Question

Which ONE of the following curves represents the equation \( y = e^{-x} \)?
Figures not to scale

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Check the sign of the derivative of \( e^{-x} \) and see whether the curve should rise or fall as x increases.

Answer 1

The Correct Option is B

Step 1: Find key points on the curve \( y = e^{-x} \).
At \( x = 0 \), \( y = e^{0} = 1 \), so the curve passes through the point \( (0,1) \), just like every option shown.

Step 2: Check how \( y \) behaves as \( x \) grows in each direction.
As \( x \to +\infty \), \( e^{-x} \to 0 \), so the curve must flatten down toward the x-axis on the right side without ever touching it. As \( x \to -\infty \), \( e^{-x} \to \infty \), so the curve must shoot up steeply on the left side. This means the curve is a smooth, always-positive, strictly decreasing curve.

Step 3: Confirm it is strictly decreasing with the derivative.
\( \dfrac{dy}{dx} = -e^{-x} \), which is negative for every real \( x \), since \( e^{-x} \) is always positive. A negative derivative everywhere means \( y \) keeps falling as \( x \) increases, with no flat spots or corners anywhere.

Step 4: Match this behaviour to the four curves shown.
Option A rises from left to right and blows up as \( x \to +\infty \), so its derivative is positive, the opposite of what is needed. Option C and option D both have a sharp corner at the origin and are not smooth exponential curves at all. Option B is high on the left, passes through \( (0,1) \), and flattens toward the x-axis on the right, which is exactly the shape found in Step 2.

Final Answer:
The curve that is steep and high on the left, passes through \( (0,1) \), and decays toward zero on the right is option B.
\[ \boxed{\text{Option B}} \]