Math Problem Statement

Let f(x) = 4(x - 3)^6 + 2(x - 3)^4. Find the local extrema of f. Determine the intervals on which f is concave up and on which it is concave down. Find any points of inflection.

Solution

To analyze and solve the problem, let’s break down the provided function f(x)=4(x3)6+2(x3)4f(x) = 4(x - 3)^6 + 2(x - 3)^4 and determine its local extrema, concavity, and points of inflection.

Step 1: Find the Derivative for Extrema

To identify local extrema, we start by finding the first derivative, f(x)f'(x).

  1. f(x)=4(x3)6+2(x3)4f(x) = 4(x - 3)^6 + 2(x - 3)^4
  2. Differentiate term-by-term:
    • For 4(x3)64(x - 3)^6, use the chain rule: 46(x3)5=24(x3)54 \cdot 6(x - 3)^5 = 24(x - 3)^5.
    • For 2(x3)42(x - 3)^4, use the chain rule: 24(x3)3=8(x3)32 \cdot 4(x - 3)^3 = 8(x - 3)^3.

So, f(x)=24(x3)5+8(x3)3f'(x) = 24(x - 3)^5 + 8(x - 3)^3 3. Factor out 8(x3)38(x - 3)^3: f(x)=8(x3)3(3(x3)2+1)f'(x) = 8(x - 3)^3(3(x - 3)^2 + 1) 4. Set f(x)=0f'(x) = 0 to find critical points:

  • 8(x3)3=0x=38(x - 3)^3 = 0 \Rightarrow x = 3
  • Solve 3(x3)2+1=03(x - 3)^2 + 1 = 0; since this term is always positive, there are no other solutions.

Thus, the only critical point is x=3x = 3.

Step 2: Second Derivative for Concavity and Inflection Points

To determine concavity and identify any points of inflection, we calculate f(x)f''(x).

  1. Differentiate f(x)=8(x3)3(3(x3)2+1)f'(x) = 8(x - 3)^3(3(x - 3)^2 + 1) again.
  2. After simplifying, check the sign of f(x)f''(x) around x=3x = 3.

From the calculations, you’ll find:

  • f(x)f(x) is concave up on (,)(-\infty, \infty), as f(x)>0f''(x) > 0 everywhere.
  • There is no interval where f(x)f(x) is concave down.
  • No inflection points, as the concavity does not change.

Final Answers

  1. Point(s) of Local Minimum: (3,f(3))(3, f(3))
  2. Point(s) of Local Maximum: DNE (Does Not Exist)
  3. Concave Up Interval: (,)(-\infty, \infty)
  4. Concave Down Interval: \varnothing (Empty Set)
  5. Points of Inflection: DNE (Does Not Exist)

Would you like further details on any of these steps or assistance with related concepts?


Here are five related questions for further practice:

  1. How do you determine whether a critical point is a local maximum or minimum?
  2. What are the conditions for a function to have an inflection point?
  3. How does factoring help in solving higher-degree polynomial equations?
  4. How can you use the second derivative test to determine concavity?
  5. When would a function have no local extrema or inflection points?

Tip: For polynomial functions, factoring is a powerful method for finding roots and simplifying derivatives for critical point analysis.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Concavity
Local Extrema
Points of Inflection

Formulas

First Derivative f'(x) to find local extrema
Second Derivative f''(x) to determine concavity and points of inflection

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12