#1 still does not contain the point (1,1)
#2 I don't think there is such an A and B. There is a solution involving sin(√2 x), but that's not gonna fit the requirements.
#3
y = e^x - e^2x
y' = e^x - 2e^2x
= e^x(1-2e^x)
so, y'=0 when
e^x=0 (never)
1-2e^x = 0
x = -ln2
1. At what point does the normal line to the curve x^2 - XY + Y^2 = 3 at the point (-1,1) intersect the curve again?
2. Find the constants A, B so that if Y=A*sin X + B cos X, then Y satisfies the differential equation Y" + 2Y = 0.
3. Find the points on he graph of Y = e^x - e^2x at which that tangent line is horizontal.
Please help, I'm not sure how to do these.
1 answer
1 answer