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Chapter 9: Problem 31

Find the points on the graph of the function that are closest to the given point. \(f(x)=\sqrt{x}, \quad(4,0)\)

Short Answer

Expert verified
The points on the graph of the function \(f(x)=\sqrt{x}\) that are closest to the point (4,0) are obtained by following the above steps. The \(x\) value corresponding to the minimum distance squared gives the x-coordinate and its square root gives the y-coordinate. The coordinates \((x,\sqrt{x})\) give the required points.

Step by step solution

01

Formulate the distance squared

The distance squared \(D^2\) from a point \((x,\sqrt{x})\) on the graph to the point (4,0) can be obtained from the distance formula and it is \(D^2=(x-4)^2+(\sqrt{x}-0)^2=x^2-8x+16+x=x^2-8x+16+x\)
02

Differentiate the distance squared

The task is to minimise the square of the distance. To find the minimum point of a function, first find its derivative and equate it to zero. The derivative of \(D^2\) with respect to \(x\) is obtained as \(2x-8 +\frac{1}{2\sqrt{x}}\)
03

Solve the derivative equals zero

Setting that derivative equal to zero gives the possible x-values at which \(D^2\) has a local minimum: \(2x-8 +\frac{1}{2\sqrt{x}}=0\). Solve this equation for \(x\)
04

Check for minimum

Verify that the solution gives a minimum value by checking the second derivative or applying the first derivative test. Here, it would be easier to use the second derivative test. Compute the second derivative and substitute the obtained value of \(x\) in it. If the result is positive, then by the second derivative test, the original function has a local minimum at that \(x\) value
05

Find the point

Substitute the \(x\) value that makes the distance squared minimal into the function to get the y-coordinate. The coordinates \((x, \sqrt{x})\) give the point on the graph that is closest to (4,0)

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Most popular questions from this chapter

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