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Вопрос от Анонимного юзера 14 января 2025 04:04

Х³-9х^2+24х-1 [3,6]

Ответ нейросети

14 января 2025 04:04

To analyze the function ( f(x) = x^3 - 9x^2 + 24x - 1 ) over the interval ([3, 6]), we can follow these steps:

  1. Evaluate the Function at the Endpoints:

    • Calculate ( f(3) )
    • Calculate ( f(6) )
  2. Find the Critical Points:

    • Take the derivative ( f'(x) ) and set it to zero to find critical points within the interval ([3, 6]).
    • Solve ( f'(x) = 0 ).
  3. Evaluate the Function at Critical Points:

    • Determine which critical points lie within ([3, 6]) and evaluate ( f ) at those points.
  4. Determine the Maximum and Minimum Values:

    • Compare the values of ( f(3) ), ( f(6) ), and the values from any critical points to find the maximum and minimum on the interval.

Let's perform these steps as follows:

Step 1: Evaluate at the Endpoints

Calculate: [ f(3) = 3^3 - 9 \cdot 3^2 + 24 \cdot 3 - 1 = 27 - 81 + 72 - 1 = 17 ] [ f(6) = 6^3 - 9 \cdot 6^2 + 24 \cdot 6 - 1 = 216 - 324 + 144 - 1 = 35 ]

Step 2: Find the Derivative

Calculate the derivative: [ f'(x) = 3x^2 - 18x + 24 ] Set the derivative to zero: [ 3x^2 - 18x + 24 = 0 ] Dividing by 3: [ x^2 - 6x + 8 = 0 ] Factoring: [ (x - 2)(x - 4) = 0 ] Thus, (x = 2) and (x = 4) are the critical points.

Step 3: Evaluate at Critical Points

Since we are only interested in the interval ([3, 6]), we evaluate: [ f(4) = 4^3 - 9 \cdot 4^2 + 24 \cdot 4 - 1 = 64 - 144 + 96 - 1 = 15 ]

Step 4: Compare Values

Now we have:

  • (f(3) = 17)
  • (f(4) = 15)
  • (f(6) = 35)

Conclusion

The maximum value of ( f(x) ) on the interval ([3, 6]) is ( 35 ) at ( x = 6 ), and the minimum value is ( 15 ) at ( x = 4 ).

Thus, the maximum and minimum values of the function on the interval ([3, 6]) are:

  • Maximum: ( 35 ) at ( x = 6 )
  • Minimum: ( 15 ) at ( x = 4 )

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х³-9х^2+24х-1 [3,6]
To analyze the function \( f(x) = x^3 - 9x^2 + 24x - 1 \) over the interval \([3, 6]\), we can follow these steps: 1. **Evaluate the Function at the Endpoints:** - Calculate \( f(3) \) - Calculate \( f(6) \) 2. **Find the Critical Points:** - Take the derivative \( f'(x) \) and set it to zero to find critical points within the interval \([3, 6]\). - Solve \( f'(x) = 0 \). 3. **Evaluate the Function at Critical Points**: - Determine which critical points lie within \([3, 6]\) and evaluate \( f \) at those points. 4. **Determine the Maximum and Minimum Values**: - Compare the values of \( f(3) \), \( f(6) \), and the values from any critical points to find the maximum and minimum on the interval. Let's perform these steps as follows: ### Step 1: Evaluate at the Endpoints Calculate: \[ f(3) = 3^3 - 9 \cdot 3^2 + 24 \cdot 3 - 1 = 27 - 81 + 72 - 1 = 17 \] \[ f(6) = 6^3 - 9 \cdot 6^2 + 24 \cdot 6 - 1 = 216 - 324 + 144 - 1 = 35 \] ### Step 2: Find the Derivative Calculate the derivative: \[ f'(x) = 3x^2 - 18x + 24 \] Set the derivative to zero: \[ 3x^2 - 18x + 24 = 0 \] Dividing by 3: \[ x^2 - 6x + 8 = 0 \] Factoring: \[ (x - 2)(x - 4) = 0 \] Thus, \(x = 2\) and \(x = 4\) are the critical points. ### Step 3: Evaluate at Critical Points Since we are only interested in the interval \([3, 6]\), we evaluate: \[ f(4) = 4^3 - 9 \cdot 4^2 + 24 \cdot 4 - 1 = 64 - 144 + 96 - 1 = 15 \] ### Step 4: Compare Values Now we have: - \(f(3) = 17\) - \(f(4) = 15\) - \(f(6) = 35\) ### Conclusion The maximum value of \( f(x) \) on the interval \([3, 6]\) is \( 35 \) at \( x = 6 \), and the minimum value is \( 15 \) at \( x = 4 \). Thus, the maximum and minimum values of the function on the interval \([3, 6]\) are: - Maximum: \( 35 \) at \( x = 6 \) - Minimum: \( 15 \) at \( x = 4 \)