graphing tip required! (1 Viewer)

[eliseliselise]

hi!!! i am currently on a quest to find a simple, time-efficient method for graphing in questions akin to the one below.

part iii) and iv) are simple, and can be done quite quickly... however with part i) and ii)- i can eventually get the answers but my technique would be inappropriate in an actual exam-- way too time consuming, and after doing what i can i would probably move on, hence only scoring a max of 1/2 marks for part i)/ii) which is kinda measley for a question 3!

i know what the answers look like, but i am interested in the method you would go about in order to know what part i) and ii) would look like & if there is some easy/simpler way to attempt these style questions, considering the other graphs [part iii) and iv)] have the same amount of marks allocated, although require much less effort [imo].

thankyou!!

 

[Slidey]

Ahhh it's been a while.

OK, for part 1)
ln(1)=0
Between x=1 and x=2, there's a y asymptote from y=0 to negative infinity.
Between x=0 and x=1, there's a y asymptote from y=0 to negative infinity.
So basically it sort of looks like a negative parabola squished between 0 and 2 with a maximum at (1,0).
Disregard from x=2 onwards (ln(-|x|) doesn't exist).
Discontinuity at x=0. From x=0 to x -> negative infinity, it basically looks like the graph of ln(-x), since from y=1 upwards it is essentially linear, and between -2<x<0 it is fractional (the asymptote to negative infinity part).

Part 2)
Ah, just flip it? :S
Eh 3unitz already posted good tips, follow them.

 

[Slidey]

A note on inversion: I forget what you call it, but the 'direction' of a piece of curve flips.

Try graphing some functions and then their inverses to see what I mean. Good examples would be y=x^2, y=ln(x).

Use Graphmatica, as linked to in stickies, to test out a lot of graphing for yourself. It's quite fun noticing the patterns that emerge, but moreover you'll learn quite a bit from it.

 

[Trebla]

A useful method for y = ln [f(x)] is to consider the features of y = ln x and simply replace x with f(x).

So features of y = ln x include:
- Defined only for x > 0
- As x --> ∞, y --> ∞
- As x --> 0 from the positive end, y --> - ∞
- When x = 1, y = 0
- For 0 < x < 1, y < 0
- For x > 1, y > 0

So for y = ln [f(x)], replace x with f(x):
- Defined only for f(x) > 0
- As f(x) --> ∞, y --> ∞
- As f(x) --> 0 from the positive end, y --> - ∞
- When f(x) = 1, y = 0
- For 0 < f(x) < 1, y < 0
- For f(x) > 1, y > 0

Depending on what f(x) actually is, you may also need to apply other restrictions. Simply use the above features to deduce the graph.

Also for most composite functions of graphs, the x-values of the turning points are conserved since:
y = g(f(x))
dy/dx = g'(f(x))f'(x)
So f'(x) = 0 satisfies dy/dx = 0

 

[Affinity]

here are some things to look out for

- vertical asymptotes become roots

this is not true.. you get an undefined point

 

[Affinity]

there isn't a root there

hmm for example " f(x) = 1/x"
means f(x) = 1/x if x is not 0 and is undefined otherwise.
so 1/f(x) = x for all non zero x's but is undefined at 0.

compare with for example the function (x^2 -9)/(x+3)

 

[Slidey]

this is not true.. you get an undefined point

Ooh, cool. I did think it sounded a bit odd.

Good summary Trebla.

 

[Affinity]

undefined points that look like roots?

pretty much .. but you should put in a hollow circle there

 

[kaz1]

here are some things to look out for

when sketching 1/f(x):
- roots become vertical asymptotes
- [EDIT: vertical asymptotes become undefined points on x-axis]
- all points where f(x) = 1 or - 1 remain
- 1/f(x) is large when f(x) is close to 0
- 1/f(x) is small when f(x) is large

when sketching ln[f(x)]:
- all points where f(x) = 1 become roots (ln 1 = 0)
- roots become vertical asymptotes
- if f(x) < 0 ln[f(x)] is undefined
- 0 < f(x) < 1 then ln[f(x)] is negative

Thx for the tips.

Btw would you get ln[f(x)] in 3 unit?

 

[eliseliselise]

thanks so much everyone!! it's all alot clearer now thankyou

 

[Slidey]

Thx for the tips.

Btw would you get ln[f(x)] in 3 unit?

It's within your ability, but since it's an actual common 4u question, I doubt it.