Which of the following functions is not an are not an injective map(s) ?

Options

A. `f(x)=|x+1|,x in [-1,oo)`
B. `g(x)=x+(1)/(x), x in (0,oo)`
C. `h(x)=x^(2)+4x-5, x in (0,oo)`
D. `k(x)=e^(-x), x in [0,oo)`

Answer

To determine which of the given functions is not an injective map, we will analyze each option based on the properties of their derivatives. A function is injective (one-to-one) if it is either strictly increasing or strictly decreasing over its entire domain.

### Step-by-Step Solution:

1. **Understanding Injective Functions**:
- A function \( f(x) \) is injective if for every \( x_1 \) and \( x_2 \) in the domain, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
- This can be determined by examining the derivative \( \frac{dy}{dx} \). If \( \frac{dy}{dx} > 0 \) for all \( x \) in the domain, the function is strictly increasing (and thus injective). If \( \frac{dy}{dx} < 0 \), the function is strictly decreasing (also injective). If the derivative changes sign, the function is not injective.

2. **Analyzing Each Function**:
- **Option 1**: \( f(x) = x \) for \( x > -1 \)
- Derivative: \( \frac{dy}{dx} = 1 \)
- Since the derivative is always positive, this function is injective.

- **Option 2**: \( f(x) = x - \frac{1}{x} \) for \( x > 0 \)
- Derivative:
\[ \frac{dy}{dx} = 1 + \frac{1}{x^2} \]
- The derivative is positive for \( x > 1 \) and negative for \( 0 < x < 1 \). Since the derivative changes sign, this function is not injective.

- **Option 3**: \( f(x) = x^2 + 4x \)
- Derivative:
\[ \frac{dy}{dx} = 2x + 4 \]
- The derivative is always positive for all \( x \) (since \( 2x + 4 > 0 \) for all \( x \)). Thus, this function is injective.

- **Option 4**: \( f(x) = -e^{-x} \)
- Derivative:
\[ \frac{dy}{dx} = -e^{-x} \]
- The derivative is always negative, indicating that the function is strictly decreasing. Therefore, this function is injective.

3. **Conclusion**:
- The only function that is not injective is from **Option 2**: \( f(x) = x - \frac{1}{x} \) for \( x > 0 \).

### Final Answer:
The function that is not an injective map is **Option 2**: \( f(x) = x - \frac{1}{x} \).