Question

Find the points on the graph of the function that are closest to the given point. \(f(x)=\sqrt{x-8}, \quad(2,0)\)

Short answer

The point on \(f(x) = \sqrt{x-8}\) that is closest to (2,0) is (8,0)

Step-by-step solution

Set Up The Distance Equation

Firstly, describe the distance between the point (2,0) and any point on the function \(f(x) = \sqrt{x-8}\). The distance, \(d\), between two points \((x_1,y_1)\) and \((x_2,y_2)\) in a coordinate plane is given by \[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]. Applying this formula, using \(x, f(x)\) for \((x_2,y_2)\) and (2,0) for \((x_1,y_1)\) we get \(d = \sqrt{(x-2)^2 + (\sqrt{x-8}-0)^2}\)

Simplify the Equation

Now, you need to simplify the distance equation in order to make it easier to differentiate. This can be done by squaring both sides to get rid of the square root on the left-hand side. The new equation will be \(d^2 = (x-2)^2 + (x-8)\)

Differentiate the distance function

Next, differentiate the distance squared function with respect to \(x\) to get its derivative. The resulting derivative is \[d'^2 = 2(x-2) +1\]

Find the critical points

The critical points of a function, points where the derivative is zero or undefined, often contain maximum and minimum values for the function. So, set the derivative to zero and solve for \(x\) to get the critical points. \[2(x-2) + 1 = 0\] which simplifies to \[x = -\frac{1}{2} +2 = \frac{3}{2}\]

Evaluate the function at the critical and endpoint values

Now evaluate \(d^2\) at \(x = \frac{3}{2}\). It's important to note that since we've made \(d^2\) positive by squaring, we can say if \(d^2\) is minimized then \(d\) is also minimized. So we're effectively checking for the minimum distance here. Additionally, since the domain of the function \(f(x) = \sqrt{x-8}\) is \([8, \infty)\), the minimum possible value for \(x\) is 8, not our usual \(-\infty\). Upon evaluating \(d^2\) at \(x=\frac{3}{2}\), and \(x=8\), it can be observed that the minimum value is at \(x = 8\)

Find the corresponding \(y\) value

Now that we know the \(x\) value that minimizes the distance, we need to find the corresponding \(y\) value. This is done by substituting \(x = 8\) into the function \(f(x)\) to evaluate \(f(8)\), which equals 0