Problem 9 - Thomas Calculus

Question

In Exercises \(9-50\) , identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x^{2}-4 x+3 \end{equation}

Step-by-Step Solution

Answer

The local and absolute minimum is at (2, -1); there are no inflection points.

Step 1: Find the First Derivative

To find the extreme points, we first need the first derivative of the function. The function is given as \( y = x^2 - 4x + 3 \). Compute the derivative: \( y' = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4 \).

Step 2: Set First Derivative to Zero

To find the critical points where the extreme values might occur, set the first derivative to zero: \( 2x - 4 = 0 \). Solve for \( x \) by adding 4 to both sides, \( 2x = 4 \), and then dividing by 2, giving \( x = 2 \). This is a critical point.

Step 3: Determine the Nature of Critical Point

To determine whether this critical point is a maximum, minimum, or neither, evaluate the second derivative. The second derivative of \( y = x^2 - 4x + 3 \) is \( y'' = 2 \), which is positive. Therefore, \( x = 2 \) is a local minima.

Step 4: Calculate the y-coordinate of Minima

Substitute \( x = 2 \) back into the original function to find the minima's y-coordinate: \( y = 2^2 - 4 \times 2 + 3 = 4 - 8 + 3 = -1 \). Thus, the local minimum is at \((2, -1)\).

Step 5: Find Intensity Points

For a quadratic function \( ax^2 + bx + c \), there are no inflection points because the second derivative is a constant \( y'' = 2 \) and doesn't change sign.

Step 6: Identify Absolute Extreme Points

For a quadratic function which opens upward (since \( a > 0 \)), the local minimum is also the absolute minimum, and there are no absolute maxima. Hence, the absolute minimum is at \((2, -1)\).

Step 7: Graph the Function

Sketch the graph of \( y = x^2 - 4x + 3 \). It's a parabola opening upwards with the vertex at \((2, -1)\). The axis of symmetry is the vertical line \( x = 2 \), and the y-intercept is \(3\) when \( x = 0 \).

Key Concepts

First Derivative

In mathematics, the first derivative of a function gives us a lot of valuable information about the function's behavior. It can help us identify extbf{critical points}, where the function's slope is zero, which are potential places for extit{extreme points} like local maxima or minima.
For the given quadratic function, \( y = x^2 - 4x + 3 \), we found the first derivative by using basic differentiation rules. The computation is:

• Differentiate \( x^2 \) to get \( 2x \)
• Differentiate \( -4x \) to get \(-4\)
• Differentiate the constant \( 3 \) which becomes zero.

This results in the first derivative: \( y' = 2x - 4 \).
Setting \( y' = 0 \) allows us to solve for the \( x \)-value of the critical point.
Thus, \( 2x - 4 = 0 \) simplifies to \( x = 2 \). This tells us where the function's graph could potentially change direction or flatten out.

Second Derivative

The second derivative of a function provides information about the function’s concavity, which informs us further about the nature of critical points. For a quadratic function like \( y = x^2 - 4x + 3 \), the process is straightforward.
The first derivative we calculated was \( y' = 2x - 4 \). The second derivative is the derivative of the first derivative.
In this case:

• Differentiating \( 2x \) gives \( 2 \)
• Differentiating \( -4 \) gives \( 0 \)

This results in the second derivative: \( y'' = 2 \).
This constant positive value indicates that the original function is concave up everywhere.This confirms that any critical point found is a local minimum, because the function is always bending upwards.

Quadratic Functions

Quadratic functions are polynomial functions of degree 2, often written in the form \( ax^2 + bx + c \). The graph of a quadratic function is a parabola.
Here, the given function \( y = x^2 - 4x + 3 \) is quadratic where \( a = 1 \), \( b = -4 \), and \( c = 3 \). The parabola opens upwards because the coefficient \( a \) is positive.
Key features of a quadratic function include:

• The vertex, which is the highest or lowest point depending on the orientation of the parabola.
• The axis of symmetry, a vertical line passing through the vertex: \( x = 2 \) in our function.
• The direction of opening, which tells us if the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).

Understanding these characteristics helps us sketch the graph and analyze the function effectively.

Critical Points

Critical points are key in calculus as they are locations where a function's slope is zero or undefined. These points can uncover local maxima, local minima, or saddle points.
In our example with \( y = x^2 - 4x + 3 \), we found a critical point by using the first derivative.To identify:

• Set the first derivative \( y' = 2x - 4 \) to zero: \( 2x - 4 = 0 \).
• Solve for \( x \), resulting in \( x = 2 \).

This \( x \)-value of 2 is our critical point. Substituting it back into the original function gives us the \( y \)-coordinate.For this function, \( y(2) = 2^2 - 4(2) + 3 = -1 \).Thus, the critical point is \((2, -1)\).
This initial operation sets the stage for deeper analysis using the second derivative.

Local Minimum

A local minimum is a point where a function's value is lower than all nearby values. It represents a valley in the graph.
To determine if a critical point is a local minimum, we typically use the second derivative.
For our function, we calculated the second derivative as \( y'' = 2 \).
Since this is positive, it indicates that the function is concave up around that point.Therefore:

• The critical point \((2, -1)\) is confirmed as a local minimum.

This point is also the absolute minimum because, for a quadratic function that opens upwards, the lowest point on the graph is the vertex.Understanding local minima is crucial as it helps in applications such as optimization where we seek the lowest cost or greatest efficiency.