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:
[tex]\begin{gathered} d=\sqrt{(x-0)^2+(x^2+1-2)^2} \\ d=\sqrt{x^2+(x^2-1)^2} \\ d=\sqrt{x^2+x^4-2x^2+1} \\ d=\sqrt{x^4-x^2+1} \end{gathered}[/tex]
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:
[tex]\begin{gathered} d^{\prime}(x)=\frac{4x^3-2x}{2\sqrt{x^4-x^2+1}} \\ \frac{2x^3-x}{\sqrt{x^4-x^2+1}}=0 \\ \frac{x(2x^2-1)}{\sqrt{x^4-x^2+1}}=0 \\ x_1=-\frac{1}{\sqrt{2}} \\ x_2=0 \\ x_3=\frac{1}{\sqrt{2}} \\ d(-\frac{1}{\sqrt{2}})=\frac{1}{2} \\ d(0)=1 \\ d(\frac{1}{\sqrt{2}})=\frac{1}{2} \end{gathered}[/tex]
Then we have:
[tex]\begin{gathered} y_1=(-\frac{1}{\sqrt{2}})^2+1=\frac{3}{2} \\ y_3=(\frac{1}{\sqrt{2}})^2+1=\frac{3}{2} \end{gathered}[/tex]
Anser:
(-0.71,1.5), (0.71,1.5)
