Question

Find the point on the graph of the function$f$ that is closest to the point $(a,b)$ by minimizing the square of the distance from the graph to the point.

$f\left(x\right)=x^2-2x+1$and the point$(1,2)$

Short answer

Ans: The point is$\left(\frac{2+\sqrt{6}}{2},1.75\right);\left(\frac{2-\sqrt{6}}{2},1.75\right)$

Step-by-step solution

Step 1.  Given information.

given expression,

$f\left(x\right)=x^2-2x+1$

Step 2. The objective is to find a point closest to the function that is closest to the point by minimizing the graph from the distance to the point.

Let, the point be $(x,y)$.

The distance between $(1,2)$) and $(x,y)$is,

$\begin{array}{r}D\left(x\right)=\sqrt{(2-y{)}^2+(1-x{)}^2}\\ D\left(x\right)=\sqrt{{\left(2-x^2+2x-1\right)}^2+(1-x{)}^2}\\ D\left(x\right)=\sqrt{{\left(-x^2+2x+1\right)}^2+1-2x+x^2}\\ D\left(x\right)=\sqrt{x^4+4x^2+1-4x^3+4x-2x^2+1-2x+x^2}\\ D\left(x\right)=\sqrt{x^4-4x^3+3x^2+2x+2}\end{array}$

Now,

$\begin{array}{r}D\left(x\right)=\sqrt{x^4-4x^3+3x^2+2x+2}\\ D^{\mathrm{\prime }}\left(x\right)=\frac{2x^3-6x^2+3x+1}{\sqrt{x^4-4x^3+3x^2+2x+2}}\end{array}$

Step 3. Again, 

$\begin{array}{r}{D}^{\mathrm{\prime }}\left(x\right)=0\\ \frac{2x^3-6x^2+3x+1}{\sqrt{x^4-4x^3+3x^2+2x+2}}=0\\ 2x^3-6x^2+3x+1=0\\ x=1,\frac{2\pm \sqrt{6}}{2}\end{array}$

Putting $\frac{2\pm \sqrt{6}}{2}\text{in}f\left(x\right)=x^2-2x+1$

$\begin{array}{l}f\left(x\right)=1.75\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[x=\frac{2+\sqrt{6}}{2}\right]\\ f\left(x\right)=1.75\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[x=\frac{2-\sqrt{6}}{2}\right]\end{array}$

Therefore, the point is$\left(\frac{2+\sqrt{6}}{2},1.75\right);\left(\frac{2-\sqrt{6}}{2},1.75\right)$