- (C) At a point of intersection, y ′ = x + y and x + y = 0. So y ′ = 0, which implies that y has a
critical point at the intersection. Since y ′′ = 1 + y ′ = 1 + (x + y) = 1 + 0 = 1, y ′′ > 0 and the
function has a local minimum at the point of intersection. [See Figure N9–5, showing the slope
field for y ′ = x + y and the curve y = ex − x − 1 that has a local minimum at (0, 0).]
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(dmanu)
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