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solving this..

I need to find if this:
http://i47.tinypic.com/6xu68h.jpg
has any stationary points but I am not sure what to do next.
Any hints? xx
Reply 1
Avatar for Simba
Simba
16
Well, we know e^(2x) > 0, and e^x > 0 for all real x, so what can you conclude about 2e^(2x)*e^x? And how does your conclusion contradict it being = 0 :smile: ?
Simba
Well, we know e^(2x) > 0, and e^x > 0 for all real x, so what can you conclude about 2e^(2x)*e^x? And how does your conclusion contradict it being = 0 :smile: ?

that it has a stationary point? Im very confused. I also need to state the coordinates of the stationary point xx
anyone?
Reply 4
Avatar for cheena
cheena
3
is that 2*e^(2x) * e^(x) =0 ?

could you write it out
Reply 5
Avatar for TheNack
TheNack
That picture isn't very clear, is this the expression you posted:

2e2xex=02e^{2x}e^x=0

If so, then the left side will simplify to:

2e3x2e^{3x}

Can you be clearer about what you're asking? Are you asking whether A) y=2e3xy=2e^{3x} has any stationary points, or B) that the original expression (2e3x)(2e^{3x}) is dy/dxdy/dx and youre trying to find stationary points by finding where it's equal to 0? In either case, think about when you have a derivative of the form aebxae^{bx} (where aa and bb are constants), how could it be equal to 0, if at all? (i.e. is exe^x ever 0?)
Reply 6
Avatar for Larry<3
Larry<3
Find dy/dx and set it equal to zero.

:smile:

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