Q. 31 - Calculus

Question

In Exercises 31–34, 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.

$f\left(x\right)=3x+1$ and the point $(-2,1)$

Step-by-Step Solution

Answer

The point on the graph of the function that is close to the point is $(-\frac{1}{5},\frac{2}{5})$.

Step 1. Given Information.

The function:

$f\left(x\right)=3x+1$

The point:

$(-2,1)$

Step 2. Write the distance between thee function and the point.

The distance between $(-2,1)$ and $(x,y)$ is:
$$\begin{gathered} D\left(x\right)=\sqrt{{(1-y)}^2+{(-2-x)}^2} \\ =\sqrt{{(1-3x-1)}^2+{(-2-x)}^2} \\ =\sqrt{10x^2+4x+4} \end{gathered}$$

Step 3. Find the derivative of the function.

$$\begin{gathered} D\left(x\right)=\sqrt{10x^2+4x+4} \\ D\text{'}\left(x\right)=\frac{1}{2}{(10x^2+4x+4)}^{-\frac{1}{2}}.(20x+4) \\ =\frac{2(10x+2)}{2(\sqrt{10x^2+4x+4)}} \\ =\frac{10x+2}{\sqrt{10x^2+4x+4}} \end{gathered}$$

Step 4. Find the point close to the function and the point.

To find the point that is closest to the function and the point,

$$\begin{gathered} D\text{'}\left(x\right)=0 \\ \frac{10x+2}{\sqrt{10x^2+4x+4}}=0 \\ 10x+2=0 \\ 10x=-2 \\ x=-\frac{2}{10} \\ =-\frac{1}{5} \end{gathered}$$

Step 5. Find y.

Substitute the value of x in the function to get y,

$$\begin{gathered} y=3x+1 \\ =3(-\frac{1}{5})+1 \\ =\frac{-3+5}{5} \\ =\frac{2}{5} \end{gathered}$$

So the closest point is $(-\frac{1}{5},\frac{2}{5})$