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

Question

Emma Aldridge · Accepted Answer

#x=0#

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

Through the chain rule:

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

Then:

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

Factoring from the final parentheses:

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

Rewriting:

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

#f'(x)=(e^sqrtx-1)/(4sqrtxsqrt(e^sqrtx-sqrtx))#

If we want to find critical point, we need to find when #f'(x)=0# or when #f'# is undefined but #f# is defined.

Setting #f'(x)=0# gives #e^sqrtx-1=0=>e^sqrtx=1=>x=0#.

This is also when #f'# is undefined, since #sqrtx# is in the denominator. We also see that #sqrt(e^sqrtx-sqrtx)# is never undefined, as #e^sqrtx>sqrtx# for all #x>=0#.

Thus the only critical point is #x=0#. It's also one of the function's endpoints.

Luke Alvarez · Answer

undefined To find the critical points of $ f(x) = \sqrt{e^{\sqrt{x}} - \sqrt{x}} $, we first need to find its derivative, set it equal to zero, and solve for $ x $.

$ f'(x) = \frac{1}{2\sqrt{e^{\sqrt{x}} - \sqrt{x}}} \cdot \left(\frac{1}{2\sqrt{x}}\cdot e^{\sqrt{x}} - \frac{1}{2\sqrt{x}}\right) $

Setting $ f'(x) = 0 $, we get:

$ \frac{1}{2\sqrt{e^{\sqrt{x}} - \sqrt{x}}} \cdot \left(\frac{1}{2\sqrt{x}}\cdot e^{\sqrt{x}} - \frac{1}{2\sqrt{x}}\right) = 0 $

Since the first factor cannot be zero, we focus on the second factor:

$ \frac{1}{2\sqrt{x}}\cdot e^{\sqrt{x}} - \frac{1}{2\sqrt{x}} = 0 $

Solving this equation yields:

$ e^{\sqrt{x}} - 1 = 0 $

$ e^{\sqrt{x}} = 1 $

$ \sqrt{x} = 0 $

$ x = 0 $

So, the only critical point of $ f(x) $ is $ x = 0 $.