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Вопрос от Анонимного юзера 03 февраля 2025 02:08

1.5x-5/3x^3;[0;2]

Ответ нейросети

03 февраля 2025 02:08

To analyze the function ( f(x) = \frac{1.5x - 5}{3x^3} ) on the interval ([0, 2]), we first rewrite the function in a more manageable form:

[ f(x) = \frac{1.5x - 5}{3x^3} = \frac{1.5}{3} \cdot \frac{x - \frac{10}{3}}{x^3} = \frac{0.5(x - \frac{10}{3})}{x^3} ]

Next, we need to examine the interval ([0, 2]) and identify any critical points, discontinuities, or end behavior.

Step 1: Domain Considerations

The function is defined for ( x \neq 0 ) because of the ( x^3 ) denominator. Therefore, the domain on the interval ([0, 2]) is any point in ((0, 2]).

Step 2: Finding Critical Points

To find the critical points, we first need to take the derivative of ( f(x) ) and set it to zero:

Using the quotient rule:

[ f'(x) = \frac{(3x^3)(1.5) - (1.5x - 5)(9x^2)}{(3x^3)^2} ]

Simplifying the numerator:

  1. ( 3x^3 \cdot 1.5 = 4.5x^3 )
  2. Distribute ( (1.5x - 5)(9x^2) ):

[ (1.5x)(9x^2) - (5)(9x^2) = 13.5x^3 - 45x^2 ]

So the derivative becomes:

[ f'(x) = \frac{4.5x^3 - (13.5x^3 - 45x^2)}{(3x^3)^2} = \frac{4.5x^3 - 13.5x^3 + 45x^2}{9x^6} ]

[ = \frac{-9x^3 + 45x^2}{9x^6} = \frac{45x^2 - 9x^3}{9x^6} ]

Setting the numerator equal to zero:

[ 45x^2 - 9x^3 = 0 \ 9x^2(5 - x) = 0 ]

This gives us critical points ( x = 0 ) (not in the domain) and ( x = 5 ) (not in the interval).

Step 3: Evaluate Endpoints in the Interval

Since ( x = 0 ) is not in the domain and there are no critical points in the interval ((0, 2]), we evaluate the function at the endpoint ( x = 2 ).

[ f(2) = \frac{1.5(2) - 5}{3(2)^3} = \frac{3 - 5}{3 \cdot 8} = \frac{-2}{24} = -\frac{1}{12} ]

Step 4: Conclusion

The function ( f(x) ) is decreasing towards ( x = 2 ) and goes to ( -\infty ) as ( x ) approaches ( 0 ) from the right. The only value we can confirm in the specified interval is at the endpoint:

  • ( f(2) = -\frac{1}{12} )

Thus, the only output of interest in the interval [0, 2] is:

  • The function approaches ( -\infty ) as ( x ) approaches ( 0^+ ),
  • ( f(2) = -\frac{1}{12} ).

If you have further questions or need additional analysis or specific points of interest, feel free to ask!

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1.5x-5/3x^3;[0;2]
To analyze the function \( f(x) = \frac{1.5x - 5}{3x^3} \) on the interval \([0, 2]\), we first rewrite the function in a more manageable form: \[ f(x) = \frac{1.5x - 5}{3x^3} = \frac{1.5}{3} \cdot \frac{x - \frac{10}{3}}{x^3} = \frac{0.5(x - \frac{10}{3})}{x^3} \] Next, we need to examine the interval \([0, 2]\) and identify any critical points, discontinuities, or end behavior. ### Step 1: Domain Considerations The function is defined for \( x \neq 0 \) because of the \( x^3 \) denominator. Therefore, the domain on the interval \([0, 2]\) is any point in \((0, 2]\). ### Step 2: Finding Critical Points To find the critical points, we first need to take the derivative of \( f(x) \) and set it to zero: Using the quotient rule: \[ f'(x) = \frac{(3x^3)(1.5) - (1.5x - 5)(9x^2)}{(3x^3)^2} \] Simplifying the numerator: 1. \( 3x^3 \cdot 1.5 = 4.5x^3 \) 2. Distribute \( (1.5x - 5)(9x^2) \): \[ (1.5x)(9x^2) - (5)(9x^2) = 13.5x^3 - 45x^2 \] So the derivative becomes: \[ f'(x) = \frac{4.5x^3 - (13.5x^3 - 45x^2)}{(3x^3)^2} = \frac{4.5x^3 - 13.5x^3 + 45x^2}{9x^6} \] \[ = \frac{-9x^3 + 45x^2}{9x^6} = \frac{45x^2 - 9x^3}{9x^6} \] Setting the numerator equal to zero: \[ 45x^2 - 9x^3 = 0 \\ 9x^2(5 - x) = 0 \] This gives us critical points \( x = 0 \) (not in the domain) and \( x = 5 \) (not in the interval). ### Step 3: Evaluate Endpoints in the Interval Since \( x = 0 \) is not in the domain and there are no critical points in the interval \((0, 2]\), we evaluate the function at the endpoint \( x = 2 \). \[ f(2) = \frac{1.5(2) - 5}{3(2)^3} = \frac{3 - 5}{3 \cdot 8} = \frac{-2}{24} = -\frac{1}{12} \] ### Step 4: Conclusion The function \( f(x) \) is decreasing towards \( x = 2 \) and goes to \( -\infty \) as \( x \) approaches \( 0 \) from the right. The only value we can confirm in the specified interval is at the endpoint: - \( f(2) = -\frac{1}{12} \) Thus, the only output of interest in the interval [0, 2] is: - The function approaches \( -\infty \) as \( x \) approaches \( 0^+ \), - \( f(2) = -\frac{1}{12} \). If you have further questions or need additional analysis or specific points of interest, feel free to ask!