What is the slope of #f(x)=xe^(x-x^2) # at #x=-1#?

Question

What is the slope of #f(x)=xe^(x-x^2)   # at #x=-1#?

Emilia Aguilar · Accepted Answer

The slope of the function at x=-1 is #-2/e^2#

In order to find the slope of a function,we have to firstly determine its derivative. so, #d/dx(xe^(x-x^2))#

Applying product rule,

#(f.g)'=f'.g+g.f'#

so, #f=x,\g=e^{x-x^2}#

#\frac{d}{dx}(x)e^{x-x^2}+\frac{d}{dx}(e^{x-x^2})x#

We know... #d/dx(x)=1#

and, #\frac{d}{dx}(e^{x-x^2})# Applying chain rule, #\frac{df(u\)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}#

let #x-x^2=u#

#\frac{d}{du}(e^u\)\frac{d}{dx}\(x-x^2)#

We get,

#(e^u)(1-2x)#

Substituting back #x-x^2=u#,

=#e^{x-x^2}(1-2x)#

= #1e^{x-x^2}+e^{x-x^2}(1-2x)x#

Simplifying it,

#e^{x-x^2}(-2x^2+x+1)#

Finally when #x=-1# # e^{(-1)-(-1)^2}[-2(-1)^2+(-1)+1]# = #e^-2 . (-2)#= #-2/e^2#

Cameron Alexander · Answer

undefined To find the slope of the function $ f(x) = xe^{x-x^2} $ at $ x = -1 $, we first need to find its derivative, then evaluate it at $ x = -1 $.

$ f'(x) = e^{x-x^2} + xe^{x-x^2}(1-2x) $

Now, evaluate $ f'(-1) $:

$ f'(-1) = e^{-1-(-1)^2} + (-1)e^{-1-(-1)^2}(1-2(-1)) $

$ f'(-1) = e^{-1-1} - e^{-1-1}(1+2) $

$ f'(-1) = e^{-2} - 3e^{-2} $

$ f'(-1) = (1 - 3)e^{-2} $

$ f'(-1) = -2e^{-2} $

Therefore, the slope of $ f(x) = xe^{x-x^2} $ at $ x = -1 $ is $ -2e^{-2} $.