Asked by Sam
Find the equation of the tangent to the curve y=2-e^-x at the point where x=1.
Can someone please do this step by step and explain?
Can someone please do this step by step and explain?
Answers
Answered by
Reiny
Point of contact :
when x=1 , y = 2 - e^-1 = 2 - 1/e
= (2e-1)/e
dy/dx = e^-x
so when x=1 , slope = e^-1 = 1/e
using y = mx + b
(2e-1)/e = (1/e)(1) + b
b = (1-e)/e
so y = (1/e)x + (1-e)/e
when x=1 , y = 2 - e^-1 = 2 - 1/e
= (2e-1)/e
dy/dx = e^-x
so when x=1 , slope = e^-1 = 1/e
using y = mx + b
(2e-1)/e = (1/e)(1) + b
b = (1-e)/e
so y = (1/e)x + (1-e)/e
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