Math Problem Statement

Consider the polynomial function f shown to the right. State x values where f'(x) = 0, f'(x) < 0, and f'(x) > 0. Additionally, state x values where f''(x) = 0, f''(x) < 0, and f''(x) > 0.

Solution

The problem is asking to analyze the graph of the polynomial function f(x)f(x) to answer various questions about its first and second derivatives, f(x)f'(x) and f(x)f''(x), based on the graphical information.

Let's break down each part of the question:

a. State the xx-values where f(x)=0f'(x) = 0.

The derivative f(x)f'(x) is zero at critical points, which occur at the local maxima and minima of the function. From the graph, the critical points are:

  • x=3x = -3 (local maximum)
  • x=1x = 1 (local minimum)

Thus, the xx-values where f(x)=0f'(x) = 0 are: x=3,1x = -3, 1

b. State the open intervals of xx where f(x)<0f'(x) < 0.

f(x)<0f'(x) < 0 where the function is decreasing. From the graph, the function is decreasing in two intervals:

  • Between the local maximum at x=3x = -3 and the local minimum at x=1x = 1.

Thus, the open interval is: (3,1)(-3, 1)

c. State the open intervals of xx where f(x)>0f'(x) > 0.

f(x)>0f'(x) > 0 where the function is increasing. From the graph, the function is increasing in two intervals:

  • From x=x = -\infty to x=3x = -3 (before the local maximum).
  • From x=1x = 1 to x=x = \infty (after the local minimum).

Thus, the open intervals are: (,3),(1,)(-\infty, -3), (1, \infty)

d. State the xx-values where f(x)=0f''(x) = 0.

f(x)=0f''(x) = 0 at points of inflection, where the concavity changes. From the graph, there is a point of inflection near x=1x = -1, where the concavity shifts from concave up to concave down.

Thus, the xx-value where f(x)=0f''(x) = 0 is approximately: x=1x = -1

e. State the open intervals of xx where f(x)<0f''(x) < 0.

f(x)<0f''(x) < 0 where the graph is concave down. This happens from the point of inflection at x=1x = -1 to the local minimum at x=1x = 1.

Thus, the open interval is: (1,1)(-1, 1)

f. State the open intervals of xx where f(x)>0f''(x) > 0.

f(x)>0f''(x) > 0 where the graph is concave up. This happens in two regions:

  • From x=x = -\infty to the point of inflection at x=1x = -1.
  • From x=1x = 1 to x=x = \infty.

Thus, the open intervals are: (,1),(1,)(-\infty, -1), (1, \infty)


Would you like further details on any of these answers or the reasoning? Here are some additional questions to deepen understanding:

  1. How would you determine concavity using the second derivative test?
  2. Why do critical points occur where f(x)=0f'(x) = 0?
  3. Can there be more than one point of inflection on a polynomial function?
  4. How is the behavior of the graph different when f(x)>0f'(x) > 0 versus when f(x)<0f'(x) < 0?
  5. Why is the second derivative important in analyzing the shape of the graph?

Tip: To find intervals of increase and decrease, always look for the critical points where the slope (derivative) is zero, then analyze the sign of the slope on either side of those points.

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Math Problem Analysis

Mathematical Concepts

Calculus
First Derivative
Second Derivative
Increasing and Decreasing Functions
Concavity
Inflection Points

Formulas

f'(x) = 0 at critical points
f''(x) = 0 at inflection points

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12