2}{3}x(x-2)

The key is, to think of the function, as you have done, in terms of limits, derivatives, concavity and stationary values/inflections.

And if you think of it in terms of separate or piecewise functions, if you draw the right half of the graph, and on the same plot, the graph of y=x, you can see that in the interval for

0<x<1

, the positive values of the function y=x in the same range, will decrease the negative values of f(x), at a greater rate the closer to zero you get, and a lesser rate nearer to x=1 (which is why there`s a slight inflection before x=1). For the interval

1<x<2

the depth of the graph will increase slightly, because as the values of x get greater - the negative aspect of y= f(x) gets greater.

Basically x=1 graph doesn`t alter
0<x<1 graph gets less negative, with a change in concavity
1<x<2 graph gets more negative (or "similar" but "further down")

(stick this code into wolfram alpha, and you`ll see your graph):

Plot[Piecewise[{{(-4 x/Pi)*ArcTan[x], x < 0}, {(2/3)*x^2*(x - 2),
x >= 0}}], {x, -4, 4}, PlotRange -> 5]

(edited 9 years ago)

OP

Original post by Hasufel

x

Ah, I see, that makes perfect sense. Basically, I just had to examine the graph for

0<x<2

more minutely, paying attention to how the values for

0<x<1

decrease depth of the function whereas

1<x<2

increased it and

x=1

had no change, 'causing some inflexionary behaviour at that point.

Thank you! :smile:

P.S: How'd you manage to guess the piecewise graphs like that? Is it just lots and lots of experience or...?

Reply 11

Hasufel

16

[QUOTE=...P.S: How'd you manage to guess the piecewise graphs like that? Is it just lots and lots of experience or...?

some experience, mostly because i really like things like the heaviside function and O.D.E.`s with laplace transforms, which can and do involve a lot of piecewise stuff.

(to begin with, if i`m honest, i suspected it might be piecewise because of it`s wierd shape!)

Sketching functions of the form y=xf(x)

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