Answer is: option2

\( 0 \) only

Solution:

Given function values:

\[ f(-1) = \frac{-1}{-1+2} = \frac{-1}{1} = -1 \]

\[ f(2) = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2} \]

Computing the mean value theorem slope:

\[ \frac{f(2) - f(-1)}{2 - (-1)} = \frac{\frac{1}{2} - (-1)}{2+1} = \frac{\frac{1}{2} + 1}{3} = \frac{\frac{3}{2}}{3} = \frac{1}{2} \]

Thus, we must solve for \( c \) such that:

\[ f'(c) = \frac{1}{2} \]

Using the quotient rule:

\[ \left( \frac{x}{x+2} \right)' = \frac{(x+2)(1) - x(1)}{(x+2)^2} = \frac{x+2-x}{(x+2)^2} = \frac{2}{(x+2)^2} \]

Thus,

\[ f'(x) = \frac{2}{(x+2)^2} \]

Setting \( f'(c) = \frac{1}{2} \):

\[ \frac{2}{(c+2)^2} = \frac{1}{2} \]

Cross multiplying:

\[ 2 \times 2 = (c+2)^2 \]

\[ 4 = (c+2)^2 \]

Taking the square root:

\[ c+2 = \pm 2 \]

\[ c = -2 \pm 2 \]

Thus, the two possible values are:

\[ c = 0 \quad \text{or} \quad c = -4 \]

Checking interval validity:

• \( c = 0 \) is in the interval \( (-1,2) \).
• \( c = -4 \) is not in the interval.

Thus, the only valid solution is \( c = 0 \).

The correct answer is: 0