Asked by Linda
Find a curve y=f(x) with the following properties (i) y"=6x and (ii) its graph passes through the point (1,1) and has a horizontal tangent there.
Answers
Answered by
Jai
y'' = 6x
Integrate:
y' = 3x^2 + c1
Since it was said that it has a horizontal tangent at (1,1), then we let y' (the slope) be equal to 0:
0 = 3x^2 + c1
0 = 3(1)^2 + c1
0 = 3 + c1
c1 = -3
Integrate y' further (and substitute c1 = -3):
y' = 3x^2 - 3
y = x^3 - 3x + c2
It passes through the point (1,1). Thus,
1 = (1)^3 - 3(1) + c2
1 = 1 - 3 + c2
1 = -2 + c2
c2 = 3
Therefore,
y = x^3 - 3x + 3
hope this helps~ `u`
Integrate:
y' = 3x^2 + c1
Since it was said that it has a horizontal tangent at (1,1), then we let y' (the slope) be equal to 0:
0 = 3x^2 + c1
0 = 3(1)^2 + c1
0 = 3 + c1
c1 = -3
Integrate y' further (and substitute c1 = -3):
y' = 3x^2 - 3
y = x^3 - 3x + c2
It passes through the point (1,1). Thus,
1 = (1)^3 - 3(1) + c2
1 = 1 - 3 + c2
1 = -2 + c2
c2 = 3
Therefore,
y = x^3 - 3x + 3
hope this helps~ `u`
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