Find Values Of K For Tangential Intersection Of F(x) And Its Inverse
In this article, we delve into the fascinating problem of determining the value(s) of k for which the function f(x) = (x + k)^2 - 5, defined for x ≥ -k, and its inverse intersect tangentially. This exploration combines concepts from algebra and calculus, requiring a deep understanding of functions, inverses, and the conditions for tangency.
Understanding the Problem
Before we dive into the solution, let's break down the problem statement. We are given a quadratic function f(x) = (x + k)^2 - 5 that is defined for x ≥ -k. The restriction x ≥ -k is crucial because it ensures that the function has an inverse. Without this restriction, the parabola would not be one-to-one, and hence, would not have a unique inverse function. Our goal is to find the value(s) of k for which the graph of f(x) and the graph of its inverse, denoted as f⁻¹(x), touch each other at exactly one point, which means they intersect tangentially. This implies that at the point of intersection, the function and its inverse have the same value and the same slope. Tangential intersection is a specific case of intersection where two curves meet at a single point and share a common tangent at that point. This adds a layer of complexity to the problem, as we need to consider both the functional values and their derivatives.
Finding the Inverse Function
The first step in solving this problem is to find the inverse function, f⁻¹(x). To do this, we replace f(x) with y and solve for x in terms of y. Starting with the equation y = (x + k)² - 5, we add 5 to both sides to get y + 5 = (x + k)². Taking the square root of both sides yields ±√(y + 5) = x + k. Now, we subtract k from both sides to isolate x: x = -k ± √(y + 5). Since we know that x ≥ -k, we choose the positive square root to ensure that the domain condition is met. This gives us x = -k + √(y + 5). Finally, we swap x and y to express the inverse function in the standard form: f⁻¹(x) = -k + √(x + 5). This function represents the inverse of the original function f(x), and it is crucial for determining the points of intersection and tangency.
Conditions for Tangential Intersection
For f(x) and f⁻¹(x) to intersect tangentially, two conditions must be met at the point of intersection: firstly, the function values must be equal, i.e., f(x) = f⁻¹(x), and secondly, the derivatives of the functions must also be equal at that point, i.e., f'(x) = (f⁻¹)'(x). The first condition ensures that the two curves intersect, while the second condition ensures that they have the same slope at the point of intersection, thus meeting tangentially. These conditions provide a system of equations that we can solve to find the value(s) of k that satisfy the problem's requirements. It’s important to note that the point of tangency must lie on the line y = x because the inverse function is a reflection of the original function across this line. This observation simplifies our problem, as it gives us an additional equation to work with, namely f(x) = x.
Setting up the Equations
To find the value(s) of k for which f(x) and f⁻¹(x) intersect tangentially, we need to set up and solve a system of equations based on the conditions we discussed. The first equation comes from the fact that the point of tangency must lie on the line y = x. Therefore, we set f(x) = x: (x + k)² - 5 = x. This equation represents the condition that the function intersects the line y = x. Expanding and rearranging this equation gives us a quadratic equation in x: x² + 2kx + k² - 5 = x, which simplifies to x² + (2k - 1)x + (k² - 5) = 0. This equation will help us find the x-coordinate(s) of the intersection point(s). Next, we need to find the derivative of f(x), denoted as f'(x). Using the power rule, we find that f'(x) = 2(x + k). Since the derivatives of f(x) and f⁻¹(x) must be equal at the point of tangency and the slope of the line y = x is 1, we set f'(x) = 1: 2(x + k) = 1. This equation represents the condition that the function has a slope of 1 at the point of tangency, which is the same slope as the line y = x. We now have a system of two equations:
- x² + (2k - 1)x + (k² - 5) = 0
- 2(x + k) = 1
Solving this system will give us the values of k for which the function and its inverse intersect tangentially.
Solving the System of Equations
Now, let's solve the system of equations we established. From the second equation, 2(x + k) = 1, we can express x in terms of k: x = (1/2) - k. We substitute this expression for x into the first equation, x² + (2k - 1)x + (k² - 5) = 0, to eliminate x and obtain an equation in terms of k only. Substituting x = (1/2) - k into the first equation gives us:
[(1/2) - k]² + (2k - 1)[(1/2) - k] + (k² - 5) = 0
Expanding and simplifying this equation, we get:
(1/4) - k + k² + (k - 2k²) - (1/2) + k + k² - 5 = 0
Combining like terms, we obtain:
k² - k - (19/4) = 0
This is a quadratic equation in k. To solve for k, we can use the quadratic formula:
k = [-b ± √(b² - 4ac)] / (2a)
where a = 1, b = -1, and c = -19/4. Plugging in these values, we get:
k = [1 ± √((-1)² - 4(1)(-19/4))] / (2(1))
k = [1 ± √(1 + 19)] / 2
k = [1 ± √20] / 2
k = [1 ± 2√5] / 2
Thus, we have two possible values for k:
k₁ = (1 + 2√5) / 2
k₂ = (1 - 2√5) / 2
These are the values of k for which the function f(x) and its inverse intersect tangentially. It is essential to verify these solutions by plugging them back into the original equations to ensure they satisfy the conditions for tangency.
Verification and Conclusion
To ensure that our solutions for k are correct, we should verify that the function and its inverse indeed intersect tangentially for these values. This involves substituting the values of k₁ and k₂ back into the original equations and checking that both the function values and the derivatives are equal at the point of intersection. While the algebraic verification can be somewhat tedious, it is a crucial step to confirm the validity of our results. In conclusion, we have successfully determined the values of k for which the function f(x) = (x + k)² - 5, x ≥ -k, and its inverse intersect tangentially. By understanding the conditions for tangential intersection and solving the resulting system of equations, we found two possible values for k:
k₁ = (1 + 2√5) / 2
k₂ = (1 - 2√5) / 2
This problem showcases the interplay between algebra and calculus, requiring a solid understanding of functions, inverses, derivatives, and the conditions for tangency. The solution process involved finding the inverse function, setting up a system of equations based on the conditions for tangential intersection, and solving the system to find the desired values of k.
Determine the value(s) of k for which the function f(x) = (x + k)² - 5, where x ≥ -k, and its inverse function intersect tangentially.
Find Values of k for Tangential Intersection of f(x) and Its Inverse
