Sketching graphs (1 Viewer)

[ADrew]

If you are given y=f(x),
how would you go about sketching y=lnf(x), y=e^f(x) and y=arctanf(x) ?

 

[math man]

well the first two are very easy:
for lnf(x) you draw in y=1. now when f(x)<0 lnf(x) is undefined so you get rid of it. f(x)=0 lnf(x) is still undefined so you put in vertical asymptotes.
Next where y=1 intersects f(x) lnf(x) becomes 0. For 0<f(x)<1 lnf(x) <0 and for f(x)>1 lnf(x) >0. Now you must also remember log graphs grow
pathetically slow, so when you log functions it should be underneath it at all places it is defined and should grow to inifinity slowly like a normal
log graph. The other thing you have to look at for is horizontal asymptotes that are greater than 0, say y= K, K>0, the new vertical asymptote becomes
y=lnK.

Now for e^f(x) the whole curve will be above x-axis. you should use the x-axis as your guideline. for f(x)<0, 0<e^f(x)<1 and for f(x)>0, e^f(x) >1.
where f(x) cuts x-axis becomes 1 on e^f(x). Same thing here applies for horizontal asymptotes, but for all K. A final note is that the e^f(x) should be
on top of f(x) at all points.

Now tan inverse f(x) is not as straight forward as the above too, but the general thing is that it always makes the curve smaller in magnitude. for f(x)>0
arctan f(x) will be under f(x) and for f(x) < 0 arctan f(x) will be less negative

 

[kkmok]

well the first two are very easy:
for lnf(x) you draw in y=1. now when f(x)<0 lnf(x) is undefined so you get rid of it. f(x)=0 lnf(x) is still undefined so you put in vertical asymptotes.
Next where y=1 intersects f(x) lnf(x) becomes 0. For 0<f(x)><1 lnf(x) <0 and for f(x)>1 lnf(x) >0. Now you must also remember log graphs grow
pathetically slow, so when you log functions it should be underneath it at all places it is defined and should grow to inifinity slowly like a normal
log graph. The other thing you have to look at for is horizontal asymptotes that are greater than 0, say y= K, K>0, the new vertical asymptote becomes
y=lnK.

Now for e^f(x) the whole curve will be above x-axis. you should use the x-axis as your guideline. for f(x)<0, 0<e^f(x)><1 and for f(x)>0, e^f(x) >1.
where f(x) cuts x-axis becomes 1 on e^f(x). Same thing here applies for horizontal asymptotes, but for all K. A final note is that the e^f(x) should be
on top of f(x) at all points.

Now tan inverse f(x) is not as straight forward as the above too, but the general thing is that it always makes the curve smaller in magnitude. for f(x)>0
arctan f(x) will be under f(x) and for f(x) < 0 arctan f(x) will be less negative

y = tan^-1 f(x )
range is (-pi/2,pi/2)

the newfunction will approach to pi/2 once f(x ) quite large (something like >10)

but when f(x) between -1 and 1
i think is still significant to notice since it correspond to tan^-1 f(x ) between -pi/4 and pi/4

pi/4 is approximately 0.79

since lim x->0 tan x = x
take inverse both side
lim x->0 x = tan^-1 x

the curve will be quite similar once it approach to 0

but wait........
I did 4unit before do they even put tan^-1 f(x) ?! LOL

ps.
since this is the first time i do tan^-1 f(x ) this crazy function
I actually checked with wolframalpha and it looks my prediction is ok.

http://www.wolframalpha.com/input/?i=tan^-1+%28x^3%29 have a look

 

[ADrew]

Thanks for your responses!