Section 1.1: Starting the Solution

To begin, let’s focus on the right side of the graph. This approach will help us eliminate the absolute value sign. Remember, in differential calculus, the derivatives of the two functions are equal at their point of tangency because they share a tangent line there. Our first step is to calculate their derivatives.

This process helps us eliminate the unknown ( k ) from the first function. Next, we will determine the x-coordinate at the tangential point.

Section 1.2: Visualizing the Problem

Let’s sketch a rough diagram to clarify our findings.

We now have the x-coordinate of the tangential point. By substituting this value into the square root function, we can find the corresponding y-coordinate.

With both x and y-coordinates established, we can input these values into the parabolic function to find the value of ( k ).

Next, let's consider the following diagram.

The shaded region represents twice the difference between the upper and lower curves, which can be expressed as an integral.

This area can be calculated using standard integration techniques. That concludes our solution! I'd love to hear what value you arrived at, so feel free to share in the comments!

Photo by Vicko Mozara on Unsplash

How intriguing! What were your thoughts during this process? Please share your insights below; I'm excited to learn from you!