The least value of `g(t) = 8t- t^(4)` on [-2, 1]` is

Options

A. `-16`
B. `-20`
C. `-32`
D. 7

Answer

To find the least value of the function \( g(t) = 8t - t^4 \) on the interval \([-2, 1]\), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of \( g(t) \).

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

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find the critical points.

\[ 8 - 4t^3 = 0 \]

Solving for \( t \):

\[ 4t^3 = 8 \implies t^3 = 2 \implies t = 2^{1/3} \]

### Step 3: Identify the critical points and endpoints
The critical point we found is \( t = 2^{1/3} \). We also need to evaluate the function at the endpoints of the interval, which are \( t = -2 \) and \( t = 1 \).

### Step 4: Evaluate the function at the critical point and endpoints
Now we will calculate \( g(t) \) at the critical point and the endpoints.

1. **At \( t = -2 \)**:
\[ g(-2) = 8(-2) - (-2)^4 = -16 - 16 = -32 \]

2. **At \( t = 1 \)**:
\[ g(1) = 8(1) - (1)^4 = 8 - 1 = 7 \]

3. **At \( t = 2^{1/3} \)**:
\[ g(2^{1/3}) = 8(2^{1/3}) - (2^{1/3})^4 = 8(2^{1/3}) - 4(2^{4/3}) = 8(2^{1/3}) - 4(2^{4/3}) = 4(2^{1/3})(2 - 1) = 4(2^{1/3}) \]
(Note: We can calculate this value if needed, but we can already see that it will be positive.)

### Step 5: Compare the values
Now we compare the values we calculated:

- \( g(-2) = -32 \)
- \( g(1) = 7 \)
- \( g(2^{1/3}) \) is positive (but we can see it is not the minimum).

### Conclusion
The least value of \( g(t) \) on the interval \([-2, 1]\) is:

\[ \boxed{-32} \]