Math Problem Statement

given the function f(x)=2x^3-6x+3. find the intervals of increasing and decreasing? Find any local maximum and local minimum points as ordered pairs? identify the intervals of concavity? Find any inflection points as ordered pairs?

Solution

Given the function f(x)=2x36x+3f(x) = 2x^3 - 6x + 3, we need to find:

  1. Intervals of increasing and decreasing
  2. Local maximum and local minimum points
  3. Intervals of concavity
  4. Inflection points

Step 1: First derivative (for increasing/decreasing intervals and critical points)

To find where the function is increasing or decreasing, we first find the first derivative f(x)f'(x):

f(x)=ddx(2x36x+3)=6x26f'(x) = \frac{d}{dx}(2x^3 - 6x + 3) = 6x^2 - 6

Now, set f(x)=0f'(x) = 0 to find the critical points:

6x26=06x^2 - 6 = 0 x2=1x^2 = 1 x=±1x = \pm 1

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

Step 2: Test for increasing/decreasing intervals

We use a sign chart for f(x)=6x26f'(x) = 6x^2 - 6 to determine where the function is increasing or decreasing.

  • For x<1x < -1, choose x=2x = -2: f(2)=6(2)26=246=18(positive, so increasing on (,1))f'(-2) = 6(-2)^2 - 6 = 24 - 6 = 18 \quad (\text{positive, so increasing on } (-\infty, -1))
  • For 1<x<1-1 < x < 1, choose x=0x = 0: f(0)=6(0)26=6(negative, so decreasing on (1,1))f'(0) = 6(0)^2 - 6 = -6 \quad (\text{negative, so decreasing on } (-1, 1))
  • For x>1x > 1, choose x=2x = 2: f(2)=6(2)26=246=18(positive, so increasing on (1,))f'(2) = 6(2)^2 - 6 = 24 - 6 = 18 \quad (\text{positive, so increasing on } (1, \infty))

Step 3: Local maximum and minimum

  • At x=1x = -1, the function changes from increasing to decreasing, so there is a local maximum at x=1x = -1.
  • At x=1x = 1, the function changes from decreasing to increasing, so there is a local minimum at x=1x = 1.

To find the values of f(x)f(x) at these points:

  • f(1)=2(1)36(1)+3=2+6+3=7f(-1) = 2(-1)^3 - 6(-1) + 3 = -2 + 6 + 3 = 7
  • f(1)=2(1)36(1)+3=26+3=1f(1) = 2(1)^3 - 6(1) + 3 = 2 - 6 + 3 = -1

So the local maximum is at (1,7)(-1, 7), and the local minimum is at (1,1)(1, -1).

Step 4: Second derivative (for concavity and inflection points)

To determine the concavity, we find the second derivative f(x)f''(x):

f(x)=ddx(6x26)=12xf''(x) = \frac{d}{dx}(6x^2 - 6) = 12x

Now, set f(x)=0f''(x) = 0 to find potential inflection points:

12x=012x = 0 x=0x = 0

So, x=0x = 0 is a possible inflection point.

Step 5: Test for concavity and inflection points

We check the sign of f(x)f''(x) on either side of x=0x = 0:

  • For x<0x < 0, choose x=1x = -1: f(1)=12(1)=12(concave down on (,0))f''(-1) = 12(-1) = -12 \quad (\text{concave down on } (-\infty, 0))
  • For x>0x > 0, choose x=1x = 1: f(1)=12(1)=12(concave up on (0,))f''(1) = 12(1) = 12 \quad (\text{concave up on } (0, \infty))

Since the concavity changes at x=0x = 0, there is an inflection point at x=0x = 0. To find the corresponding yy-value:

f(0)=2(0)36(0)+3=3f(0) = 2(0)^3 - 6(0) + 3 = 3

So the inflection point is at (0,3)(0, 3).

Summary of results:

  1. Intervals of increasing: (,1)(-\infty, -1) and (1,)(1, \infty)
  2. Intervals of decreasing: (1,1)(-1, 1)
  3. Local maximum: (1,7)(-1, 7)
  4. Local minimum: (1,1)(1, -1)
  5. Intervals of concavity:
    • Concave down on (,0)(-\infty, 0)
    • Concave up on (0,)(0, \infty)
  6. Inflection point: (0,3)(0, 3)

Would you like further details on any of these steps?


Here are 5 related questions:

  1. What happens if we change the coefficients in the cubic function?
  2. How do you determine if a point is a local maximum or minimum without using the first derivative test?
  3. Can a function have more than one inflection point?
  4. How does the shape of a cubic function change with different coefficients?
  5. How do you find critical points for more complex functions like trigonometric or logarithmic?

Tip: When analyzing concavity, always remember that inflection points occur where the concavity changes from up to down or vice versa.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Concavity
Inflection Points

Formulas

First derivative: f'(x) = 6x^2 - 6
Second derivative: f''(x) = 12x

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus)