Is #f(x)=1/e^x# increasing or decreasing at #x=0#?

Question

Emilia Aguilar · Accepted Answer

Decreasing

First, recognize that #f(x)# can be written as #f(x)=e^-x# We use the sign of the first derivative to determine whether this is increasing or decreasing at a given point. Now, to find the derivative, we will use the chain rule. In the case of an exponential function with base #e#, the chain rule states that #d/dx(e^u)=e^u*u'# Here, #u=-x#, so #f'(x)=e^-x*d/dx(-x)=e^-x*(-1)=-e^-x# Find the sign of the derivative at #x=0#: #f'(0)=-e^-0=-e^0=-1# Recall that anything (other than #0#) to the #0# power is #1#. Since #-1<0#, the function is decreasing at #x=0#. A graph of the original function can be examined here: chart{e^-x [-10, 15.31, -4.05, 8.6]}

Emma Aldridge · Answer

undefined To determine if the function $ f(x) = \frac{1}{e^x} $ is increasing or decreasing at $ x = 0 $, we need to examine the sign of its derivative at that point.

First, let's find the derivative of $ f(x) = \frac{1}{e^x} $ using the chain rule:

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

Now, evaluate $ f'(0) $:

$$ f'(0) = -\frac{1}{e^{2 \cdot 0}} = -\frac{1}{e^0} = -1 $$

Since $ f'(0) = -1 $, which is negative, the function $ f(x) = \frac{1}{e^x} $ is decreasing at $ x = 0 $.