Daily Graph 28-12-08

But function is zero where x=ln(-x) ... i.e for some negative x ≈-.567

Doing it without transformatons...first time...

lyl=lx+e^xl

y=x+e^x
Dom=R
Range=R
y'=1+e^x

But e^x>0 always
So y'>0
=>y is increasing function for all xεR

y"=e^x>0 always
So concave graph

X intercept => y=0 => e^x=-x
We cant predict the accurate value so,we can predict its range of values
For x>0 e^x≠-x becoz e^x,x>0
So it will lie in (-∞,0)
for xε(-∞,-1) Range of values of y=x >1
So intersection must take place b/w xε(-1,0)
Y intercept => x=0 =>y=1

So graph y=x+e^x will be..

Now y=lx+e^xl
We know expression x+e^x vanishes somewhere b/w xε(-1,0)
So for xε(-∞-1) y=-(x+e^x)
xε(-1,+∞) y=x+e^x

Analysing for this case, y=-(x+e^x)
y'<0
y"<0
so convex decreasing

So graph of y=lx+e^xl will be

Now continuing for lyl=lx+e^xl
For y<0 -(y)=lx+e^xl
y>0 y=lx+e^xl

From this four cases arise,
y<0 xε(-∞-1) y=x+e^x
y<0 xε(-1,+∞) y=-(x+e^x)
y>0 xε(-∞-1) y=-(x+e^x)
y>0 xε(-1,+∞) y=x+e^x

So the final graph will be..

edited

there are more issues... when x--> -∞ y-->∞ which you havent shown

write the sign scheme of (e^x+x)

and that will help

I only use it plot good quality graphs with good details..

never use in getting answer..

I had this software frm the begining but no one has ever seen this. in the forum as i never used it...

Its Microsoft Maths

But function is zero somewhere

where x=ln(-x)

...contd

try for negative....

Non calculus method...

First y=|x+e^x|

for x≥0

it is the sum of graph of y=x
and y=e^x

here e^x will dominate the nature of curve.... :)

so for y=|x+e^x| for x≥0
it is almost similar to e^x
with a slight difference

blue is e^x+x
green is e^x
and pink is x

so for |y|=|e^x+x| for x≥0 it is clear( mirror image about y=0...)

...contd..

yup that one in negative that almost straight line will have slope nearly 1...

and positive part almost similar to e^x

and a rough touch between where f(x)=0 and x=0..

i mean roughly mila do... dodno ko

:)

Wait i am now posting a little nice picture with some more detalis :)

i am very doubtful if i am correct ...