Find the absolute maximum and minimum values of g(t) = 8t - t^4 on [-2, 1]
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
- 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:
$$ g'(t) = \frac{d}{dt}(8t - t^4) = 8 - 4t^3 $$
- 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 ):
$$ 4t^3 = 8 $$
$$ t^3 = 2 $$
$$ t = \sqrt[3]{2} $$
- 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}) ):
$$ = 8(\sqrt[3]{2}) - 2^{\frac{4}{3}} $$
- 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 )
- 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.
