Math Problem Statement

The graph of the derivative f'(x) of a continuous function f(x) is shown. Find the following: (a) On what intervals is f increasing? Decreasing? (b) Locate the x-values where f has a local maximum and minimum. (c) On what intervals is f concave upward? Concave downward? (d) Locate the x-values where f has an inflection point.

Solution

Let’s examine each part of the question one by one based on the provided graph of f(x)f'(x), the derivative of f(x)f(x).

(a) On what intervals is f(x)f(x) increasing? Decreasing?

For f(x)f(x) to be increasing, f(x)f'(x) must be positive (above the x-axis). Similarly, f(x)f(x) is decreasing where f(x)f'(x) is negative (below the x-axis).

From the graph:

  • f(x)f(x) is increasing on the interval (2,1)(2,)(-2, -1) \cup (2, \infty) where f(x)>0f'(x) > 0.
  • f(x)f(x) is decreasing on the interval (,2)(1,2)(-\infty, -2) \cup (-1, 2) where f(x)<0f'(x) < 0.

(b) Locate the xx-value(s) where f(x)f(x) has a local maximum and minimum.

Local maxima and minima of f(x)f(x) occur where f(x)=0f'(x) = 0 (crosses the x-axis) and changes sign:

  • At x=2x = -2: f(x)f'(x) changes from negative to positive, indicating a local minimum of f(x)f(x).
  • At x=1x = -1: f(x)f'(x) changes from positive to negative, indicating a local maximum of f(x)f(x).
  • At x=2x = 2: f(x)f'(x) changes from negative to positive, indicating another local minimum of f(x)f(x).

Thus:

  • Local maximum at x=1x = -1
  • Local minima at x=2x = -2 and x=2x = 2

(c) On what intervals is f(x)f(x) concave upward? Concave downward?

Concavity of f(x)f(x) depends on the derivative f(x)f'(x):

  • f(x)f(x) is concave upward where f(x)f'(x) is increasing (positive slope).
  • f(x)f(x) is concave downward where f(x)f'(x) is decreasing (negative slope).

From the graph:

  • f(x)f(x) is concave upward on the intervals (,0)(1,)(-\infty, 0) \cup (1, \infty) where f(x)f'(x) has a positive slope.
  • f(x)f(x) is concave downward on the interval (0,1)(0, 1) where f(x)f'(x) has a negative slope.

(d) Locate the xx-value(s) where f(x)f(x) has an inflection point.

Inflection points occur where the concavity changes, which is where the slope of f(x)f'(x) changes from positive to negative or vice versa.

From the graph:

  • The concavity changes at x=0x = 0 and x=1x = 1.

Thus, the inflection points of f(x)f(x) are at x=0x = 0 and x=1x = 1.


Would you like further clarification on any part? Here are some related questions that expand on these concepts:

  1. What does the graph of f(x)f(x) look like based on the information given about f(x)f'(x)?
  2. How would changes in f(x)f'(x)'s behavior affect the second derivative f(x)f''(x) and, consequently, the concavity of f(x)f(x)?
  3. Can we determine the relative sizes of the maximum and minimum values of f(x)f(x) at the identified points?
  4. How would the intervals where f(x)f(x) is increasing or decreasing change if f(x)f'(x) was shifted vertically?
  5. What impact would adding a constant to f(x)f(x) have on the locations of maxima, minima, and inflection points?

Tip: Remember that inflection points in a function f(x)f(x) occur where the second derivative f(x)f''(x) changes sign, corresponding to changes in the slope of f(x)f'(x) in this context.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Concavity
Inflection Points
Increasing and Decreasing Intervals

Formulas

If f'(x) > 0, f(x) is increasing
If f'(x) < 0, f(x) is decreasing
Concavity determined by f'(x) slope
Inflection points where concavity changes

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12 or early college calculus