Math Problem Statement

We are interested in a function f, but the only graph provided is of its derivative, f', as shown in the image. Analyze the graph of f' to verify which of the given statements about f are true.

Solution

Based on the images you uploaded, let us analyze the graph of f(x)f'(x) and verify the given statements about f(x)f(x):

Observations:

  1. Increasing/Decreasing behavior of f(x)f(x):

    • f(x)f(x) increases where f(x)>0f'(x) > 0 and decreases where f(x)<0f'(x) < 0.
    • From the graph of f(x)f'(x), we see f(x)>0f'(x) > 0 on the intervals: (,6)and(0,6).(-\infty, -6) \quad \text{and} \quad (0, 6).
    • f(x)<0f'(x) < 0 on the intervals: (6,0)and(6,).(-6, 0) \quad \text{and} \quad (6, \infty).
  2. Concavity of f(x)f(x):

    • f(x)f(x) is concave up where f(x)=f(x)f''(x) = f'(x) is increasing (f(x)f'(x) has a positive slope).
    • f(x)f(x) is concave down where f(x)=f(x)f''(x) = f'(x) is decreasing (f(x)f'(x) has a negative slope).
    • From the graph, f(x)f'(x) is:
      • Increasing on (23,23)(-2\sqrt{3}, 2\sqrt{3}), so f(x)f(x) is concave up on this interval.
      • Decreasing on (,23)(23,)(-\infty, -2\sqrt{3}) \cup (2\sqrt{3}, \infty), so f(x)f(x) is concave down on these intervals.
  3. Critical Points of f(x)f(x):

    • Critical points occur where f(x)=0f'(x) = 0, which corresponds to x=6,0,6x = -6, 0, 6.
  4. Inflection Points of f(x)f(x):

    • Inflection points occur where the concavity changes (i.e., where f(x)f'(x) changes from increasing to decreasing or vice versa).
    • This occurs at x=23x = -2\sqrt{3} and x=23x = 2\sqrt{3}.

Statement Validation:

  1. f(x)f(x) is increasing on (,6)(0,6)(-\infty, -6) \cup (0, 6):

    • True. f(x)>0f'(x) > 0 on these intervals.
  2. f(x)f(x) is decreasing on (,6)(0,6)(-\infty, -6) \cup (0, 6):

    • False. f(x)f(x) is decreasing on (6,0)(6,)(-6, 0) \cup (6, \infty), not the given intervals.
  3. f(x)f(x) is concave up on (23,23)(-2\sqrt{3}, 2\sqrt{3}):

    • True. f(x)f'(x) is increasing in this interval.
  4. f(x)f(x) is concave down on (23,23)(-2\sqrt{3}, 2\sqrt{3}):

    • False. f(x)f(x) is concave down on (,23)(23,)(-\infty, -2\sqrt{3}) \cup (2\sqrt{3}, \infty).
  5. The critical points of f(x)f(x) are x=6,x=0,x=6x = -6, x = 0, x = 6:

    • True. These are the points where f(x)=0f'(x) = 0.
  6. The inflection points of f(x)f(x) are x=6,x=0,x=6x = -6, x = 0, x = 6:

    • False. Inflection points occur at x=23x = -2\sqrt{3} and x=23x = 2\sqrt{3}.

Correct Answer:

The correct set of statements is: (1),(3),(5)\boxed{(1), (3), (5)}

Would you like further details or clarifications on any step? Here are some additional questions for deeper understanding:

  1. Why does f(x)f(x) increase when f(x)>0f'(x) > 0?
  2. How do we determine concavity from the graph of f(x)f'(x)?
  3. What distinguishes a critical point from an inflection point?
  4. Why is f(x)f(x) concave up specifically on (23,23)(-2\sqrt{3}, 2\sqrt{3})?
  5. How can we confirm the exact intervals of increasing and decreasing for f(x)f(x) using a test point method?

Tip: Always analyze the behavior of the derivative and its slope to confirm intervals of concavity and

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Math Problem Analysis

Mathematical Concepts

Derivative Analysis
Critical Points
Concavity
Intervals of Increase and Decrease

Formulas

f(x) increases where f'(x) > 0
f(x) is concave up where f''(x) > 0

Theorems

Critical Point Theorem
Inflection Point Definition

Suitable Grade Level

Grades 11-12