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

Question

Christopher Agresta · Accepted 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.

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#.

Charles Arocha · Answer

undefined To find the critical points of $ f(x) = \frac{x}{e^{\sqrt{x}}} $, we first find the derivative of $ f(x) $ and then set it equal to zero to solve for $ x $. The derivative is $ f'(x) = \frac{e^{\sqrt{x}} - \frac{x}{2\sqrt{x}}e^{\sqrt{x}}}{e^{2\sqrt{x}}} $. Setting this equal to zero and simplifying gives us $ e^{\sqrt{x}}(2 - \frac{x}{2\sqrt{x}}) = 0 $. This simplifies to $ 2e^{\sqrt{x}} - \frac{x}{\sqrt{x}}e^{\sqrt{x}} = 0 $, which further simplifies to $ 2e^{\sqrt{x}} - xe^{\sqrt{x}} = 0 $. Factoring out an $ e^{\sqrt{x}} $ gives $ e^{\sqrt{x}}(2 - x) = 0 $. Setting each factor to zero gives us $ e^{\sqrt{x}} = 0 $ (which has no real solutions) and $ 2 - x = 0 $, which gives $ x = 2 $. Therefore, the critical point of $ f(x) $ is $ x = 2 $.