Math Problem Statement

Найти экстремумы функции

Найти экстремумы функции

Solution

To find the extrema of the function f(x)=x36x23f(x) = \sqrt[3]{x^3 - 6x^2}, we will follow these steps:

  1. Find the derivative f(x)f'(x).
  2. Set f(x)=0f'(x) = 0 to find critical points.
  3. Analyze the critical points and intervals using the first or second derivative test to classify extrema.

Let’s calculate step by step.

Step 1: Derivative of f(x)f(x)

The derivative of a cube root function g(x)3\sqrt[3]{g(x)} is: ddxg(x)3=13g(x)2/3g(x)\frac{d}{dx} \sqrt[3]{g(x)} = \frac{1}{3} g(x)^{-2/3} \cdot g'(x) Here, g(x)=x36x2g(x) = x^3 - 6x^2 and g(x)=3x212xg'(x) = 3x^2 - 12x.

So, f(x)=13(x36x2)2/3(3x212x)f'(x) = \frac{1}{3} (x^3 - 6x^2)^{-2/3} \cdot (3x^2 - 12x)

Simplify: f(x)=(3x212x)3(x36x2)2/3f'(x) = \frac{(3x^2 - 12x)}{3 \cdot (x^3 - 6x^2)^{2/3}}

Step 2: Solve f(x)=0f'(x) = 0

The numerator 3x212x3x^2 - 12x must equal zero: 3x(x4)=03x(x - 4) = 0 Thus, x=0x = 0 or x=4x = 4.

The denominator (x36x2)2/30(x^3 - 6x^2)^{2/3} \neq 0 for these points since it only vanishes at x=0x = 0 or x=6x = 6, but cube roots are defined everywhere.

Step 3: Analyze Critical Points

We will examine x=0x = 0 and x=4x = 4 using the first derivative test.

First Derivative Test

  • For x<0x < 0, choose x=1x = -1: f(x)>0f'(x) > 0 (positive slope).
  • For 0<x<40 < x < 4, choose x=2x = 2: f(x)<0f'(x) < 0 (negative slope).
  • For x>4x > 4, choose x=5x = 5: f(x)>0f'(x) > 0 (positive slope).

Behavior:

  • At x=0x = 0, f(x)f(x) changes from increasing to decreasing. This is a local maximum.
  • At x=4x = 4, f(x)f(x) changes from decreasing to increasing. This is a local minimum.

Extrema:

  • Local maximum at x=0x = 0, f(0)=036023=0f(0) = \sqrt[3]{0^3 - 6 \cdot 0^2} = 0.
  • Local minimum at x=4x = 4, f(4)=436423=64963=323=2f(4) = \sqrt[3]{4^3 - 6 \cdot 4^2} = \sqrt[3]{64 - 96} = \sqrt[3]{-32} = -2.

Final Answer:

  • Local maximum: (0,0)(0, 0)
  • Local minimum: (4,2)(4, -2)

Would you like further details or explanations?


Related Questions:

  1. How do you find inflection points for this function?
  2. What is the second derivative f(x)f''(x) for further analysis?
  3. How do we graph f(x)=x36x23f(x) = \sqrt[3]{x^3 - 6x^2}?
  4. What are the domain and range of f(x)f(x)?
  5. How does the behavior of f(x)f(x) change as x±x \to \pm\infty?

Tip:

When dealing with cube roots, always pay attention to the behavior near zero, as the derivative can have unique properties.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Extrema Analysis

Formulas

Derivative of cube root: d/dx(√[3]{g(x)}) = (1/3)g(x)^(-2/3) * g'(x)
Critical points: f'(x) = 0 or undefined
First Derivative Test

Theorems

Critical Point Theorem
First Derivative Test

Suitable Grade Level

Grades 11-12 and College Level