Problem 9 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{2}-4 x+3$$

Step-by-Step Solution

Answer

Local and absolute minimum at (2, -1); no inflection points.

Step 1: Find the first derivative

To find critical points where local extrema might occur, we first need the first derivative of the function. Differentiate the function \( y = x^2 - 4x + 3 \) with respect to \( x \). This gives us: \[ \frac{dy}{dx} = 2x - 4. \]

Step 2: Set the first derivative to zero

Set the first derivative \( 2x - 4 = 0 \) to find the critical points. Solve the equation: \[ 2x - 4 = 0 \] \[ 2x = 4 \] \[ x = 2. \] So, \( x = 2 \) is a critical point.

Step 3: Determine the second derivative

To identify if the critical point is a local minimum or maximum, we check the concavity using the second derivative. Differentiate \( 2x - 4 \) to get: \[ \frac{d^2y}{dx^2} = 2. \]

Step 4: Evaluate the second derivative at the critical point

Evaluate the second derivative \( 2 \) at the critical point \( x = 2 \). Since \( \frac{d^2y}{dx^2} = 2 > 0 \), the function is concave up, indicating a local minimum at \( x = 2 \).

Step 5: Find the function value at the critical point

Substitute \( x = 2 \) into the original function to find the y-coordinate of the local minimum: \[ y = 2^2 - 4 \times 2 + 3 = 4 - 8 + 3 = -1. \] The local minimum is at \( (2, -1) \).

Step 6: Check for absolute extrema and inflection points

Since the quadratic function is a parabola opening upwards, the point \( (2, -1) \) is also the absolute minimum, and there are no maximum points. For inflection points, check where the second derivative is zero. Since \( \frac{d^2y}{dx^2} = 2 \) is constant and never zero, there are no inflection points.

Key Concepts

Critical Points

Critical points are locations on a graph where the function's derivative is either zero or undefined. Identifying these points is important because they may be locations of local maximums or minimums but simply represent locations where the slope of the function is zero or changes sign. In our exercise, we consider the function \( y = x^2 - 4x + 3 \). To find the critical points, the first derivative \( \frac{dy}{dx} = 2x - 4 \) is set to zero. By solving \( 2x - 4 = 0 \), we find the critical point at \( x = 2 \).

• Critical point calculation helps in predicting structure and layout of curves.
• They are candidates for local extrema.

Understanding critical points helps build foundational calculus skills.

Second Derivative Test

The second derivative test is a handy tool to determine the concavity of the function at the critical points found using the first derivative. It helps indicate whether the critical point is a local minimum, a local maximum, or neither. To use the test, we first find the second derivative of our function \( y = x^2 - 4x + 3 \), which gives us \( \frac{d^2y}{dx^2} = 2 \). Evaluating this at the critical point \( x = 2 \), we see that \( \frac{d^2y}{dx^2} = 2 \) is greater than zero, indicating the function is concave up at this point.

• A concave-up test result means a local minimum.
• This test streamlines identifying the type of extrema.

The second derivative is positive, confirming that the function holds a local minimum at \( x = 2 \).

Extrema

Extrema refer to points of maximum and minimum values a function might have, divided between local (relative) and absolute (global) extrema. In our example function \( y = x^2 - 4x + 3 \), the parabola opens upward, suggesting the existence of a minimum point but not a maximum.The local minimum previously discovered through calculus at \( x = 2 \) leads us to the point \( (2, -1) \), which is calculated by substituting \( x = 2 \) back into the function.

• Local extrema occur within a neighborhood of a function.
• Absolute extrema represent the highest and lowest function values over the entire domain.

Since this parabola is concave upwards and extends infinitely, the point \( (2, -1) \) is both a local and absolute minimum for the function. This comprehensive understanding ensures that students grasp both theoretical and practical aspects of extrema.