Problem 32 - Thomas Calculus

Question

In Exercises \(29-36 :\) a. Identify the function's local extreme values in the given domain, and say where they are assumed. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$ g(x)=-x^{2}-6 x-9, \quad-4 \leq x<\infty $$

Step-by-Step Solution

Answer

Local max at \( x = -3, g(-3) = 0 \); it's also an absolute max.

Step 1: Find the derivative of the function

To locate the critical points of the function where local extrema may occur, first find the derivative of the function, \( g(x) = -x^2 - 6x - 9 \). The derivative is: \[ g'(x) = \frac{d}{dx}(-x^2 - 6x - 9) = -2x - 6. \]

Step 2: Solve for critical points

Set the derivative equal to zero and solve for \( x \) to find critical points. \[ -2x - 6 = 0 \] Solving for \( x \): \[ -2x = 6 \] \[ x = -3. \] Thus, there is a critical point at \( x = -3 \).

Step 3: Determine if critical point is a local extremum

Since this is a continuous quadratic function, examine around the critical point \( x = -3 \) within the domain \(-4 \leq x < \infty \). Use the first derivative test or the nature of the function. Since the parabola opens downwards (the coefficient of \( x^2 \) is negative), this point will be a local maximum.

Step 4: Calculate the function's value at the critical point

Now substitute \( x = -3 \) back into the original function to find the value of \( g(x) \) at this local maximum. \[ g(-3) = -(-3)^2 - 6(-3) - 9 = -9 + 18 - 9 = 0. \] So, the function has a local maximum at \( x = -3 \) and \( g(-3) = 0 \).

Step 5: Determine absolute extrema in the given domain

Since the domain \(-4 \leq x < \infty\) extends indefinitely to the right, check the end behavior of the function. As \( x \to \infty \), \[ g(x) = -x^2 - 6x - 9 \to -\infty \]. Evaluate the function at the endpoint \( x = -4 \): \[ g(-4) = -(-4)^2 - 6(-4) - 9 = -16 + 24 - 9 = -1. \] In comparing \( g(-3) = 0 \) (local max) and \( g(-4) = -1 \), since the value \( 0 \) is greater than \( -1 \) and \( g(x) \to -\infty \) as \( x \to \infty \), \( g(-3) = 0 \) is also an absolute maximum within the domain.

Step 6: Use a graphing tool for support

Plot the function \( g(x) = -x^2 - 6x - 9 \) using a graphing calculator or software. By visual examination, observe the parabola opens downwards and confirms that the highest point, located at \( x = -3 \), corresponds to the value 0, demonstrating it as an absolute maximum.

Key Concepts

Derivative

The derivative is a fundamental tool in calculus used to analyze the behavior of functions. In simple terms, it represents the rate at which a function is changing at any given point. For a quadratic function like \( g(x) = -x^2 - 6x - 9 \), finding its derivative helps identify where the function reaches its local extrema (peaks or troughs). To find the derivative, differentiate each term in the polynomial:

• The derivative of \(-x^2\) is \(-2x\).
• The derivative of \(-6x\) is \(-6\).
• The derivative of the constant \(-9\) is 0.

Therefore, the derivative of the function \( g(x) \) is \( g'(x) = -2x - 6 \). By setting this derivative equal to zero, you can solve for \( x \) to pinpoint where critical points, and thus potential local extrema, are located.

Quadratic Functions

Quadratic functions are polynomial functions of degree 2, generally represented as \( f(x) = ax^2 + bx + c \). These functions describe a parabola when graphed on the Cartesian plane. The coefficient \( a \) determines the parabola's direction:

• If \( a > 0 \), the parabola opens upwards, resembling a U-shape.
• If \( a < 0 \), the parabola opens downwards, forming an inverted U-shape.

In the exercise, the quadratic function is \( g(x) = -x^2 - 6x - 9 \), with \( a = -1 \). This means the parabola opens downwards, suggesting a maximum rather than a minimum. Quadratic functions have a symmetrical axis, known as the axis of symmetry, which divides the parabola into two mirror images. Identifying this axis can help find the vertex, which is the highest or lowest point on the graph, depending on the direction the parabola opens.

Critical Points

Critical points occur where the derivative of a function is zero or undefined, indicating potential locations for local extrema. In the given function \( g(x) = -x^2 - 6x - 9 \), the derivative \( g'(x) = -2x - 6 \) is set to zero to find the critical points. Solving the equation \( -2x - 6 = 0 \) yields \( x = -3 \), which is a critical point. At a critical point, further analysis is necessary to determine whether it represents a local maxima, minima, or a saddle point. For quadratic functions, if the parabola opens downwards (i.e., the coefficient of \( x^2 \) is negative), the critical point will represent a local maximum. Thus, at \( x = -3 \), you find the function's value to identify the local maximum: \( g(-3) = 0 \). This value analysis ensures you identify not just a local maximum, but with proper domain boundary evaluation, also potentially an absolute maximum within the given constraints.