math

Function Overview

• The function in consideration is given by:
  $f(x) = -24x + 5$

• We are analyzing this function in the closed interval from $[-3, 6]$.

• Key point: There are no discontinuities since there are no denominators involved, meaning the function is continuous.

Finding Critical Points

• Since the function is continuous, we are assured of the existence of both a maximum and a minimum within the interval.

• We need to identify critical points by using the derivative of the function.

• The derivative of the function is:
  $f'(x) = 3x^2 - 6x - 24$

• To find critical points, we must set the derivative equal to zero:
  $3x^2 - 6x - 24 = 0$

Factoring the Derivative

• Factoring out a 3 from the derivative:
  $3(x^2 - 2x - 8) = 0$

• The equation simplifies to:
  $x^2 - 2x - 8 = 0$
  This factors into:
  $(x - 4)(x + 2) = 0$

• Critical points identified:

  • $x = 4$

  • $x = -2$

Evaluating Critical Points in the Domain

• Important step: Verify if the critical points are within the specified interval $[-3, 6]$.

• Both critical points $-2$ and $4$ are within the domain.

Evaluating the Function at Endpoints and Critical Points

• We evaluate the function $f(x)$ at the endpoints and critical points:

  1. At $x = -3$:

  • Calculation:

    • $f(-3) = -24(-3) + 5 = 72 + 5 = 77$

  • Result: Point is $( -3, 77 )$.

  1. At $x = -2$:

  • Calculation:

    • $f(-2) = -24(-2) + 5 = 48 + 5 = 53$

  • Result: Point is $( -2, 53 )$.

  1. At $x = 4$:

  • Calculation:

    • $f(4) = -24(4) + 5 = -96 + 5 = -91$

  • Result: Point is $( 4, -91 )$.

  1. At $x = 6$:

  • Calculation:

    • $f(6) = -24(6) + 5 = -144 + 5 = -139$

  • Result: Point is $( 6, -139 )$.

Summary of Evaluated Points

• The evaluated points yield:

  • $(-3, 77)$

  • $(-2, 53)$

  • $(4, -91)$

  • $(6, -139)$

Conclusion of Maximum and Minimum Values

• Finding the Maximum:

  • Highest y-value is 77 (from $x = -3$).

  • Maximum point identified:

    • Point: $(-3, 77)$

• Finding the Minimum:

  • Lowest y-value is -139 (from $x = 6$).

  • Minimum point identified:

    • Point: $(6, -139)$

Key Concept Clarifications

• Remember that the derivative is primarily used to determine critical points. Once critical points are identified, evaluations are carried out using the original function.

• Endpoints and valid critical points (within the restricted domain) must be considered for finding absolute maximum and minimum values.

• Scenarios where critical points fall outside the restricted domain mean relying solely on endpoint evaluations to find maximum or minimum values.

Continuous Functions in Context

• For another function example, $f(x) = \frac{x}{x^2 + 2}$ analyzed from $[0, 4]$:

  • Noteworthy point: The function is continuous as the denominator $x^2 + 2$ does not permit zero.

  • Critical points derived through the quotient rule and evaluated similarly to the previous example are essential for finding the absolute extrema on given intervals.

  Quotient Rule Definition:

  • The derivative using the quotient rule is given as:
    $f'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^{2}}$ where $f(x)$ is the numerator and $g(x)$ the denominator.

Final Thoughts

• Highest and lowest values ultimately depend on careful evaluations, showing the importance of derivative analysis and critical point checks within specified intervals.

• Utilizing calculators is acceptable for evaluating possibly complex expression outputs, especially under logarithmic scenarios.

• Understanding the range of functions within specified domains enhances our mathematical comprehension and real-world applications.