Finding Stationary Points And Inflection Points For Y=1/3x³-2x²+3x

by THE IDEN 67 views

In this comprehensive guide, we will delve into the process of determining the stationary points and inflection points for the equation y = (1/3)x³ - 2x² + 3x. Understanding these points is crucial in calculus as they provide valuable insights into the behavior and shape of a curve. Stationary points, also known as critical points, are locations where the gradient of the curve is zero, indicating a potential local maximum, local minimum, or saddle point. Inflection points, on the other hand, mark the locations where the curve changes its concavity, transitioning from curving upwards to curving downwards or vice versa.

1. Determining Stationary Points

Stationary points, also known as critical points, are points on the graph of a function where the derivative is equal to zero or undefined. These points are crucial for understanding the behavior of a function, as they can indicate local maxima, local minima, or saddle points. In our case, to find the stationary points of the function y = (1/3)x³ - 2x² + 3x, we need to follow these steps:

1.1. Find the First Derivative

The first step in finding stationary points is to determine the first derivative of the function. The first derivative, denoted as dy/dx or y', represents the slope of the tangent line to the curve at any given point. To find the first derivative of our equation, we apply the power rule of differentiation:

  • d/dx (xⁿ) = nxⁿ⁻¹

Applying this rule to each term in the equation y = (1/3)x³ - 2x² + 3x, we get:

  • dy/dx = (1/3) * 3x² - 2 * 2x + 3
  • dy/dx = x² - 4x + 3

1.2. Set the First Derivative to Zero

The key to finding stationary points is recognizing that they occur where the slope of the tangent line is zero. This means that the first derivative, dy/dx, must be equal to zero at these points. Therefore, we set our calculated first derivative equal to zero:

  • x² - 4x + 3 = 0

1.3. Solve for x

Now, we need to solve the quadratic equation x² - 4x + 3 = 0 for x. This can be done by factoring, completing the square, or using the quadratic formula. In this case, the equation can be easily factored:

  • (x - 1)(x - 3) = 0

This gives us two possible solutions for x:

  • x = 1
  • x = 3

These values of x represent the x-coordinates of the stationary points.

1.4. Find the Corresponding y Values

To fully determine the stationary points, we need to find the corresponding y-coordinates for each x-value we found. We do this by substituting the x-values back into the original equation, y = (1/3)x³ - 2x² + 3x.

For x = 1:

  • y = (1/3)(1)³ - 2(1)² + 3(1)
  • y = 1/3 - 2 + 3
  • y = 4/3

For x = 3:

  • y = (1/3)(3)³ - 2(3)² + 3(3)
  • y = 9 - 18 + 9
  • y = 0

Therefore, the stationary points are (1, 4/3) and (3, 0).

1.5. Determine the Nature of Stationary Points

To fully understand the behavior of the curve at the stationary points, we need to determine whether they are local maxima, local minima, or saddle points. There are two common methods for doing this: the first derivative test and the second derivative test. Let's use the second derivative test for this example.

1.5.1. Find the Second Derivative

The second derivative, denoted as d²y/dx² or y'', represents the rate of change of the slope of the curve. To find the second derivative, we differentiate the first derivative, dy/dx = x² - 4x + 3, again:

  • d²y/dx² = 2x - 4

1.5.2. Evaluate the Second Derivative at Each Stationary Point

Now, we evaluate the second derivative at each of the x-values of the stationary points:

For x = 1:

  • d²y/dx² = 2(1) - 4
  • d²y/dx² = -2

For x = 3:

  • d²y/dx² = 2(3) - 4
  • d²y/dx² = 2

1.5.3. Interpret the Results

The sign of the second derivative at a stationary point tells us about the concavity of the curve at that point:

  • If d²y/dx² > 0, the curve is concave up (U-shaped), and the stationary point is a local minimum.
  • If d²y/dx² < 0, the curve is concave down (inverted U-shaped), and the stationary point is a local maximum.
  • If d²y/dx² = 0, the test is inconclusive, and further analysis is needed.

Based on our calculations:

  • At (1, 4/3), d²y/dx² = -2, which is negative. Therefore, this point is a local maximum.
  • At (3, 0), d²y/dx² = 2, which is positive. Therefore, this point is a local minimum.

2. Determining Inflection Points

Inflection points are points on the curve where the concavity changes. This means the curve transitions from being concave up (U-shaped) to concave down (inverted U-shaped) or vice versa. To find the inflection points of the function y = (1/3)x³ - 2x² + 3x, we need to follow these steps:

2.1. Find the Second Derivative

As we already calculated in the previous section, the second derivative of our function is:

  • d²y/dx² = 2x - 4

2.2. Set the Second Derivative to Zero

Inflection points occur where the concavity changes, which means the second derivative must be equal to zero at these points. Therefore, we set the second derivative equal to zero:

  • 2x - 4 = 0

2.3. Solve for x

Solving the equation 2x - 4 = 0 for x, we get:

  • 2x = 4
  • x = 2

This value of x is a potential x-coordinate of an inflection point. However, we need to confirm that the concavity actually changes at this point.

2.4. Find the Corresponding y Value

To find the y-coordinate of the potential inflection point, we substitute x = 2 back into the original equation, y = (1/3)x³ - 2x² + 3x:

  • y = (1/3)(2)³ - 2(2)² + 3(2)
  • y = 8/3 - 8 + 6
  • y = 2/3

So, the potential inflection point is (2, 2/3).

2.5. Verify the Change in Concavity

To confirm that (2, 2/3) is indeed an inflection point, we need to verify that the concavity changes at this point. We can do this by examining the sign of the second derivative to the left and right of x = 2.

2.5.1. Choose Test Values

Let's choose two test values: one less than 2 (e.g., x = 1) and one greater than 2 (e.g., x = 3).

2.5.2. Evaluate the Second Derivative at the Test Values

For x = 1:

  • d²y/dx² = 2(1) - 4
  • d²y/dx² = -2 (negative, concave down)

For x = 3:

  • d²y/dx² = 2(3) - 4
  • d²y/dx² = 2 (positive, concave up)

2.5.3. Interpret the Results

Since the second derivative changes sign from negative to positive as we move from x = 1 to x = 3, the concavity changes at x = 2. This confirms that (2, 2/3) is indeed an inflection point.

3. Summary

In summary, for the equation y = (1/3)x³ - 2x² + 3x, we have found:

  • Stationary Points:
    • Local Maximum: (1, 4/3)
    • Local Minimum: (3, 0)
  • Inflection Point: (2, 2/3)

These points provide a comprehensive understanding of the behavior and shape of the curve represented by the given equation. By finding and analyzing stationary points and inflection points, we gain valuable insights into the function's increasing and decreasing intervals, concavity, and overall graph.

This detailed analysis of stationary points and inflection points demonstrates how calculus techniques can be used to understand the behavior of functions and their graphical representations. Understanding these concepts is crucial for various applications in mathematics, physics, engineering, and other fields.