Question: In this problem, you will find the point on the parabola y = x2 + 2, that is closest to the origin. The formula for

 In this problem, you will find the point on the parabolay = x2 + 2, that is closest to the origin. The

formula for the squared distance of a point (x, y) to theorigin is given by 506,30 = Upon substituting y into the squared

In this problem, you will find the point on the parabola y = x2 + 2, that is closest to the origin. The formula for the squared distance of a point (x, y) to the origin is given by 506,30 = Upon substituting y into the squared distance function S(x, y) we get Q06) = Find the x component of the critical point of Q x: Is this critical point a minimum or a maximum (You have 1 attempt for this part of the problem) local maximum, since the second derivative of Q at this point is positive local minimum, since the second derivative of Q at this point is positive local maximum, since the second derivative of Q at this point is negative local minimum, since the second derivative of Q at this point is negative What is the minimal distance to the origin? Consider the function and its derivatives: ex (x2)-ex .- " (x24-x+6)-ex f(XJ= 2, f(X)= f (X): x x3 x4 a) What is the domain of f in interval notation? b} Find the set of all critical points of f. c) Find the intervals of increase and decrease for f. f is increasing on: f is decreasing on: d) Find the intervals of concavity for f. f is concave up on: f is concave down on: e) Find the set of all xvalues of the inflection points of f

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