Math Problem Statement

Use a graphing utility to graph f left parenthesis x right parenthesis equals x cubed minus 3 x plus 2f(x)=x3−3x+2 on the interval left bracket negative 2 comma 2 right bracket[−2,2] and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing.

Solution

To analyze the function f(x)=x33x+2f(x) = x^3 - 3x + 2, we can break down the tasks into the following steps:

  1. Graph the Function: We'll use a graphing tool to plot f(x)=x33x+2f(x) = x^3 - 3x + 2 over the interval [2,2][-2, 2].
  2. Find Local Maximum and Minimum Values: We can approximate the local extrema by finding the derivative f(x)f'(x), setting it to zero to locate critical points, and then analyzing the function's behavior around those points.
  3. Determine Intervals of Increase and Decrease: By examining the sign of f(x)f'(x), we can determine where f(x)f(x) is increasing or decreasing.

Step 1: Compute the Derivative

The derivative of f(x)=x33x+2f(x) = x^3 - 3x + 2 is: f(x)=3x23f'(x) = 3x^2 - 3

Setting f(x)=0f'(x) = 0 gives: 3x23=0x2=1x=±13x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1

So, the critical points are at x=1x = 1 and x=1x = -1.

Step 2: Determine Local Extrema

To confirm whether these critical points are local maxima or minima, we can evaluate f(x)f(x) at x=1x = -1 and x=1x = 1 and consider the sign of f(x)f'(x) around these points.

  • At x=1x = -1: f(1)=(1)33(1)+2=1+3+2=4f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4

  • At x=1x = 1: f(1)=(1)33(1)+2=13+2=0f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0

Using these results, we approximate that there is a local maximum at (1,4)(-1, 4) and a local minimum at (1,0)(1, 0).

Step 3: Intervals of Increase and Decrease

To determine the intervals where f(x)f(x) is increasing or decreasing, we analyze f(x)=3x23f'(x) = 3x^2 - 3:

  • When x<1x < -1, f(x)>0f'(x) > 0 (increasing).
  • When 1<x<1-1 < x < 1, f(x)<0f'(x) < 0 (decreasing).
  • When x>1x > 1, f(x)>0f'(x) > 0 (increasing).

Thus, f(x)f(x) is:

  • Increasing on (,1)(-\infty, -1) and (1,)(1, \infty).
  • Decreasing on (1,1)(-1, 1).

Summary

  • Local Maximum: At (1,4)(-1, 4)
  • Local Minimum: At (1,0)(1, 0)
  • Intervals of Increase: (,1)(1,)(-\infty, -1) \cup (1, \infty)
  • Intervals of Decrease: (1,1)(-1, 1)

Would you like a detailed graph or further explanation?


Follow-Up Questions

  1. What is the second derivative of f(x)=x33x+2f(x) = x^3 - 3x + 2, and what does it tell us about concavity?
  2. How would the behavior of f(x)f(x) change if we modified the constant term?
  3. What are the points of inflection for this function, if any?
  4. How does the behavior of f(x)f(x) compare to that of g(x)=x3g(x) = x^3?
  5. What effect does the coefficient of xx have on the local extrema of similar cubic functions?

Tip: When analyzing function behavior, check critical points using both first and second derivatives to understand the shape of the curve.

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Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Intervals of Increase and Decrease

Formulas

Derivative f'(x) = 3x^2 - 3
Critical points: f'(x) = 0
Function evaluation f(x)

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12