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# Find the point on the curve y = sqrt(x) that is closest to the point (3, 0)

**Solution:**

Given, the curve is y = √x --- (1)

If D is the distance from (x, y) to the point (3, 0), by Pythagorean distance formula

D^{2} = (x - 3)^{2} + y^{2}

= (x - 3)^{2} + (√x)^{2}

= x^{2 }- 6x + 9 + x

D^{2} = x^{2} - 5x + 9

On differentiating,

2D.D’ = 2x - 5

D’ = \(\frac{2x-5}{2D}\)

\(D'=\frac{2x-5}{2\sqrt{x^{2}-5x+9}}\)

\(D'=(2x-5).\frac{1}{2(x^{2}-5x+9)^{\frac{1}{2}}}\)

Now, \((2x-5).\frac{1}{2(x^{2}-5x+9)^{\frac{1}{2}}}=0\)

⇒ 2x - 5 = 0

⇒ 2x = 5

⇒ x = 5/2

This is known as the critical value and it represents the x-value for which the function is minimised.

Put x = 5/2 in (1)

y = √5/2

y = 1.58

Therefore, the point on the curve y = √x closest to the point (3, 0) is (5/2, 1.58).

## Find the point on the curve y = sqrt(x) that is closest to the point (3, 0)

**Summary:**

The point on the curve y = √x that is closest to the point (3, 0) is (5/2, 1.58).

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