Find Nature of Stationary Point of y=e^(x/2)-ln(x)

[icystrike]

Homework Statement

Show that there is only one stationary point of the curve y=e^{x/2} - ln (x), where x>0 and determine the nature of the stationary point.


My approach:

dy/dx = 0.5e^{x/2} - 1/x

When dy/dx=0 For stationary point.

Thus, through algebraic manipulation,

ln(2)-0.5x=ln(x)

Since,ln(2)-0.5x is a deacreasing function ,

and ln(x) is a increasing function with an horizontal asymptote of x=0.

Therefore , there is only an intersection at coordinate (1.13429 , 0.126007) by Newton-raphson method.

Correct me if I am wrong or please show me a prove that is more elegant.

 

[CFDFEAGURU]

You are trying to find a zero slope (where the derivative vanishes) not a root. I haven't worked this out yet myself, but your methodology looks correct to me.

Thanks
Matt

 

[icystrike]

You are trying to find a zero slope (where the derivative vanishes) not a root. I haven't worked this out yet myself, but your methodology looks correct to me.

Thanks
Matt

A million thanks (=
Have a great day!

 

[CFDFEAGURU]

Remember you can always graph your original function before you perform any differentiation to see where there might be a slope of zero. This can be very helpful if you have a difficult function to diferentiate.

Thanks
Matt

 

[icystrike]

Remember you can always graph your original function before you perform any differentiation to see where there might be a slope of zero. This can be very helpful if you have a difficult function to diferentiate.

Thanks
Matt

Noted (=

 

[Mentallic]

Oh wow... this looks quite elegant, if it's legitimate
I am completely lost on what you've done, and after seeing that for this equation, solving for x when dy/dx=0 cannot be done as simply as through my usual routine, I'm curious to what you've done.
If you don't mind, could you please elaborate on what the "through algebraic manipulation" actually involved and thus how you resulted in the next line?

dy/dx = 0.5e^{x/2} - 1/x


When dy/dx=0 For stationary point.

Thus, through algebraic manipulation,

ln(2)-0.5x=ln(x)