Find the absolute maximum and minimum values of g(t) = 8t - t^4 on [-2, 1]

Question image

Understand the Problem

The question is asking to find the absolute maximum and minimum values of the function g(t) = 8t - t^4 on the interval [-2, 1]. This involves evaluating the function at critical points and the endpoints of the interval.

Answer

The absolute maximum is approximately $7.56$ and the absolute minimum is $-32$.
Answer for screen readers

The absolute maximum value of ( g(t) = 8t - t^4 ) on the interval ([-2, 1]) is approximately $7.56$ at $t = \sqrt[3]{2}$, and the absolute minimum value is $-32$ at $t = -2$.

Steps to Solve

  1. Find the derivative of the function

To locate critical points, we first need to find the derivative of the function ( g(t) = 8t - t^4 ).

The derivative is given by:

$$ g'(t) = \frac{d}{dt}(8t - t^4) = 8 - 4t^3 $$

  1. Set the derivative to zero

Next, we set the derivative equal to zero to find the critical points:

$$ 8 - 4t^3 = 0 $$

Solving for ( t ):

$$ 4t^3 = 8 $$

$$ t^3 = 2 $$

$$ t = \sqrt[3]{2} $$

  1. Evaluate the function at the endpoints and critical points

Now, we evaluate ( g(t) ) at the endpoints, ( t = -2 ) and ( t = 1 ), and at the critical point ( t = \sqrt[3]{2} ).

  • For ( t = -2 ):

$$ g(-2) = 8(-2) - (-2)^4 = -16 - 16 = -32 $$

  • For ( t = 1 ):

$$ g(1) = 8(1) - (1)^4 = 8 - 1 = 7 $$

  • For ( t = \sqrt[3]{2} ):

$$ g(\sqrt[3]{2}) = 8(\sqrt[3]{2}) - (\sqrt[3]{2})^4 $$

Calculating ( g(\sqrt[3]{2}) ):

$$ = 8(\sqrt[3]{2}) - 2^{\frac{4}{3}} $$

  1. Compare the values

After calculating ( g(\sqrt[3]{2}) ), we need to compare all values to find absolute maximum and minimum.

Assuming ( g(\sqrt[3]{2}) \approx 8(1.26) - 2.52 ):

  • Calculation: ( g(\sqrt[3]{2}) \approx 10.08 - 2.52 \approx 7.56 )

Now, we have:

  • ( g(-2) = -32 )
  • ( g(1) = 7 )
  • ( g(\sqrt[3]{2}) \approx 7.56 )
  1. Identify the maximum and minimum

From the values computed:

  • Absolute maximum ( \approx 7.56 ) at ( t = \sqrt[3]{2} )
  • Absolute minimum ( -32 ) at ( t = -2 )

The absolute maximum value of ( g(t) = 8t - t^4 ) on the interval ([-2, 1]) is approximately $7.56$ at $t = \sqrt[3]{2}$, and the absolute minimum value is $-32$ at $t = -2$.

More Information

This problem involves finding absolute extrema on a closed interval, which is important in calculus for understanding the behavior of functions. Evaluating these points helps us compare values definitively.

Tips

  • Forgetting to check endpoints could lead to missing the absolute extrema.
  • Incorrectly computing the function value at the critical points or endpoints.
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