Find point of f(x)

Question

Find point on f(x)=x^2 that is closest to point (4,0).

Answer 1

Take any point on f(x) = x² as (x, x²).

The line through the closest point (x, x²) to (4, 0) will have slope

equal to (x² − 0)/(x − 4) or x²/(x − 4).

The line through (x, x²) tangent to the curve f(x) = x² will have slope

(or first derivative) equal to 2x and will be perpendicular to the line

through the closest point (x, x²) to (4, 0). That is to say, 2x and

x²/(x − 4) are negative reciprocals or x²/(x − 4) = -1/(2x).

Then x²/(x − 4) = -1/(2x) can be taken to 4 − x = 2x³.

Next write 2x³ + x − 4 = 0 and use Newton's Method Of Root Approximation.

Build the expression x − (2x³ + x − 4)/(6x² + 1) with 6x² + 1 the first derivative of

2x³ + x − 4.

A programmable graphing calculator shows that a line from (1, 1) to (4,0) appears nearly

perpendicular (or normal) to the curve of f(x) = x².

You would then program the calculator to evaluate the expression x − (2x³ + x − 4)/(6x² + 1)

starting with x = 1. Each value of the expression obtained is fed back into the expression as

"the new x" until the results start to repeat. Here, values will not change past 1.128173898,

which squares to 1.272776344.

(1.128173898, 1.272776344) is then the point sought.