Question

True or False If \(f(x)\) is a solution to \(d y / d x=2 x\) , then \(f^{-1}(x)\) is a solution to \(d y / d x=2 y .\) Justify your answer.

Short answer

False. If \(f(x)\) is a solution to \(dy/dx = 2x\), then \(f^{-1}(x)\) is not a solution to \(dy/dx = 2y\). This is because \(f^{-1}(x) ≠ ±1/2\), and a function cannot have two distinct values for the same \(x\).

Step-by-step solution

Understand the problem

We are given that \(f(x)\) is a solution to the differential equation \(dy/dx = 2x\). We need to check whether the inverse of this function \(f^{-1}(x)\), is a solution to the differential equation \(dy/dx = 2y\). This involves finding the derivative of the inverse function and substituting \(y = f^{-1}(x)\) into \(dy/dx = 2y\) to check if the equation holds true.

Differentiate the inverse function

Using the formula for the derivative of an inverse function, which is \(1 / f'(f^{-1}(x))\), and knowing that \(f'(x) = 2x\), we can find that \(f^{-1}'(x) = 1 / (2f^{-1}(x))\).

Substitution

We replace \(y = f^{-1}(x)\) and \(dy/dx = f^{-1}'(x)\) in the equation \(dy/dx = 2y\). Upon doing this, we get \(1 / (2f^{-1}(x)) = 2f^{-1}(x)\).

Solving the equation

Solving the equation \(1 / (2f^{-1}(x)) = 2f^{-1}(x)\), we get \(f^{-1}(x) = ±1/2.\) However, \(f^{-1}(x)\) is a function, and a function cannot have two distinct values for the same \(x\). Therefore, \(f^{-1}(x)\) cannot be a solution to the differential equation \(dy/dx = 2y\).

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are central to many areas of mathematics and science, from the prediction of population growth in biology to the dynamics of planetary motion in astronomy.

In the context of calculus, a differential equation like the one presented in our exercise, \( dy/dx = 2x \), represents a rate of change—specifically, how the rate of change of one variable, y, is related to another variable, x. The solutions to such equations are functions that satisfy the relationship stipulated by the equation. When we are given a function \( f(x) \) and asked if its inverse \( f^{-1}(x) \) is also a solution, we are essentially exploring how the rate of change of the inverse function is related to its variable. In simpler terms, it's like asking: if a car is speeding up at a certain rate, does that tell us anything about the rate at which it should slow down to reach a certain point going backward?

The study of differential equations not only helps in understanding change but also in forecasting future behavior by modeling real-world phenomena. For instance, given the current rate of spread of an infectious disease, a differential equation could predict the future number of affected individuals.

Derivative of Inverse Function

The derivative of an inverse function provides information about the slope of the inverse function at a given point. Given a function \( f(x) \) and its inverse \( f^{-1}(x) \), the relationship between their derivatives is given by the formula \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \).

In plain language, this means that the rate at which the inverse function changes—its 'steepness', so to speak—is the reciprocal of the rate at which the original function itself changes at the corresponding point. This relationship is central when dealing with inverse functions in calculus.

Understanding Through the Exercise

Given that the derivative of \( f(x) \) is \( 2x \), we apply that to our formula to find the derivative of the inverse: \( (f^{-1})'(x) = \frac{1}{2f^{-1}(x)} \). This tells us how quickly or slowly \( f^{-1}(x) \) is changing with respect to x. To determine whether \( f^{-1}(x) \) satisfies a given differential equation, we use this derivative and examine if it meets the necessary conditions imposed by the equation.

Justifying Problem Solutions

Justifying problem solutions in mathematics is a crucial process that goes beyond merely finding an answer. It involves a detailed explanation as to why the answer obtained is correct, in alignment with the principles and theorems of mathematics.

To justify a solution, one must show that it logically follows from the premises and adheres to mathematical rules. This may include proving that the solution satisfies all the conditions set by the problem, and that no other solution can exist. For the exercise in question, to justify the claim that \( f^{-1}(x) \) is not a solution to \( dy/dx = 2y \) we must delve into the process of solving the differential equation and critically analyze the implications of each step.

The Justification Process

In our specific problem, we see that solving the equation \( 1/(2f^{-1}(x)) = 2f^{-1}(x) \) yields \( f^{-1}(x) = ±1/2 \). This suggests that for a particular x, the inverse function would need to have two different values, which is not possible for a function by definition – a function must assign exactly one output for each input. This contradiction serves as a justification for why \( f^{-1}(x) \) cannot be the solution to \( dy/dx = 2y \). Thus, we achieve not just a solution, but an understanding of why that solution is the correct one.