You need to have patience and not post the same stuff umpteen times. The tutors who concentrate on this type of math are not online yet.
Please do not keep posting the same question over and over. It's considered spamming and could get you banned from posting here.
at (1 , 1) dy/dx = slope = 1*1/2 = .5
y = .5 x + b is tangent for some b
put in (1 , 1 )
1 = .5 + b
b = .5
so tangent at (1,1) is
y = .5 x + .5
at x = 1.2
y = .5(1.2) + .5 = 1.1
dy/y = (1/2) x dx
ln y = (1/4) x^2 + C
y = k e^(x^2/4)
1 = k e^(1/4)
1 = 1.28 k
k = .779
y = .779 e^(x^2/4)
at x = 1.2
y = .779 e^(1.44/4)
y = .779 * 1.433
y = 1.116
Do not panic. Plug and chug.
I think the whole non panicing ship sailed a long time ago haha. Sorry and thanks for the help!
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