Math Problem Statement

F(x)=x^3-3x^2+2x find tge local maxima and minima

Solution

To find the local maxima and minima of the function F(x)=x33x2+2xF(x) = x^3 - 3x^2 + 2x, we need to follow these steps:

Step 1: Find the first derivative of F(x)F(x)

The first derivative will help us find the critical points where the slope is zero.

F(x)=ddx(x33x2+2x)F'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2x) F(x)=3x26x+2F'(x) = 3x^2 - 6x + 2

Step 2: Set the first derivative equal to zero to find critical points

3x26x+2=03x^2 - 6x + 2 = 0

Solve this quadratic equation using the quadratic formula: x=(6)±(6)24(3)(2)2(3)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} x=6±36246x = \frac{6 \pm \sqrt{36 - 24}}{6} x=6±126x = \frac{6 \pm \sqrt{12}}{6} x=6±236x = \frac{6 \pm 2\sqrt{3}}{6} x=1±33x = 1 \pm \frac{\sqrt{3}}{3}

Thus, the critical points are: x1=1+33,x2=133x_1 = 1 + \frac{\sqrt{3}}{3}, \quad x_2 = 1 - \frac{\sqrt{3}}{3}

Step 3: Use the second derivative to classify the critical points

The second derivative will help us determine whether each critical point is a local maximum or minimum.

F(x)=ddx(3x26x+2)F''(x) = \frac{d}{dx}(3x^2 - 6x + 2) F(x)=6x6F''(x) = 6x - 6

Now, evaluate F(x)F''(x) at each critical point:

  1. For x1=1+33x_1 = 1 + \frac{\sqrt{3}}{3}: F(x1)=6(1+33)6=6+236=23F''(x_1) = 6\left(1 + \frac{\sqrt{3}}{3}\right) - 6 = 6 + 2\sqrt{3} - 6 = 2\sqrt{3} Since F(x1)>0F''(x_1) > 0, x1x_1 is a local minimum.

  2. For x2=133x_2 = 1 - \frac{\sqrt{3}}{3}: F(x2)=6(133)6=6236=23F''(x_2) = 6\left(1 - \frac{\sqrt{3}}{3}\right) - 6 = 6 - 2\sqrt{3} - 6 = -2\sqrt{3} Since F(x2)<0F''(x_2) < 0, x2x_2 is a local maximum.

Conclusion:

  • Local minimum at x1=1+33x_1 = 1 + \frac{\sqrt{3}}{3}.
  • Local maximum at x2=133x_2 = 1 - \frac{\sqrt{3}}{3}.

Would you like more details or further clarification?

Here are 5 related questions to explore:

  1. How do you find inflection points of a function?
  2. What is the significance of the second derivative in classifying critical points?
  3. Can local maxima and minima occur at endpoints of a function's domain?
  4. What are the global maximum and minimum values of a function?
  5. How does the behavior of cubic functions differ from quadratic functions in terms of critical points?

Tip: Always check the second derivative at critical points to confirm whether they are maxima or minima.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Second Derivative Test

Formulas

First derivative: F'(x) = 3x^2 - 6x + 2
Second derivative: F''(x) = 6x - 6
Quadratic formula: x = [-b ± √(b² - 4ac)] / 2a

Theorems

Second Derivative Test: Determines whether a critical point is a local maximum or minimum by evaluating the second derivative at that point

Suitable Grade Level

Grades 11-12