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

Question

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

Isabella Ackerman · Accepted Answer

The slope at #x=-1# is #-2e^2# The slope of #f(x)# at #x=-1# is determined by differentiating the function at #x=-1# that is by computing #color(red)(f'(-1))# Differentiate #f(x)# is determined by using the quotient rule differentiation Let #u(x)=x+2 and v(x)=e^(x-x^2)# Then #f(x)=(u(x))/(v(x))# #color(red)(f'(x)=(color(blue)(u'(x))*v(x)-color(green)(v'(x))*u(x))/(v^2(x))# Let us determine #color(blue)(u'(x))# and #color(green)(v'(x))# #u(x)=x+2 rArr color(blue)(u'(x)=1)# # v(x)=e^(x-x^2)# is a composite of two function ,the exponential and polynomial so its derivative is determined by using chain rule. Let #color(brown)(g(x)=e^x and h(x)=x-x^2)# #color(brown)(v(x)=g(h(x)))# Then the derivative of the composite function is#color(green)(v'(x)=g'(h(x))*h'(x))# #g(x)=e^xrArrg'(x)=e^x# #g'(h(x))=e^(h(x))# #color(green)(g'(h(x))=e^(x-x^2))# # h(x)=x-x^2rArrcolor(green)(h'(x)=1-2x)# #color(green)(v'(x)=e^(x-x^2)*(1-2x))# #color(red)(f'(x)=(color(blue)(u'(x))*v(x)-color(green)(v'(x))*u(x))/(v^2(x))# #color(red)(f'(x))=(color(blue) 1(e^(x-x^2))-color(green)(e^(x-x^2)(1-2x)) *(x+2))/ (e^(x-x^2))^2# #color(red)(f'(x))=((e^(x-x^2))-(e^(x-x^2))(1-2x) *(x+2))/ (e^(x-x^2))^2# #color(red)(f'(x))=(e^(x-x^2)[1-(1-2x) *(x+2)])/ (e^(x-x^2))^2# Simplifying the quotient by #e^(x-x^2)# #color(red)(f'(x))=(1-(1-2x) *(x+2))/ (e^(x-x^2))# Then, #color(red)(f'(-1))=(1-(1-2(color(red)(-1))) *(color(red)(-1)+2))/ (e^(color(red)(-1)-color(red)((-1))^2))# #color(red)(f'(x))=(1-(1+2) *(+1))/ (e^(-2))# #color(red)(f'(x))=(-2)/ (e^(-2))# #color(red)(f'(x))=-2e^2#

Charles Arocha · Answer

undefined To find the slope of the function $ f(x) = \frac{x + 2}{e^{x - x^2}} $ at $ x = -1 $, you need to compute the derivative of the function and then evaluate it at $ x = -1 $. The derivative of the function $ f(x) $ with respect to $ x $ is:

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

Now, substitute $ x = -1 $ into $ f'(x) $ to find the slope at $ x = -1 $:

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

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

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

$$ f'(-1) = \frac{3e^{-2} - 3e^{-2}}{e^{-4}} $$

$$ f'(-1) = \frac{0}{e^{-4}} $$

$$ f'(-1) = 0 $$

Therefore, the slope of the function $ f(x) = \frac{x + 2}{e^{x - x^2}} $ at $ x = -1 $ is $ 0 $.