What are the critical points of #f(x) = sqrt(sqrt(x)e^(sqrtx)-sqrtx)-sqrtx#?

1 Answer
Aug 23, 2017

Answer:

There are none.

Explanation:

#f(x)=(x^(1/2)e^(x^(1/2))-x^(1/2))^(1/2)-x^(1/2)#

#f'(x)=1/2(x^(1/2)e^(x^(1/2))-x^(1/2))^(-1/2)d/dx(x^(1/2)e^(x^(1/2))-x^(1/2))-1/2x^(-1/2)#

#color(white)(f'(x))=1/2(x^(1/2)e^(x^(1/2))-x^(1/2))^(-1/2)(1/2x^(-1/2)e^(x^(1/2))+x^(1/2)e^(x^(1/2))(1/2x^(-1/2))-1/2x^(-1/2))-1/2x^(-1/2)#

Factoring #1/2x^(-1/2)# from the first term:

#f'(x)=1/4x^(-1/2)(x^(1/2)e^(x^(1/2))-x^(1/2))^(-1/2)(e^(x^(1/2))+x^(1/2)e^(x^(1/2))-1)-1/2x^(-1/2)#

Note that #(x^(1/2)e^(x^(1/2))-x^(1/2))^(-1/2)=(x^(1/2)(e^(x^(1/2))-1))^(-1/2)=x^(-1/4)(e^(x^(1/2))-1)^(-1/2)#.

#f'(x)=1/4x^(-3/4)(e^(x^(1/2))-1)^(-1/2)(e^(x^(1/2))+x^(1/2)e^(x^(1/2))-1)-1/2x^(-1/2)#

#color(white)(f'(x))=(e^(x^(1/2))+x^(1/2)e^(x^(1/2))-1)/(4x^(3/4)(e^(x^(1/2))-1))-1/(2x^(1/2))#

#color(white)(f'(x))=(e^(x^(1/2))+x^(1/2)e^(x^(1/2))-1)/(4x^(3/4)(e^(x^(1/2))-1))-(2x^(1/4)(e^(x^(1/2))-1))/(4x^(3/4)(e^(x^(1/2))-1))#

#color(white)(f'(x))=(e^sqrtx+sqrtxe^sqrtx-1-2root4xe^sqrtx-2root4x)/(4root4(x^3)(e^sqrtx-1))#

Graph this to find its #0#s.

graph{(e^(x^(1/2))+x^(1/2)e^(x^(1/2))-1)/(4x^(3/4)(e^(x^(1/2))-1))-(2x^(1/4)(e^(x^(1/2))-1))/(4x^(3/4)(e^(x^(1/2))-1)) [-10, 10, -5, 5]}

There are no #0#s and the domain of #f'# is the same as that of #f#, so the function has no critical values.