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

1 Answer
Oct 19, 2017

Answer:

Use the quotient rule and the definition of the derivative of the #e^x# function (and/or the chain rule). x=4 is the only critical pt.

Explanation:

The critical points will occur where #f'(x) = 0#. To find f'(x) we will need the quotient rule and the definition of the derivative of #e^x#. The former states that given #f(x)=(g(x))/(h(x)), f'(x) = (g'(x)h(x) - g(x)h'(x))/(h^2(x))#. Using that here...

#g(x) = x, g'(x) = 1, h(x) = e^sqrt(x), h'(x) = d/dx (e^sqrt(x))#

The definition of #d/dxe^x=dx*e^x#. THat in mind, we have #e^sqrt(x) = e^(x^(1/2))#, and #d/dx x^(1/2) = 1/(2x^(1/2)) = 1/(2sqrtx)#. Then #d/dx e^sqrtx = 1/(2 sqrtx)e^sqrtx#, and thus...

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

Recall that #(a^b)^c = a^(b*c)...#

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

#= (e^sqrtx (1 - sqrtx/2))/e^(2sqrtx)#

This can be further simplified if we wish, since #a^b/a^c = a^(b-c)...#

#= (1 - sqrtx/2)/e^sqrtx#.

To find the critical points, we set this equal to 0...

#(1-sqrtx/2)/e^sqrtx = 0 -> 1 - sqrtx/2 = 0 -> 1 = sqrtx/2 -> sqrtx = 2 -> x = 4#

Thus, there is one real-valued critical point, and it occurs at #x=4#

graph{x/(e^sqrtx) [-10, 10, -5, 5]}

The graph bears this out, as the maximum for the function appears to occur at #x=4#.