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Chapter 12: Problem 63

The following functions have exactly one isolated peak or one isolated depression (one local maximum or minimum). Use a graphing utility to approximate the coordinates of the peak or depression. $$h(x, y)=1-e^{-\left(x^{2}+y^{2}-2 x\right)}$$

Short Answer

Expert verified
Question: Determine the approximate coordinates of the peak of the function \(h(x,y) = 1 - e^{-(x^2 + y^2 - 2x)}\) using a graphing utility. Answer: The peak of the function is approximately at the coordinates (1, 0).

Step by step solution

01

Understand peaks and depressions

A function has a peak at a point if it has a local maximum at that point, and it has a depression if it has a local minimum at that point. So, we are looking for a point \((x,y)\) such that \(h(x,y)\) has either a local maximum or a local minimum.
02

Graph the function

Use a graphing utility like Desmos or Geogebra to graph the function \(h(x,y) = 1 - e^{-(x^2 + y^2 - 2x)}\). Note that since we have a function of two variables, we need to use a 3D graphing tool to visualize it properly.
03

Find the coordinates of the peak or depression

Once you have graphed the function, inspect the graph visually to find the peak or depression. You will notice it has a peak shape. Since the function is continuous, you can declare the highest point of the peak as the coordinate requested. To help estimate the coordinates of this point, you can rotate the graph to align with one of the axes, or add contour lines which show points of equal heights. Formally and algebraically, we could have used the second derivative test to confirm the peak nature of the coordinate.
04

Record the approximate coordinates

With the graph properly visualized, you can now use your cursor or the tools provided by the graphing utility to find the coordinates of the peak. In this case, the approximate coordinates are \((1, 0)\). So, we can conclude that the function has a peak at \(\approx (1, 0)\).

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