Least value of the function , `f(x)=2^(x^2)-1+2/(2^(x^2)+1)` is :

Options

A. 0
B. `(3)/(2)`
C. `(2)/(3)`
D. 1

Answer

To find the least value of the function \( f(x) = 2^{x^2} - 1 + \frac{2}{2^{x^2} + 1} \), we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative \( f'(x) \) to locate the critical points. The derivative of \( 2^{x^2} \) can be found using the formula:
\[ \frac{d}{dx}(a^{g(x)}) = a^{g(x)} \cdot \ln(a) \cdot g'(x) \]
where \( g(x) = x^2 \) and \( a = 2 \). The derivative of \( \frac{2}{2^{x^2} + 1} \) can be found using the quotient rule.

### Step 2: Apply the derivative rules
Differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2^{x^2}) - 0 + \frac{d}{dx}\left(\frac{2}{2^{x^2} + 1}\right) \]
Using the chain rule for the first term:
\[ \frac{d}{dx}(2^{x^2}) = 2^{x^2} \cdot \ln(2) \cdot (2x) \]
For the second term, using the quotient rule:
\[ \frac{d}{dx}\left(\frac{2}{2^{x^2} + 1}\right) = -\frac{2 \cdot (2^{x^2} \cdot \ln(2) \cdot (2x))}{(2^{x^2} + 1)^2} \]
Combining these gives:
\[ f'(x) = 2^{x^2} \cdot 2x \ln(2) - \frac{4x \cdot 2^{x^2} \ln(2)}{(2^{x^2} + 1)^2} \]

### Step 3: Set the derivative to zero
To find critical points, set \( f'(x) = 0 \):
\[ 2^{x^2} \cdot 2x \ln(2) - \frac{4x \cdot 2^{x^2} \ln(2)}{(2^{x^2} + 1)^2} = 0 \]
Factoring out \( 2^{x^2} \cdot x \ln(2) \):
\[ 2^{x^2} \cdot x \ln(2) \left( 2 - \frac{4}{(2^{x^2} + 1)^2} \right) = 0 \]
This gives us two cases:
1. \( x = 0 \)
2. \( 2 - \frac{4}{(2^{x^2} + 1)^2} = 0 \)

### Step 4: Solve the second case
From the second case:
\[ 2 = \frac{4}{(2^{x^2} + 1)^2} \implies (2^{x^2} + 1)^2 = 2 \implies 2^{x^2} + 1 = \sqrt{2} \]
This leads to \( 2^{x^2} = \sqrt{2} - 1 \), which is not possible since \( 2^{x^2} \) is always non-negative.

### Step 5: Evaluate at the critical point
The only critical point is \( x = 0 \). Now, we evaluate \( f(0) \):
\[ f(0) = 2^{0^2} - 1 + \frac{2}{2^{0^2} + 1} = 1 - 1 + \frac{2}{1 + 1} = 0 + 1 = 1 \]

### Step 6: Confirm it's a minimum
To confirm that this is a minimum, we can check the second derivative \( f''(x) \) at \( x = 0 \). If \( f''(0) > 0 \), then \( x = 0 \) is indeed a local minimum.

### Conclusion
Thus, the least value of the function \( f(x) \) is:
\[ \boxed{1} \]