I need help sketching y=1/[f(x)] graphs its so confusing right now! (Spiralflex!) (1 Viewer)

[bleakarcher]

thanks guys.

 

[SpiralFlex]

thanks guys.

I am not in the mood for a lengthy explanation haha. What do you need?

However if I have time tomorrow morning I will do a lengthy explanation. Someone else can do it, if not I will get back to you tomorrow.

 

[bleakarcher]

I can sketch it when it gives me the function but not when it doesnt, I need help with that dude. Its alright if you dont want to do it now. I need help with the ones that include trig functions so essentially sketching reciprocal trig functions.

 

[4025808]

Try experimenting with an original graph.

Note that the f(x) is the y value of that original graph. Substitute that y value into 1/f(x), and then you will see that 1/f(x) is a reciprocal function.
Also, note that if at a point, f(x) = 1, then 1/f(x) -> 1/1 = 1.
If f(x) = 0 , then 1/f(x) -> 1/0 which is undefined.
If f(x) = infinity, then 1/f(x) -> 1/infinity = 0

Also take into account that if there is an asymptote on an original graph, drawing the reciprocal function [1/f(x)] would result in drawing an open circle at that point when the graph hits y = 0

 

[RivalryofTroll]

Something like this? Basically, if you look at the points, you'll notice where the reciprocal function points are (original function y = 1 then reciprocal function is y = 1 too, OF y = 2 then RF y = 1/2)

EDIT: The explanation above is really good!

 

[zeebobDD]

for curves like f(x)=(x-3)(x-5)(x+4) , and then y=1/f(x) , i usually just draw that guide graph of f(x) first, then you should automatically know that all the maxima's will become minima's and vice versa, Always test if its above or below the orignal maxima/minima by subing in the Y value into 1/(f(X)

Marking out the asymptotes, then simply using the guide graph to see where each region will be, and then just sketching, though if i had latex and a graphing program my explanation wouldnt have been as bad:/

 

[goobi]

Try experimenting with an original graph.

Note that the f(x) is the y value of that original graph. Substitute that y value into 1/f(x), and then you will see that 1/f(x) is a reciprocal function.
Also, note that if at a point, f(x) = 1, then 1/f(x) -> 1/1 = 1.
If f(x) = 0 , then 1/f(x) -> 1/0 which is undefined.
If f(x) = infinity, then 1/f(x) -> 1/infinity = 0

Also take into account that if there is an asymptote on an original graph, drawing the reciprocal function [1/f(x)] would result in drawing an open circle at that point when the graph hits y = 0

What if f(x)=1/g(x) where g(x) is a polynomial? Do we still need to draw an open circle at that point?

 

[4025808]

What if f(x)=1/g(x) where g(x) is a polynomial? Do we still need to draw an open circle at that point?

no we dont, in fact, a polynomial will never have an asymptote itself (from what I recall)

 

[IamBread]

Roots become asymptotes, asymptotes become roots, max becomes min, min becomes max etc.

 

[study-freak]

Roots become asymptotes, asymptotes become roots, max becomes min, min becomes max etc.

NO.

 

[RANK 1]

substitute the points in one by one

 

[Carrotsticks]

Roots become asymptotes, asymptotes become roots, max becomes min, min becomes max etc.

NO.

They become 'roots' but with open circles.

 

[IamBread]

They become 'roots' but with open circles.

That^
Is what I meant

 

[goobi]

no we dont, in fact, a polynomial will never have an asymptote itself (from what I recall)

Yeah I know, that's why I was wondering

 

[bleakarcher]

I am not in the mood for a lengthy explanation haha. What do you need?

However if I have time tomorrow morning I will do a lengthy explanation. Someone else can do it, if not I will get back to you tomorrow.

Spiralflex?

 

[Trebla]

To sketch y = 1/f(x)

- This should be obvious but note that the sign of f(x) is the same sign as 1/f(x). So for a given domain, no part of the graph of f(x) should suddenly switch to the other side of the x-axis when graphing 1/f(x)

- The function is undefined when f(x) = 0 (which leads to the x-intercepts graphically) which often implies discontinuity often in the form of an asymptote if f(x) is smooth and continuous about that point

- As f(x) --> ±∞ then 1/f(x) --> 0 which means that for large magnitudes of f(x), the x-axis becomes an asymptote in 1/f(x). This also implies that there are no x-intercepts for y = 1/f(x)

- Note that d(1/f(x))/dx = - f '(x)/[f(x)]². This means several things:
* If f '(x) > 0 then d(1/f(x))/dx < 0 => wherever f(x) is increasing, 1/f(x) is decreasing
* If f '(x) < 0 then d(1/f(x))/dx > 0 => wherever f(x) is decreasing, 1/f(x) is increasing
* If f '(x) = 0 then d(1/f(x))/dx = 0, which means that the particular x-values which make f(x) stationary also make 1/f(x) stationary. This implies that the x-values of the stationary points are conserved (obviously the y-values change). Also, note that as a consequence of the other two results, a first derivative test tells us that the maximum turning points of f(x) become minimum turning points for 1/f(x) and vice versa.

 

[SpiralFlex]

-Reserved for posting later.

 

[bleakarcher]

Thanks everyone especially Trebla and thank you Spiralflex in advance.

 

[SpiralFlex]

Will probably be done by 10 AM tomorrow. I will stay up tonight as I have other things to do first.

 

[bleakarcher]

Spiralflex, as you know from my PM to you I'd rather you just go through y=[f(x)]^2 and y=[f(x)]^3. I understand y=1/[f(x)] graphs now. Just to save you the effort.