Question

The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

• A. \( e^{-3} \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( -e^3 \)

Hint:

For any strictly decreasing function in \( [a, b] \), the least value is always \( f(b) \).
For any strictly increasing function in \( [a, b] \), the least value is always \( f(a) \).

Answer 1

The Correct Option is A

Concept:
• For a continuous and monotonic function, the absolute maximum and minimum values in a closed interval \( [a, b] \) occur at the endpoints or at critical points.
• An exponential function \( e^{kx} \) is strictly monotonic.
• If \( k < 0 \), the function is strictly decreasing.

Step 1: Determine the nature of the function by finding its derivative
The given function is: \[ f(x) = e^{-x} \] Differentiating with respect to \( x \): \[ f'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x} \] Since \( e^{-x} > 0 \) for all real \( x \), we have: \[ f'(x) = -e^{-x} < 0 \text{ for all } x \in [0, 3] \]

Step 2: Evaluate the function at the boundary points of the interval
Because \( f'(x) < 0 \), the function \( f(x) \) is strictly decreasing on the interval \( [0, 3] \).
This implies the maximum value is at the left endpoint and the minimum value is at the right endpoint. Value at the left endpoint \( x = 0 \): \[ f(0) = e^{-(0)} = 1 \] Value at the right endpoint \( x = 3 \): \[ f(3) = e^{-3} \]

Step 3: Identify the least value
Since the function is strictly decreasing: \[ f(3) < f(0) \implies e^{-3} < 1 \] Thus, the least value of the function in the interval \( [0, 3] \) is \( e^{-3} \).