differentiation of exponentials question (1 Viewer)

[kloudsurfer]

Hey,
A couple of questions like this have popped up and i dont know how to do them. Could anyone help me?

Find the stationary point on the curve y=xe^x and determine its nature.

Thanks in advance.

 

[imaginarylife]

y = x e^x

using product rule, dy/dx = x e^x + e^x
stationary point when dy/dx = 0
e^x(x +1) = 0
e^x cannot = 0, therefore stat. pt. when x = -1

to determine its nature use second derivative or test points either side
by product rule,
y"= e^x(x+1) + e^x
= e^x(x+2)
for x= -1, y" is positive
therefore it is concave up
therefore min. turning point at (-1, -e^-1)

 

[kloudsurfer]

Ohhh I get it now. Thanks.

cept ive got one question

e^x cannot = 0

why not?

 

[ellen.louise]

Because x=0 is the assymptote for the curve: no value you ever put in for x in y=e^x will ever equal 0 or a neg. number.

 

[jb_nc]

why not?

As x -> minus infinity, y -> a tiny positive number. It's never 0.

 

[aussiechica7]

hehe i think something like that question was in my methods exam (2003)

 

[kloudsurfer]

Because x=0 is the assymptote for the curve: no value you ever put in for x in y=e^x will ever equal 0 or a neg. number.

Oh right...

Ok really I have no idea.

Why is x=0 the asymptote again?

I think I have to go over some yr 11 stuff...

 

[idling fire]

Because x=0 is the assymptote for the curve

Uhm... no... incorrect.
The second part was right though:

no value you ever put in for x in y=e^x will ever equal 0 or a neg. number.

x CAN equal zero. Domain for x is all real numbers.
y CANNOT equal zero, because no value of x can ever make the RHS zero.
Therefore asymptote is y=0 (the x-axis) for y=e^x.

 

[ellen.louise]

Oh right...

Ok really I have no idea.

Why is x=0 the asymptote again?

I think I have to go over some yr 11 stuff...

because it just is...
um...
oooh! got it!!!!

because if y=e^x, x=logey. and.... and.... someone help here!!!

NO I WAS WRONG!!! it's for the log curve that x=0 is the assymptote. (I probably should have checked my trusty template earlier)

for y=e^x, y=0 is the assymptote, because x=log e y and you cannot find the log of a negative number, and you can't multiply anything by a power that comes to 0.

 

[kloudsurfer]

for y=e^x, y=0 is the assymptote, because x=log e y and you cannot find the log of a negative number, and you can't multiply anything by a power that comes to 0.

Still confused.

Will this make sense further into the topic? Because we havent even learnt about logs yet, we are only up to differentiation of exponentials.

 

[bos1234]

You need to learn logs and it will be MUCh easier

 

[kloudsurfer]

You need to learn logs and it will be MUCh easier

Ahhh ok. I knew there was something I hadnt learnt in there yet.

Thanks

 

[bos1234]

To find the x-intercept sub in y=0 and solve

y=e^x
0=e^x


if you draw the exponential curve y = e^x you will see that it never touches or goes below the x-axis. So the x-axis is an asymptote so the above equation cannot be solved

-------------------------------------------
Also you can do this by above to see it has no soln.

----------------------------------------------

hope u understood

thanks bye