Problem 37 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test or the Second Derivative Test to determine the relative extreme values, if any, of the function. Then sketch the graph of the function. $$ f(x)=x^{2}+8 x+12 $$

Step-by-Step Solution

Answer

The function has a relative minimum at \((-4, -4)\). Its graph is a parabola opening upwards with this vertex.

Step 1: Find the First Derivative

The given function is \( f(x) = x^2 + 8x + 12 \). First, we need to find the derivative of this function. The derivative \( f'(x) \) is found using the power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\). Thus, \( f'(x) = 2x + 8 \).

Step 2: Set the First Derivative to Zero

To find the critical points where relative extrema might occur, set the first derivative \( f'(x) = 2x + 8 \) equal to zero. Solve for \( x \): \( 2x + 8 = 0 \). This simplifies to \( 2x = -8 \), which gives \( x = -4 \).

Step 3: Apply the First Derivative Test

We have the critical point at \( x = -4 \). To determine the nature of this critical point, use the First Derivative Test by checking the sign of \( f'(x) \) before and after \( x = -4 \).- Test a point before \( x = -4 \), say \( x = -5 \): \( f'(-5) = 2(-5) + 8 = -10 + 8 = -2 \), which is negative.- Test a point after \( x = -4 \), say \( x = -3 \): \( f'(-3) = 2(-3) + 8 = -6 + 8 = 2 \), which is positive.Since \( f'(x) \) changes from negative to positive at \( x = -4 \), \( f(x) \) has a relative minimum at \( x = -4 \).

Step 4: Find the Y-Coordinate of the Critical Point

Substitute \( x = -4 \) back into the original function to find the y-coordinate of the relative minimum: \( f(-4) = (-4)^2 + 8(-4) + 12 = 16 - 32 + 12 = -4 \). Therefore, the relative minimum point is \( (-4, -4) \).

Step 5: Sketch the Graph

To sketch the graph of the function, note that \( f(x) = x^2 + 8x + 12 \) is a quadratic function which opens upwards because the coefficient of \( x^2 \) is positive. The vertex or the minimum point is at \( (-4, -4) \). The y-intercept is \( f(0) = 12 \). Thus, the graph is a parabola opening upwards with its vertex at \( (-4, -4) \) and crossing the y-axis at \( (0, 12) \).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial that takes the form \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). In this equation, the term \( ax^2 \) gives the function its characteristic "U" shape, known as a parabola.
Quadratic functions can open upwards or downwards depending on the value of \( a \). If \( a > 0 \), the parabola opens upwards, like a valley. If \( a < 0 \), it opens downwards like a hill.
For the function \( f(x) = x^2 + 8x + 12 \), we note that the coefficient \( a = 1 \), which is positive. This means the parabola opens upwards. The highest or lowest point of this parabola is known as the vertex, which can be found using various methods, including completing the square or using derivatives.

Relative Extrema

In the context of quadratic functions, relative extrema refer to the highest or lowest points on the graph. These points are also known as local maxima or minima.
Since a quadratic function is a smooth, continuous curve, it has only one type of extremum. If the parabola opens upwards, it has a lowest point which is a minimum. If it opens downwards, it has a highest point, which is a maximum.
For \( f(x) = x^2 + 8x + 12 \), the parabola opens upwards, indicating a relative minimum. By finding this minimum, you can determine where the curve changes its direction from decreasing to increasing.

Critical Points

Critical points of a function are values of \( x \) where the derivative \( f'(x) \) is zero or undefined. These points are potential locations of relative extrema.
To find the critical points of the function \( f(x) = x^2 + 8x + 12 \), we first took its derivative, resulting in \( f'(x) = 2x + 8 \).
Setting \( f'(x) \) to zero, \( 2x + 8 = 0 \), allows us to solve for \( x \) and find the critical point at \( x = -4 \).
By placing this critical point into the original function, \( f(-4) = -4 \), we discover the location of the relative minimum is at \( (-4, -4) \).

First Derivative

The first derivative of a function, denoted as \( f'(x) \), provides critical insights into the function's behavior. It shows the rate at which the function's value is changing.
For quadratic functions, the first derivative is a linear function. In this case, the derivative of \( f(x) = x^2 + 8x + 12 \) is \( f'(x) = 2x + 8 \).
The First Derivative Test helps us determine whether a critical point is a relative maximum, minimum, or neither. By observing the sign changes of \( f'(x) \) around critical points, we can infer the nature of these points.
At \( x = -4 \), \( f'(x) \) changes from negative to positive, indicating that the function transitions from decreasing to increasing, hence signaling a relative minimum at that point.