- #1
icystrike
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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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