91Ó°ÊÓ

Derivatives are fundamental in calculus for determining how a function changes at any point. They tell us the rate of change or slope of the function's graph. In simpler terms, think of it as finding how fast something is changing at a precise point.

For a given function like our quadratic function, which is expressed as \[ f(x) = x^2 + 4x + 5 \], we need the derivative \( f'(x) \) to locate important features like peaks and valleys. As you might have learned, the power rule helps us to find derivatives quickly. Here, the power rule tells us that the derivative of \( x^2 \) is \( 2x \), and the derivative of \( 4x \) is 4. Thus, the derivative, \( f'(x) \), becomes \( 2x + 4 \). This resulting expression gives us the slope of the tangent to the curve at any point \( x \).

• Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Linear Changes: Derivatives simplify understanding changes in linear parts of the function.

Critical points are where we look for changes in the function's behavior. They are the x-values where the derivative equals zero or is undefined. These points are essential since they can indicate where the function might have a maximum or minimum value.

For the quadratic function \( f(x) = x^2 + 4x + 5 \), we found the derivative to be \( 2x + 4 \). By setting this equation to zero \[ 2x + 4 = 0 \], we solved for \( x = -2 \). This means that \( x = -2 \) is a critical point for our function. In essence:

• Zero Derivative: Solve \( f'(x) = 0 \) to find potential extrema.
• Changes in Slope: Critical points mark where the function shifts from increasing to decreasing, or vice versa.

To determine the nature of a critical point found from the first derivative, we use the second derivative test. This test tells us if the point is a local minimum, local maximum, or neither.

The second derivative, denoted as \( f''(x) \), provides a way to measure the concavity, or the curved direction, of the function. If \( f''(x) > 0 \), the function is concave up at that point, implying a local minimum. Conversely, if \( f''(x) < 0 \), the function is concave down, indicating a local maximum.

For our function, the second derivative \( f''(x) = 2 \) is positive. Since it remains constant and positive, it confirms a local minimum at the critical point \( x = -2 \).

• Positive Second Derivative: Confirms a local minimum.
• Negative Second Derivative: Indicates a local maximum.

Quadratic functions are polynomial functions of degree 2, generally written as \( ax^2 + bx + c \). They graph as parabolas. The characteristics of these functions are easy to identify, making them significant in calculus.

For our specific problem, the function \( f(x) = x^2 + 4x + 5 \) represents a standard quadratic function.

• Parabolic Shape: The function's graph is a parabola that opens upward because the coefficient of \( x^2 \) (here, 1) is positive.
• Vertex Formula: The vertex, which for our function is at \( (-2, 1) \), can be found using \(x = -\frac{b}{2a}\) for the x-coordinate, and substituting back into the function to find the y-coordinate.

Quadratics often represent scenarios like projectile motion or optimization problems in mathematics, and understanding their basic shape and properties is crucial.