Find the absolute maximum and minimum of the function g(t) = 8t - t^4 in the interval [-2, 1]. Find the absolute maximum and minimum of the function g(x) = x^(2/3) in the interval... Find the absolute maximum and minimum of the function g(t) = 8t - t^4 in the interval [-2, 1]. Find the absolute maximum and minimum of the function g(x) = x^(2/3) in the interval [-2, 3].
Understand the Problem
The question is asking to find the absolute maximum and minimum values of two given functions within specified intervals. This involves applying calculus concepts to determine critical points and evaluating the functions at those points and at the endpoints of the given intervals.
Answer
For $g(t) = 8t - t^4$: Absolute max = 7, min = -32. For $g(x) = x^{2/3}$: Absolute max $\approx 2.080$, min = 0.
Answer for screen readers
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For $g(t) = 8t - t^4$:
- Absolute Maximum: $g(1) = 7$
- Absolute Minimum: $g(-2) = -32$
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For $g(x) = x^{2/3}$:
- Absolute Maximum: $g(3) \approx 2.080$
- Absolute Minimum: $g(0) = 0$
Steps to Solve
- Find the derivative of the first function
To find critical points for $g(t) = 8t - t^4$, we need to calculate the derivative:
$$ g'(t) = 8 - 4t^3 $$
- Set the derivative to zero and solve for critical points
We set the derivative equal to zero:
$$ 8 - 4t^3 = 0 $$
Solving for $t$ gives:
$$ 4t^3 = 8 \implies t^3 = 2 \implies t = \sqrt[3]{2} $$
- Evaluate the endpoints and critical points for the first function
We need to evaluate $g(t)$ at the endpoints $t = -2$ and $t = 1$, and 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 $t = \sqrt[3]{2}: g(\sqrt[3]{2}) = 8(\sqrt[3]{2}) - (\sqrt[3]{2})^4$
Next, compute:
$$ g(\sqrt[3]{2}) = 8(\sqrt[3]{2}) - 2^{4/3} $$
- Calculate the values
Plug in $g(\sqrt[3]{2})$ approximately: $$ g(\sqrt[3]{2}) \approx 8(1.26) - 2.52 \approx 10.08 - 2.52 \approx 7.56 $$
- Determine the maximum and minimum for the first function
We compare the values:
- $g(-2) = -32$
- $g(1) = 7$
- $g(\sqrt[3]{2}) \approx 7.56$
The absolute maximum is $g(1) = 7$ and the absolute minimum is $g(-2) = -32$.
- Find the derivative of the second function
Now, for $g(x) = x^{2/3}$:
$$ g'(x) = \frac{2}{3}x^{-1/3} $$
- Set the derivative to zero and solve for critical points
Setting the derivative equal to zero:
$$ \frac{2}{3}x^{-1/3} = 0 $$
This has no solution; however, we note that the derivative is undefined at $x = 0$.
- Evaluate the endpoints and critical points for the second function
Evaluate at the endpoints $x = -2$ and $x = 3$, as well as at the critical point $x = 0$.
- For $x = -2: g(-2) = (-2)^{2/3} = \sqrt[3]{4} \approx 1.587$
- For $x = 0: g(0) = 0^{2/3} = 0$
- For $x = 3: g(3) = 3^{2/3} = \sqrt[3]{9} \approx 2.080$
- Determine the maximum and minimum for the second function
Comparing the values:
- $g(-2) \approx 1.587$
- $g(0) = 0$
- $g(3) \approx 2.080$
The absolute maximum is $g(3) \approx 2.080$ and the absolute minimum is $g(0) = 0$.
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For $g(t) = 8t - t^4$:
- Absolute Maximum: $g(1) = 7$
- Absolute Minimum: $g(-2) = -32$
-
For $g(x) = x^{2/3}$:
- Absolute Maximum: $g(3) \approx 2.080$
- Absolute Minimum: $g(0) = 0$
More Information
The problem requires identifying absolute extrema values of functions over specified intervals, which is a common task in calculus involving derivatives, critical points, and endpoint evaluations. The methods used here are fundamental in optimization problems.
Tips
- Forgetting to evaluate the function at critical points and only focusing on endpoints.
- Miscalculating the values at critical points or endpoints.
- Not considering points where the derivative is undefined.
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