Find the function whose tangent line has a slope e^(−2x) for each x and passes through the point (0, −3).?
so this is a review question for a quiz and on the review sheet, the answer was f(x) = ((-1/2)(e)^(-2x))-(5/2)
I just want to know how my prof got this answer
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Given: y' = e^(-2x), (x, y) = 0, -3)
Integrating:
y = ∫y'dx = ∫e^(-2x)dx = -e^(-2x)/2 + C
Solving for C with (x,y) = (0, -3)
-3 = -e^(0)/2 + C = -1/2 + C
C = -3 + 1/2 = -5/2
y = -(1/2)e^(-2x) - 5/2