The difference between the maximum and minimum values of the function `f(x)=sin^(3)x-3sinx, AA x in [0,(pi)/(6)]` is

Options

A. 2
B. `(1)/(2)`
C. `(11)/(8)`
D. `(7)/(6)`

Answer

To find the difference between the maximum and minimum values of the function \( f(x) = \sin^3 x - 3\sin x \) for \( x \) in the interval \( [0, \frac{\pi}{6}] \), we will follow these steps:

### Step 1: Substitute \( \sin x \) with a new variable
Let \( t = \sin x \). The range of \( t \) for \( x \) in \( [0, \frac{\pi}{6}] \) is from \( \sin(0) = 0 \) to \( \sin(\frac{\pi}{6}) = \frac{1}{2} \). Therefore, \( t \) varies in the interval \( [0, \frac{1}{2}] \).

### Step 2: Rewrite the function in terms of \( t \)
Now, we can rewrite the function \( f(x) \) as:
\[ f(t) = t^3 - 3t \]

### Step 3: Find the derivative of \( f(t) \)
To find the critical points, we take the derivative of \( f(t) \):
\[ f'(t) = 3t^2 - 3 \]
Setting the derivative equal to zero to find critical points:
\[ 3t^2 - 3 = 0 \implies t^2 = 1 \implies t = \pm 1 \]
Since \( t \) must be in the interval \( [0, \frac{1}{2}] \), we ignore \( t = -1 \) and \( t = 1 \) is outside our interval.

### Step 4: Evaluate the function at the endpoints
Since there are no critical points in the interval, we evaluate \( f(t) \) at the endpoints of the interval \( [0, \frac{1}{2}] \).

1. At \( t = 0 \):
\[ f(0) = 0^3 - 3(0) = 0 \]

2. At \( t = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) = \frac{1}{8} - \frac{3}{2} = \frac{1}{8} - \frac{12}{8} = -\frac{11}{8} \]

### Step 5: Find the maximum and minimum values
From the evaluations:
- Maximum value \( f_{\text{max}} = f(0) = 0 \)
- Minimum value \( f_{\text{min}} = f\left(\frac{1}{2}\right) = -\frac{11}{8} \)

### Step 6: Calculate the difference
The difference between the maximum and minimum values is:
\[ f_{\text{max}} - f_{\text{min}} = 0 - \left(-\frac{11}{8}\right) = \frac{11}{8} \]

### Final Answer
Thus, the difference between the maximum and minimum values of the function is:
\[ \frac{11}{8} \]

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