Let `f(x)=2tan^(3)x-6tan^(2)x+1+sgn(e^(x)),AA x in [-(pi)/(4),(pi)/(4)],` Then the positive difference between the least value and the local maximum value of the function is (where sgn (f(x)) represents the signum function)

Options

A. 7
B. 8
C. 9
D. 10

Answer

To solve the problem, we need to analyze the function \( f(x) = 2\tan^3 x - 6\tan^2 x + 1 + \text{sgn}(e^x) \) for \( x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

### Step 1: Understand the Signum Function
The signum function \( \text{sgn}(e^x) \) is always 1 for all \( x \) since \( e^x > 0 \) for all real \( x \). Therefore, we can simplify our function:
\[ f(x) = 2\tan^3 x - 6\tan^2 x + 1 + 1 = 2\tan^3 x - 6\tan^2 x + 2. \]

### Step 2: Define \( y = \tan x \)
Let \( y = \tan x \). The range of \( y \) when \( x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \) is \( [-1, 1] \). Now we can rewrite \( f(x) \) in terms of \( y \):
\[ f(y) = 2y^3 - 6y^2 + 2. \]

### Step 3: Find the Critical Points
To find the local maxima and minima, we need to compute the derivative \( f'(y) \):
\[ f'(y) = 6y^2 - 12y. \]
Setting the derivative equal to zero to find critical points:
\[ 6y^2 - 12y = 0 \implies 6y(y - 2) = 0. \]
Thus, \( y = 0 \) or \( y = 2 \). Since \( y \) is constrained to \( [-1, 1] \), we only consider \( y = 0 \).

### Step 4: Evaluate \( f(y) \) at Critical Points and Endpoints
Now we evaluate \( f(y) \) at the critical point and the endpoints of the interval:
1. At \( y = -1 \):
\[ f(-1) = 2(-1)^3 - 6(-1)^2 + 2 = -2 - 6 + 2 = -6. \]
2. At \( y = 0 \):
\[ f(0) = 2(0)^3 - 6(0)^2 + 2 = 2. \]
3. At \( y = 1 \):
\[ f(1) = 2(1)^3 - 6(1)^2 + 2 = 2 - 6 + 2 = -2. \]

### Step 5: Determine the Least Value and Local Maximum
From our evaluations:
- The least value of \( f(y) \) is \( -6 \) (at \( y = -1 \)).
- The local maximum value of \( f(y) \) is \( 2 \) (at \( y = 0 \)).

### Step 6: Calculate the Positive Difference
The positive difference between the least value and the local maximum value is:
\[ \text{Positive Difference} = 2 - (-6) = 2 + 6 = 8. \]

### Final Answer
Thus, the positive difference between the least value and the local maximum value of the function is \( \boxed{8} \).