Math Problem Statement

Discuss the curve y = 3x^4 - 4x^3 with respect to concavity, points of inflection, and local maxima and minima.

Solution

To analyze the curve given by y=3x44x3y = 3x^4 - 4x^3 with respect to concavity, points of inflection, and local maxima/minima, let's proceed with the following steps:

Step 1: Find the first derivative to analyze critical points.

The first derivative, yy', will help identify where the slope of the curve is zero (potential local maxima or minima).

y=ddx(3x44x3)=12x312x2y' = \frac{d}{dx}(3x^4 - 4x^3) = 12x^3 - 12x^2

Step 2: Set y=0y' = 0 to find critical points.

Solve 12x312x2=012x^3 - 12x^2 = 0 by factoring:

12x2(x1)=012x^2(x - 1) = 0

This gives the critical points x=0x = 0 and x=1x = 1.

Step 3: Second derivative to analyze concavity and inflection points.

The second derivative, yy'', will help determine concavity and points of inflection.

y=ddx(12x312x2)=36x224xy'' = \frac{d}{dx}(12x^3 - 12x^2) = 36x^2 - 24x

Step 4: Set y=0y'' = 0 to find inflection points.

Solve 36x224x=036x^2 - 24x = 0 by factoring:

12x(3x2)=012x(3x - 2) = 0

This gives the potential inflection points x=0x = 0 and x=23x = \frac{2}{3}.

Step 5: Determine concavity and classify critical points.

  • For concavity:
    • If y>0y'' > 0, the curve is concave up.
    • If y<0y'' < 0, the curve is concave down.

Evaluate yy'' at test points around the critical points x=0x = 0 and x=1x = 1.

  • For x=1x = -1: y(1)=36(1)+24=60y''(-1) = 36(1) + 24 = 60 → Concave up.
  • For x=12x = \frac{1}{2}: y(12)=36(14)24(12)=912=3y''\left(\frac{1}{2}\right) = 36\left(\frac{1}{4}\right) - 24\left(\frac{1}{2}\right) = 9 - 12 = -3 → Concave down.
  • For x=2x = 2: y(2)=36(4)24(2)=14448=96y''(2) = 36(4) - 24(2) = 144 - 48 = 96 → Concave up.

Step 6: Classify critical points x=0x = 0 and x=1x = 1.

  • At x=0x = 0:
    • y(0)=0y'(0) = 0 and y(0)=0y''(0) = 0 → Possible inflection point.
  • At x=1x = 1:
    • y(1)=0y'(1) = 0 and y(1)=12y''(1) = 12 → Local minimum.

Summary of Analysis:

  • Critical Points:
    • x=0x = 0: Possible inflection point.
    • x=1x = 1: Local minimum.
  • Inflection Points: x=0x = 0, x=23x = \frac{2}{3}.
  • Concavity:
    • Concave up for x<0x < 0 and x>1x > 1.
    • Concave down between 0<x<10 < x < 1.

You can now sketch the curve using this information.

Would you like to see further details or clarification?

Here are 5 related questions:

  1. How do you interpret the behavior of the curve at each critical point?
  2. Can you confirm where the concavity changes based on the second derivative?
  3. How can you determine whether x=0x = 0 is a true inflection point?
  4. What role does the second derivative play in classifying maxima and minima?
  5. How would the graph differ if the coefficients of x4x^4 and x3x^3 were altered?

Tip: When sketching the curve, remember that inflection points indicate changes in concavity, and local extrema correspond to peaks and valleys.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Curve Analysis
Derivatives
Inflection Points
Concavity
Maxima and Minima

Formulas

First derivative to find critical points
Second derivative to find concavity and inflection points

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12