Question 1155282
Find the slope of the line tangent to the following at point, where x=2
f(x)=e^(-0.1x^2+2x-3).
--------------
f(x)=e^(-0.1x^2+2x-3)
---
f'(x) = {{{e^(-0.1x^2+2x-3)*(-0.2x + 2)}}}
Sub 2 for x
===================
---
f'(2) = {{{e^(-0.4 + 4 -3)*(-0.8 + 2)}}}
f'(2) = {{{1.2e^(0.6)}}}