Problem 40 - Calculus

Question

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. If \(f(1)=5,\) then (1,5) could be a critical point of \(f\)

Step-by-Step Solution

Answer

The statement is false; (1,5) is not a critical point because the derivative at this point is -3, not 0.

Step 1: Define Critical Points

A critical point of a function corresponds to where its derivative, with respect to its variable, equals zero. In other words, for a function \(y=f(x)\), a point \((x_0, y_0)\) is a critical point if \(\frac{dy}{dx} = 0\) at that point.

Step 2: Analyze Given Derivative

We are given the derivative \(\frac{dy}{dx} = 2x - y\). To have a critical point at \((1, 5)\), the derivative must be zero at that point. So, substitute \(x = 1\) and \(y = 5\) into the derivative to check this condition.

Step 3: Substitute and Solve

Substitute \(x = 1\) and \(y = 5\) into the expression \(\frac{dy}{dx} = 2x - y\). This gives us: \[ \frac{dy}{dx} = 2(1) - 5 = 2 - 5 = -3. \]

Step 4: Conclusion

Since the derivative at \((1, 5)\) is \(-3\), which is not equal to zero, \((1, 5)\) is not a critical point for the function \(f(x)\). For \(f(x)\) to have a critical point, the derivative must be zero, which is not the case here.

Key Concepts

Derivatives

Derivatives are an essential concept in calculus, used to describe how a function changes at any given point. In mathematical terms, the derivative of a function, typically denoted as \( \frac{dy}{dx} \), measures the rate at which the dependent variable changes relative to the independent variable. Imagine you're riding a rollercoaster; the derivative tells you how fast you're going at any given moment. If that rate is zero, you’re at the peak or trough of the ride where the speed is momentarily paused.

For the given problem, the derivative is expressed as \( \frac{dy}{dx} = 2x - y \). This formula gives us the rate of change of the variable \( y \) with respect to \( x \). When solving for critical points, we are essentially finding where this rate of change is zero. This helps identify points where the function's behavior changes significantly, such as turning points, which may potentially be maximums, minimums, or points of inflection. Understanding derivatives is crucial for analyzing the behavior and features of functions.

Solution Analysis

Solution analysis involves stepping through each part of the problem to understand what the solution is telling us. Let's break down the given exercise and solution step by step.

• First, to determine if a point \((x_0, y_0)\) is a critical point for a function \(f(x)\), we solve the condition \( \frac{dy}{dx} = 0 \).
• Next, the exercise gives us \((1, 5)\). To check if this is a critical point, we substitute these values into the equation \( \frac{dy}{dx} = 2x - y \).
• After substitution, the expression becomes \( 2(1) - 5 = -3 \). The result is not zero, indicating that the rate of change is not zero here.

Therefore, the analysis shows that at \((1, 5)\) the function does not reach a critical point since the derivative is not zero. This step-by-step verification is critical in understanding how the function behaves around given points.

Function Analysis

Function analysis allows us to explore a function's behavior over its domain, which includes identifying critical points where the function's derivative equals zero. Such points typically signal where a curve has peaks, valleys, or possibly an inflection point.

In the context of the problem \( \frac{dy}{dx} = 2x - y \), analyzing the function involves checking various values of \( x \) and \( y \) to solve for situations where the derivative becomes zero. For example, you can set \( 2x = y \) to find where the derivative is zero, which you can further investigate to determine maxima, minima, or neither. Understanding where these changes happen allows us to describe how \( f(x) \) behaves and anticipate its graphical representation.

Thus, function analysis is not just about finding critical points but also examining how these points affect the shape and direction of the graph, making it invaluable for understanding and predicting the nature of the function's changes.