Math Problem Statement

Use a graphing utility to graph

f left parenthesis x right parenthesis equals x cubed minus 3 x plus 4f(x)=x3−3x+4

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+4f(x) = x^3 - 3x + 4, let's break down the process into several steps:

1. Graphing the Function on the Interval [2,2][-2, 2]

We'll create a graph of f(x)=x33x+4f(x) = x^3 - 3x + 4 over the interval [2,2][-2, 2]. This graph will help us identify any local maxima and minima visually.

2. Finding Local Maximum and Minimum Values

To locate the local maximum and minimum values of f(x)f(x), we need to:

  • Take the first derivative of f(x)f(x):
    f(x)=3x23f'(x) = 3x^2 - 3

  • Set f(x)=0f'(x) = 0 to find critical points: 3x23=03x^2 - 3 = 0 x2=1x^2 = 1 x=±1x = \pm 1

Thus, the critical points are x=1x = -1 and x=1x = 1.

  • Evaluate f(x)f(x) at these critical points to determine if they correspond to local maxima or minima.

    • f(1)=(1)33(1)+4=1+3+4=6f(-1) = (-1)^3 - 3(-1) + 4 = -1 + 3 + 4 = 6
    • f(1)=(1)33(1)+4=13+4=2f(1) = (1)^3 - 3(1) + 4 = 1 - 3 + 4 = 2

From this, we find:

  • A local maximum at x=1x = -1 with f(1)=6f(-1) = 6.
  • A local minimum at x=1x = 1 with f(1)=2f(1) = 2.

3. Determining Where the Function is Increasing and Decreasing

Using the first derivative f(x)=3x23f'(x) = 3x^2 - 3, we can test intervals around the critical points to see where f(x)f(x) is increasing or decreasing:

  • For x<1x < -1, let x=2x = -2: f(2)=3(2)23=123=9>0f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9 > 0 So, f(x)f(x) is increasing on (,1)(-\infty, -1).

  • For 1<x<1-1 < x < 1, let x=0x = 0: f(0)=3(0)23=3<0f'(0) = 3(0)^2 - 3 = -3 < 0 So, f(x)f(x) is decreasing on (1,1)(-1, 1).

  • For x>1x > 1, let x=2x = 2: f(2)=3(2)23=123=9>0f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 > 0 So, f(x)f(x) is increasing on (1,)(1, \infty).

Summary of Results

  • Local Maximum: f(1)=6f(-1) = 6 at x=1x = -1.
  • Local Minimum: f(1)=2f(1) = 2 at x=1x = 1.
  • Intervals:
    • f(x)f(x) is increasing on (,1)(1,)(-\infty, -1) \cup (1, \infty).
    • f(x)f(x) is decreasing on (1,1)(-1, 1).

Would you like further clarification or a specific graph generated for this function?


Follow-Up Questions

  1. What are inflection points, and how do they affect the shape of the graph?
  2. How would this function behave outside of the interval [2,2][-2, 2]?
  3. Could you explain how the second derivative test works for determining concavity?
  4. Why is it important to examine where the derivative is zero or undefined?
  5. How would transformations like shifts or stretches affect the critical points?

Tip: When analyzing the behavior of a function, always check for intervals of increase and decrease by testing points around critical values.

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Math Problem Analysis

Mathematical Concepts

Calculus
Graphing
Derivative
Critical Points
Increasing and Decreasing Functions

Formulas

f(x) = x^3 - 3x + 4
First Derivative: f'(x) = 3x^2 - 3

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12