Quick Graph Question (1 Viewer)

[nrlwinner]

Hi.

I'm not really sure about the theory behind graphing functions that look like
f(x) = f abs(x)

eg. f(x) = ln abs(x)

Can someone please clarify.

 

[spammy679]

f(abs(x)) functions are just reflected against the y-axis

it's because the x values would be always positive; so whatever the graph looks like x>=0, it would be the same on the left of the y-axis

 

[adomad]

yeah, remove every thing for X<0 and the reflect the RHS onto the LHS. its sorta logical IMO. LOL spammy, i laugh everythime i see that gif in your sig. sooo funny!~~!~!~!

 

[spammy679]

hahaha

 

[nrlwinner]

Thank you very much.

 

[nrlwinner]

Can somebody give me some tips on multiplying and dividing fuctions

Eg. lnx / x

 

[The Nomad]

Can somebody give me some tips on multiplying and dividing fuctions

Eg. lnx / x

In my opinion, best way is to decide on the domain and range, then find turning points and intercepts.

 

[jet]

General method for all graphing questions:

1. Intercepts (x- and y-)
2. Asymptotes (Vertical, Horizontal and Diagonal) and other discontinuities (includes domain and range)
3. Turning Points
4. Points of Inflection (if needed)

Then mark these features. Sketch around these features. Join them together

E.g. f(x) = ln(x)/x

1. No y-intercepts
x - intercept x = 1
2. Asymptotes: x = 0, y --> -infinity since ln(x) --> -infinity quicker than x --> 0
Horizontal: y = 0, since x ---> infinity quicker than ln(x)

Domain: x > 0

3. Turning points:

dy/dx = (ln(x) - 1)/x^2

therefore a stationary point occurs at x = e, y = 1/e

Sketch it, and it will look like ln(x) with a maximum at (e, 1/e) and then curves back down to the x-axis

 

[nrlwinner]

Thanks. That really helps me. But I'm not sure how to find the limit a y approaches infinity. I know how to do the x one (please check for me)

 

[study-freak]

Thanks. That really helps me. But I'm not sure how to find the limit a y approaches infinity. I know how to do the x one (please check for me)

x--> +-2, y-->infinity

(denominator=/=0)

 

[jet]

The time when y --> infinity will generally just be when the denominator = 0. That's pretty much all you need to know. Just don't forget the general features of ln(x) etc when it comes to that

 

[nrlwinner]

Thanks for the general tips on solving graphs. It has REALLY helped me in this section. Now I'm up to graphs of composite functions and I'm not really sure what to do.

eg. Graph y=ln(cosx)

 

[jet]

Follow the features of both functions. You know cos(x) oscillates, so the composite function will oscillate too.

Max value of cos(x) is 1, minimum value is -1. You know that ln(x) is only defined for x > 0, so the composite function will only be defined for -(pi/2) ≤ x < (pi/2); (3pi/2) < x < (5pi/2) etc etc (including negative domains)

You know ln(x) --> infinity when x ---> 0, so whenever there is a zero of cos(x), there will be an asymptote in the composite function.

Max value of cos = 1 means that the function is always ≤ 0, as ln(1) = 0.

Someone might have to correct me on this, though I'm quite sure maxima and minima are also maintained.

 

[Trebla]

Thanks for the general tips on solving graphs. It has REALLY helped me in this section. Now I'm up to graphs of composite functions and I'm not really sure what to do.

eg. Graph y=ln(cosx)

The features of an ordinary y = ln x are:
- Defined for x > 0
- As x --> ∞, ln x --> ∞
- As x --> 0, ln x --> - ∞
- When x = 1, ln x = 0

Hence the features of y = ln (cos x) are:
- Defined for cos x > 0
- As cos x --> ∞, ln cos x --> ∞ is useless because cos x never approaches infinity
- As cos x --> 0, ln cos x --> - ∞
- When cos x = 1, ln cos x = 0

So from your usual y = cos x, only consider parts where cos x > 0. Around the neighbourhood of the x-intercepts of y = cos x, the log transformation forces it to approach - ∞ asymptotically. When cos x = 1 (i.e. its maximum turning point) then the log transformation makes it into an x-intercept. From that information, you should be able to deduce what the graph should look like.

 

[nrlwinner]

so the composite function will only be defined for -(pi/2) ≤ x < (pi/2); (3pi/2) < x < (5pi/2) etc etc (including negative domains)

I'm not sure how you got to that conclusion.

 

[jet]

so the composite function will only be defined for -(pi/2) ≤ x < (pi/2); (3pi/2) < x < (5pi/2) etc etc (including negative domains)

I'm not sure how you got to that conclusion.

Ok, ln(x) is defined for x > 0, therefore ln(cos(x)) is defined for cos(x) > 0. Solve this equation.

 

[Trebla]

so the composite function will only be defined for -(pi/2) ≤ x < (pi/2); (3pi/2) < x < (5pi/2) etc etc (including negative domains)

I'm not sure how you got to that conclusion.

In the equation y = ln (cos x), it is cos x that must be positive NOT x. If you stare at a graph of y = cos x, the domains in which cos x is positive (i.e. y values) are -π/2 < x < π/2 etc

 

[nrlwinner]

OH RIGHT. I get it now.

So basically, there are no general guides as to how to do these graphs. Rather, you just need to follow the features of both graphs. Right?

 

[jet]

OH RIGHT. I get it now.

So basically, there are any general guides as to how to do these graphs. Rather, you just need to follow the features of both graphs. Right?

Yes.

 

[nrlwinner]

Trebla, when you said

- As cos x --> ∞, ln cos x --> ∞ is useless because cos x never approaches infinity
- As cos x --> 0, ln cos x --> - ∞

I'm not sure how you got to that conclusion. Could you please explain it for me.