sketching exps. and logs. (1 Viewer)

[hatty]

hey
in the curve sketching section
how do u sketch them e^(x) and ln(x)
they r so damn hard, does n e 1 have n e tips or n e thing that mite helpful

thanks

 

[George W. Bush]

Draw the asymptotes, intercepts and then plot a point x>0 for e^x (try like x=1) or plot a point x>1 for ln(x) (say x=e) and just draw the curve in - generally, unless the shape of your curve is way off, you won't be deducted marks. Marks generally allocated for intercepts, asymptotes, general shape.

 

[kimmeh]

read the y values for the graph.

for logs, you cannot log negative numbers, thus all negative in your graph will be gone. then read your x intercept (where y = 0) . log of 0 will give you undefined .: theres and asymptote at all x intercepts. if you log infinity, then you will still get infinity, but "lower" (what i mean by this is log any big number ln 9999999 = 18). my teacher refers to this function as the "lazy" function because if you look at a normal log function, then it increases very slowly.

for exp: the exp graph is increasin at a fast rate, so e^infinity, will give you infinity , but higher (e^9 = 8103) so thus you start you grpah above that point, higher than that of your original curve. as x -> - infinity, then y-> 0 and e^0 = 1

 

[enter~space~cap]

yeahh...plot the points of several in(x) values, and then e^ln(x) at those points would simply be x

im assuming that u wanna graph y=e^f(x) and f(x) = ln(x)

 

[aud]

Buy a Mathemat... put it over your page and swoop with your pencil, and you have an exp graph... then flip it, do the same, and it's a ln graph

 

[hatty]

sorry aud
i meant e^[f(x)]
like e^x^3+4x+7
some shit like that

 

[George W. Bush]

Well as f(x) -> - infinity, e^f(x) -> 0
When f(x) = 0, e^f(x) = 1
As f(x) -> infinity, e^f(x) -> infinity rising faster than f(x)
etc. look at these main points for both graphs and then just sketch the curve, your sketch doesn't have to be very exact

 

[aud]

Originally posted by hatty
sorry aud
i meant e^[f(x)]
like e^x^3+4x+7
some shit like that

Eww... why would you want to do that? I'd probably just do it point by point, unless directed to show something like stat points or inflections... I know, I'm hopeless...

By the way, I love your fishing sig

 

[Xayma]

When it is like that y=e^f(x) find any zeros of f(x) at that point y=1, when it is < 0, y is less then 1 etc. Stat points will exist when f'(x)=0 (as per the normal) finding points of inflexion is harder.

As you will have
e^f(x)*f"(x)+(f'(x))²e^f(x)

then draw curves between the stat points, and 1's using the inflexion points

and for the log ones it is just reflected around y=x

 

[kimmeh]

hey xayma how did u get the superscripts?

 

[Heinz]

Originally posted by kimmeh
hey xayma how did u get the superscripts?

^text change the [] into < > you can also use sub for subscripts

quick quote xaymas post to see what he did

 

[hatty]

thanks fellas

test is over and i fucked it up
lol

 

[Xayma]

Originally posted by Xayma
As you will have
e^f(x)*f"(x)+(f'(x))²e^f(x)

Just thinking that can be simplified down further to e^f(x)(f"(x)+[f'(x)]²)
So you will have a possible point of inflexion when:
f"(x) is the negative square root of f'(x).

There will also be no need to check for left and right with e^f(x) because e is +ve e^f(x) will always be positive.