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Задача Задача №1629533
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25 сентября 2024 20:16 · публичный вопрос
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F(x)=1/5х^5-5/3х^3+6х

Нейросеть25 сентября 2024 20:16

To analyze the function \( f(x) = \frac{1}{5}x^5 - \frac{5}{3}x^3 + 6x \), we can find its critical points, determine intervals of increase and decrease, and identify any local maxima or minima. Here's how you can approach it:

Step 1: Find the First Derivative

The first step is to find the derivative of the function, \( f'(x) \).

\[ f'(x) = \frac{d}{dx}\left(\frac{1}{5}x^5\right) - \frac{d}{dx}\left(\frac{5}{3}x^3\right) + \frac{d}{dx}(6x) \]

\[ f'(x) = x^4 - 5x^2 + 6 \]

Step 2: Set the First Derivative to Zero

Next, set the first derivative equal to zero to find critical points.

\[ x^4 - 5x^2 + 6 = 0 \]

Let \( u = x^2 \), then the equation becomes:

\[ u^2 - 5u + 6 = 0 \]

Step 3: Factor or Use the Quadratic Formula

This factors to:

\[ (u - 2)(u - 3) = 0 \]

Thus, \( u = 2 \) or \( u = 3 \).

Now reverting back to \( x \):

\[ x^2 = 2 \quad \Rightarrow \quad x = \pm\sqrt{2} \] \[ x^2 = 3 \quad \Rightarrow \quad x = \pm\sqrt{3} \]

Step 4: Identify Intervals of Increase and Decrease

Now, we need to determine the sign of \( f'(x) \) in the intervals defined by our critical points \( -\sqrt{3}, -\sqrt{2}, \sqrt{2}, \sqrt{3} \).

  1. Choose a test point from each interval:

    • \( (-\infty, -\sqrt{3}) \)
    • \( (-\sqrt{3}, -\sqrt{2}) \)
    • \( (-\sqrt{2}, \sqrt{2}) \)
    • \( (\sqrt{2}, \sqrt{3}) \)
    • \( (\sqrt{3}, \infty) \)
  2. Evaluate \( f'(x) \) for each test point in the intervals.

Step 5: Find Second Derivative (Optional for Concavity)

To determine the concavity and locate inflection points, find the second derivative \( f''(x) \):

\[ f''(x) = 4x^3 - 10x \]

Setting \( f''(x) = 0 \) helps find points of inflection:

\[ 4x(x^2 - \frac{5}{2}) = 0 \]

This yields:

  • \( x = 0 \)
  • \( x = \pm\sqrt{\frac{5}{2}} \)

Summary of Results

  • Critical Points: \( x = \pm\sqrt{2}, \pm\sqrt{3} \)
  • Local Maxima/Minima: Evaluate \( f(x) \) at critical points and use the first derivative test.
  • Intervals of Increase/Decrease: Based on the sign of \( f'(x) \).

Further Analysis

You can proceed by calculating values at the critical points to determine local maxima and minima, plotting \( f(x) \), and examining the behavior at the endpoints if you're interested in global behavior.

f(x)=1/5х^5-5/3х^3+6х
To analyze the function \( f(x) = \frac{1}{5}x^5 - \frac{5}{3}x^3 + 6x \), we can find its critical points, determine intervals of increase and decrease, and identify any local maxima or minima. Here's how you can approach it: ### Step 1: Find the First Derivative The first step is to find the derivative of the function, \( f'(x) \). \[ f'(x) = \frac{d}{dx}\left(\frac{1}{5}x^5\right) - \frac{d}{dx}\left(\frac{5}{3}x^3\right) + \frac{d}{dx}(6x) \] \[ f'(x) = x^4 - 5x^2 + 6 \] ### Step 2: Set the First Derivative to Zero Next, set the first derivative equal to zero to find critical points. \[ x^4 - 5x^2 + 6 = 0 \] Let \( u = x^2 \), then the equation becomes: \[ u^2 - 5u + 6 = 0 \] ### Step 3: Factor or Use the Quadratic Formula This factors to: \[ (u - 2)(u - 3) = 0 \] Thus, \( u = 2 \) or \( u = 3 \). Now reverting back to \( x \): \[ x^2 = 2 \quad \Rightarrow \quad x = \pm\sqrt{2} \] \[ x^2 = 3 \quad \Rightarrow \quad x = \pm\sqrt{3} \] ### Step 4: Identify Intervals of Increase and Decrease Now, we need to determine the sign of \( f'(x) \) in the intervals defined by our critical points \( -\sqrt{3}, -\sqrt{2}, \sqrt{2}, \sqrt{3} \). 1. Choose a test point from each interval: - \( (-\infty, -\sqrt{3}) \) - \( (-\sqrt{3}, -\sqrt{2}) \) - \( (-\sqrt{2}, \sqrt{2}) \) - \( (\sqrt{2}, \sqrt{3}) \) - \( (\sqrt{3}, \infty) \) 2. Evaluate \( f'(x) \) for each test point in the intervals. ### Step 5: Find Second Derivative (Optional for Concavity) To determine the concavity and locate inflection points, find the second derivative \( f''(x) \): \[ f''(x) = 4x^3 - 10x \] Setting \( f''(x) = 0 \) helps find points of inflection: \[ 4x(x^2 - \frac{5}{2}) = 0 \] This yields: - \( x = 0 \) - \( x = \pm\sqrt{\frac{5}{2}} \) ### Summary of Results - **Critical Points**: \( x = \pm\sqrt{2}, \pm\sqrt{3} \) - **Local Maxima/Minima**: Evaluate \( f(x) \) at critical points and use the first derivative test. - **Intervals of Increase/Decrease**: Based on the sign of \( f'(x) \). ### Further Analysis You can proceed by calculating values at the critical points to determine local maxima and minima, plotting \( f(x) \), and examining the behavior at the endpoints if you're interested in global behavior.