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Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. All the inflection points of \(f\) lie on the line \(y=2 x-2\)

Short Answer

Expert verified
True, all inflection points lie on the line \( y = 2x - 2 \).

Step by step solution

01

Find Second Derivative

To determine the points of inflection, we need to find the second derivative of the function. We begin with the given differential equation \( \frac{dy}{dx} = 2x - y \). We need to find \( \frac{d^2y}{dx^2} \). Differentiating both sides with respect to \( x \), we get \( \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(2x - y) \). This gives us \( \frac{d^2y}{dx^2} = 2 - \frac{dy}{dx} \). Since \( \frac{dy}{dx} = 2x - y \), substitute back to get \( \frac{d^2y}{dx^2} = 2 - (2x - y) = y - 2x + 2 \).
02

Identify Inflection Points Condition

An inflection point occurs where the second derivative changes sign, which corresponds to \( \frac{d^2y}{dx^2} = 0 \). Use the expression we found: \( y - 2x + 2 = 0 \). Simplifying, we get \( y = 2x - 2 \). This equation represents the line on which the inflection points of \( f \) must lie.
03

Compare to the Given Statement

The problem statement claims that all inflection points of \( f \) lie on the line \( y = 2x - 2 \). From Step 2, we've also derived that for the second derivative to be zero, the condition is \( y = 2x - 2 \). This matches the given line in the problem statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
The second derivative is a powerful tool in calculus, often used to gain a deeper understanding of the behaviour of a function. Here, we deal with the second derivative in the context of differential equations and inflection points.

The given differential equation is \( \frac{dy}{dx} = 2x - y \). This equation describes the rate of change of \( y \) with respect to \( x \).

To find the second derivative, \( \frac{d^2y}{dx^2} \), we differentiate this equation once more with respect to \( x \).

Here’s the step-by-step process:
  • Differentiating both sides: \( \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(2x - y) \)
  • The second derivative simplifies to: \( \frac{d^2y}{dx^2} = y - 2x + 2 \)
Every time we apply differentiation, we dig deeper into understanding how the function behaves.

Thus, the second derivative tells us how the rate of change of \( y \) is changing—a concept critical for determining the concavity of a function and finding inflection points.
Inflection Points
Inflection points are crucial when studying the graph of a function. They signal where the function's concavity changes, i.e., where it shifts from being concave upwards to concave downwards or vice versa.

To find inflection points, we focus on the second derivative \( \frac{d^2y}{dx^2} \).

An inflection point occurs when:
  • The second derivative is zero (\( \frac{d^2y}{dx^2} = 0 \))
  • The sign of \( \frac{d^2y}{dx^2} \) changes around a particular value of \( x \)
This problem shows that when \( \frac{d^2y}{dx^2} = y - 2x + 2 = 0 \), solving for \( y \) gives us the line \( y = 2x - 2 \).

An important takeaway is that all inflection points of our function \( f \) should fall along this line, satisfying the equation derived from the second derivative.
Sign Change
Understanding sign changes in the context of the second derivative is key to identifying inflection points.

For an inflection point to be confirmed,
  • The second derivative, \( \frac{d^2y}{dx^2} \), must change its sign around the point. That means it transitions from negative to positive or positive to negative.
  • This change in sign indicates a change in concavity, which defines an inflection point.
In our scenario, the derived line \( y = 2x - 2 \) not only sets the condition for the second derivative being zero but is also where the sign change occurs.

Without sign changes through this line, inflection points cannot truly manifest as per their definition.

Thus, identifying where the sign change occurs in relation to \( \frac{d^2y}{dx^2} \) is integral in confirming true inflection points on the curve.

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Most popular questions from this chapter

Write a differential equation whose solution is the temperature as a function of time of a bottle of orange juice taken out of a \(40^{\circ} \mathrm{F}\) refrigerator and left in a \(65^{\circ} \mathrm{F}\) room. Solve the equation and graph the solution.

Explain what is wrong with the statement. There is a differential equation that has \(y=x\) as one of its solutions and a slope field with a slope of 0 at the point (1,1)

Explain what is wrong with the statement. The line \(y=2\) is an equilibrium solution to the differential equation \(d y / d x=y^{3}-4 x y\).

An item is initially sold at a price of \(p\text{dollars}\) per unit. Over time, market forces push the price toward the equilibrium price, \(p \text{dollars}^{*},\) at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, \(p\text{dollars},\) is proportional to the difference between the market price and the equilibrium price. (a) Write a differential equation for \(p\) as a function of \(t\) (b) Solve for \(p\) (c) Sketch solutions for various different initial prices, both above and below the equilibrium price. (d) What happens to \(p\) as \(t \rightarrow \infty ?\)

Before Galileo discovered that the speed of a falling body with no air resistance is proportional to the time since it was dropped, he mistakenly conjectured that the speed was proportional to the distance it had fallen. (a) Assume the mistaken conjecture to be true and write an equation relating the distance fallen, \(D(t),\) at time \(t,\) and its derivative. (b) Using your answer to part (a) and the correct initial conditions, show that \(D\) would have to be equal to 0 for all \(t,\) and therefore the conjecture must be wrong.

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