Question

question 8
find the point on the curve ( y = sqrt{x} ) that is closest to the point ( (4, 0) ). round answers to 2 decimal places.\
note: ( d = sqrt{(x - x_0)^2 + (y - y_0)^2} )
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Explanation:

Step1: Express the distance formula

Let the point on the curve \(y = \sqrt{x}\) be \((x,\sqrt{x})\). Using the distance formula \(D=\sqrt{(x - 4)^{2}+(\sqrt{x}-0)^{2}}=\sqrt{(x - 4)^{2}+x}\). To simplify calculations, we can minimize \(D^{2}=f(x)=(x - 4)^{2}+x\). Expand \(f(x)\): \(f(x)=x^{2}-8x + 16+x=x^{2}-7x + 16\).

Step2: Find the derivative of \(f(x)\)

Differentiate \(f(x)\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(f^\prime(x)=2x-7\).

Step3: Find the critical points

Set \(f^\prime(x) = 0\), so \(2x-7 = 0\). Solve for \(x\): \(x=\frac{7}{2}=3.5\).

Step4: Check the second - derivative (optional, for confirmation of minima)

Differentiate \(f^\prime(x)\) to get \(f^{\prime\prime}(x)=2>0\). Since the second - derivative is positive, \(x = 3.5\) gives a minimum of \(f(x)\).

Step5: Find the \(y\) - value

When \(x = 3.5\), \(y=\sqrt{3.5}\approx1.87\).

Answer:

\((3.50,1.87)\)