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The first derivative of a function provides information on the slope of the tangent line to the graph at any point. It is a powerful tool to determine where a function is increasing or decreasing. Let’s take the function:

• \( y = x^4 + 32x + 2 \)

To find the first derivative, differentiate each term:

• The derivative of \( x^4 \) is \( 4x^3 \).
• The derivative of \( 32x \) is \( 32 \).
• The derivative of a constant \( 2 \) is \( 0 \).

Thus, the first derivative, represented as \( y' \), is:

• \( y' = 4x^3 + 32 \)

This first derivative will be used to identify critical points where the slope is zero, indicating potential local maxima or minima.

The second derivative gives insights into the concavity of the function, showing how the rate of change of the slope is behaving. It is calculated by differentiating the first derivative. For our function, \\( y' = 4x^3 + 32 \), we find the second derivative to determine how the function curves:

• Differentiate \( 4x^3 \) to get \(12x^2 \).

Thus, the second derivative \( y'' \) can be expressed as:

• \( y'' = 12x^2 \)

By evaluating the second derivative at critical points, we can determine whether these points are local minima or maxima. A positive second derivative indicates that the function is concave up at that point. This suggests a local minimum.

Critical points occur where the first derivative is zero or undefined, marking potential local maximum or minimum points.To find these for our function, solve the equation:

• \( y' = 4x^3 + 32 = 0 \)
• Solving \( 4x^3 + 32 = 0 \), we get:
• \( 4x^3 = -32 \) leading to \( x^3 = -8 \)
• This gives us \( x = -2 \) as a critical point.

We will further analyze the critical point by utilizing the second derivative to understand its significance regarding the function's behavior.

Local minimums are points where the function has the lowest value in a certain interval. The second derivative test helps us verify if a critical point is a local minimum. If \( y''(x) > 0 \), the point is a local minimum since the function curves upwards.

For our critical point \( x = -2 \):

• Calculate \( y''(-2) = 12(-2)^2 = 48 \).
• Since \( 48 > 0 \), the function is concave up at \( x = -2 \), indicating a local minimum.

To find the actual value of this minimum, substitute \( x = -2 \) back into the original function:

• \( y = (-2)^4 + 32(-2) + 2 = -46 \)

Therefore, the local minimum is located at the point \((-2, -46)\).

Graphing calculators can visually confirm the analysis performed on functions by plotting the graph and identifying features like local minima, maxima, and intersection points.

• They allow for immediate visual checks of algebraic calculations.
• Use the calculator to sketch the graph of \( y = x^4 + 32x + 2 \).
• Identify the local minimum at \((-2, -46)\) on the graph.

This powerful tool assists students in verifying their manual calculations and understanding function behavior through graphical representations.