Let `g(x)` be a polynomial of degree one and `f(x)` is defined by `f(x)={g(x)`, `xleq0` and `((1+x)/(2+x))^(1/x)`,`xgt0`} Find `g(x)` such that `f(x)` is continuous and `f'(1)=f(-1)`

Answer

To solve the problem, we need to find the polynomial \( g(x) \) of degree one such that the function \( f(x) \) is continuous and satisfies the condition \( f'(1) = f(-1) \).

### Step 1: Define the polynomial \( g(x) \)
Let \( g(x) = ax + b \), where \( a \) and \( b \) are constants.

### Step 2: Define the function \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} g(x) & \text{if } x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & \text{if } x > 0 \end{cases} \]

### Step 3: Ensure continuity at \( x = 0 \)
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]
Calculating the left-hand limit:
\[ \lim_{x \to 0^-} f(x) = g(0) = b \]
Calculating the right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} = \left(\frac{1}{2}\right)^{\infty} = 0 \]
Thus, we have:
\[ b = 0 \]

### Step 4: Update \( g(x) \)
Now, we can update \( g(x) \):
\[ g(x) = ax + 0 = ax \]

### Step 5: Find \( f(-1) \)
Now, we calculate \( f(-1) \):
\[ f(-1) = g(-1) = a(-1) = -a \]

### Step 6: Find \( f'(1) \)
For \( x > 0 \), we need to differentiate \( f(x) \):
\[ f(x) = \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} \]
Taking the natural logarithm:
\[ \ln(f(x)) = \frac{1}{x} \ln\left(\frac{1+x}{2+x}\right) \]
Differentiating both sides using the product rule:
\[ \frac{f'(x)}{f(x)} = -\frac{1}{x^2} \ln\left(\frac{1+x}{2+x}\right) + \frac{1}{x} \cdot \frac{1}{\frac{1+x}{2+x}} \cdot \left(\frac{(2+x) - (1+x)}{(2+x)^2}\right) \]
Evaluating at \( x = 1 \):
\[ f'(1) = f(1) \cdot \left(-\frac{1}{1^2} \ln\left(\frac{2}{3}\right) + \frac{1}{1} \cdot \frac{1}{\frac{2}{3}} \cdot \frac{1}{(2)^2}\right) \]
Calculating \( f(1) \):
\[ f(1) = \left(\frac{2}{3}\right)^{1} = \frac{2}{3} \]
Thus, we have:
\[ f'(1) = \frac{2}{3} \left(-\ln\left(\frac{2}{3}\right) + \frac{3}{2}\cdot\frac{1}{4}\right) \]

### Step 7: Set up the equation \( f'(1) = f(-1) \)
We need to set:
\[ f'(1) = -a \]
This gives us the equation:
\[ \frac{2}{3} \left(-\ln\left(\frac{2}{3}\right) + \frac{3}{8}\right) = -a \]
Thus:
\[ a = -\frac{2}{3} \left(-\ln\left(\frac{2}{3}\right) + \frac{3}{8}\right) \]

### Step 8: Final expression for \( g(x) \)
Substituting \( a \) back into \( g(x) \):
\[ g(x) = -\frac{2}{3} \left(-\ln\left(\frac{2}{3}\right) + \frac{3}{8}\right)x \]

### Conclusion
The polynomial \( g(x) \) that makes \( f(x) \) continuous and satisfies \( f'(1) = f(-1) \) is:
\[ g(x) = -\frac{2}{3} \left(-\ln\left(\frac{2}{3}\right) + \frac{3}{8}\right)x \]