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Find the points on the curve y=x^2+1 closest to the point (0,2). Round to the nearest two decimal places. Write the point with the smaller x value first.

Find the points on the curve y=x^2+1 closest to the point (0,2). Round to the nearest-example-1

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We want to find the points on the curve y = x² + 1 that are closest to the point (0,2).

For any given value x, its correspondent point on the curve will be (x,x² + 1)

The distance between (0,2) and (x,x² + 1) is given by:


\begin{gathered} d=√((x-0)^2+(x^2+1-2)^2) \\ d=√(x^2+(x^2-1)^2) \\ d=√(x^2+x^4-2x^2+1) \\ d=√(x^4-x^2+1) \end{gathered}

Now, to find the values of x associated to the closest point of (0,2), we must derivate d and find x for d'(x) = 0:


\begin{gathered} d^(\prime)(x)=(4x^3-2x)/(2√(x^4-x^2+1)) \\ (2x^3-x)/(√(x^4-x^2+1))=0 \\ (x(2x^2-1))/(√(x^4-x^2+1))=0 \\ x_1=-(1)/(√(2)) \\ x_2=0 \\ x_3=(1)/(√(2)) \\ d(-(1)/(√(2)))=(1)/(2) \\ d(0)=1 \\ d((1)/(√(2)))=(1)/(2) \end{gathered}

Then we have:


\begin{gathered} y_1=(-(1)/(√(2)))^2+1=(3)/(2) \\ y_3=((1)/(√(2)))^2+1=(3)/(2) \end{gathered}

Anser:

(-0.71,1.5), (0.71,1.5)

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