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## Homework Statement

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

My approach:

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

When dy/dx=0 For stationary point.

Thus, through algebraic manipulation,

[tex] ln(2)-0.5x=ln(x) [/tex]

Since,[tex] ln(2)-0.5x [/tex] is a deacreasing function ,

and [tex] ln(x) [/tex] 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.

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