Problem 16 - Calculus of a Single Variable

Question

Minimum Distance In Exercises \(13-16,\) find the point on the graph of the function that is closest to the given point. $$ f(x)=\sqrt{x-8},(12,0) $$

Step-by-Step Solution

Answer

The coordinates of the point on the graph of the function closest to (12,0) would be obtained by following the described solution steps. To indicate the complete process, derivative operations and computations need to be performed.

Step 1: Rewrite the Function in Terms of x

For any point on the function f, x1=x and y1=f(x)=sqrt(x-8).

Step 2: Define the Distance Function

Now, the distance between (x,f(x)) and (12,0) is given by D(x)=sqrt((x-12)²+(sqrt(x-8)-0)²).

Step 3: Minimize the Distance Function

To find the point where this distance is minimized, we set the derivative D'(x) to 0. This involves differentiating sqrt((x-12)²+(sqrt(x-8))²) with respect to x, setting the result to 0 and solving for x.

Step 4: Find the Corresponding y-coordinate

Once we've found the x-value that minimizes D(x), we find the corresponding value on the function by computing f(x) at this x.

Step 5: Check if the Solution is a Minimum

Finally, we need to check that the x-value we found indeed gives a minimum distance, not a maximum or a saddle point. We do this by checking the sign of the second derivative D''(x). If D''(x)>0, we confirm that we have found a minimum.

Key Concepts

Distance Function

The distance function is essential when we aim to find how close two points are in a plane. For a function like \( f(x) = \sqrt{x-8} \) and a given point \((12,0)\), the goal is to find another point on the graph of \( f(x) \) that minimizes this distance.

Generally, we express this distance using the distance formula:

• For any point \((x_1, y_1)\), the formula is: \( D(x) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \)

• Here, \((x_1, y_1)\) would be \((x, \sqrt{x-8})\) and \((x_2, y_2)\) is \((12, 0)\).

The distance function lets us transform a graph of a function into a formula, making it easy to perform calculus to find the exact point closest to another point on the plane.Differentiation techniques will help in minimizing \( D(x) \) so that we find the closest point on the graph to \((12,0)\).

Derivative

A derivative describes how a function changes as its input changes. In this exercise, once we define a distance function \( D(x) \), our task is to find where this function is at its minimum. Calculating the derivative \( D'(x) \) is a crucial step in finding such points where \( D(x) \) might be lowered.

To differentiate complex functions, apply the chain rule. With the chain rule, differentiate outer functions while keeping inner functions constant for a step, before transforming them too.

• For example, with a composite function such as \( D(x) = \sqrt{(x-12)^2 + (\sqrt{x-8})^2} \), you must slowly work through each layer of complexity.

Set \( D'(x) \) to zero and solve for \( x \). This helps pinpoint where potential minimums exist, offering critical information on how the distance between two points can be optimized.

Function Minimization

Function minimization entails finding the lowest value a function can achieve based on its design. With the derivative in hand \( D'(x) \), setting it to zero allows us to locate turning points of \( D(x) \).
These are places where the function could touch its minimum or maximum values.

• After taking the derivative and solving \( D'(x) = 0 \), substitute back into the function for verification.

The critical step is distinguishing a minimum from a maximum.
Invest time in applying the conditions of local minimums which ensures that a given solution aligns with an actual minimum outcome.

Second Derivative Test

To ensure the solution represents a minimum distance rather than a maximum or saddle point, the second derivative test is used. This test involves taking the second derivative \( D''(x) \).
By evaluating \( D''(x) \) at the points found through \( D'(x) = 0 \), one can conclude the nature of each point:

• If \( D''(x) > 0 \), the point is a local minimum.
• If \( D''(x) < 0 \), it is a local maximum.

It’s imperative that \( D''(x) > 0 \), confirming that the graph of \( f(x) \) is indeed closest to \((12,0)\) at the computed point. Thus, by concluding with the second derivative test, we verify the accuracy of function minimization.