Analyzing the Function $f(x) = \frac{1}{\alpha} e^{- |\frac{x}{\alpha}|}$

To identify the correct graph, let's analyze the properties of the function $f(x) = \frac{1}{\alpha} e^{- |\frac{x}{\alpha}|}$, where $\alpha$ is typically a positive constant.

• Symmetry: The function involves $|\frac{x}{\alpha}|$. Since $|-x| = |x|$, we have $f(-x) = \frac{1}{\alpha} e^{- |\frac{-x}{\alpha}|} = \frac{1}{\alpha} e^{- |\frac{x}{\alpha}|} = f(x)$. This indicates the function is even and its graph is symmetric about the y-axis. Graphs 1 and 3 exhibit this symmetry.
• Value at $x=0$: $f(0) = \frac{1}{\alpha} e^{- |\frac{0}{\alpha}|} = \frac{1}{\alpha} e^0 = \frac{1}{\alpha}$. The graph must have a peak or pass through the point $(0, \frac{1}{\alpha})$.
• Behavior as $|x| \to \infty$: As $|x|$ increases, $|\frac{x}{\alpha}|$ increases, so $e^{- |\frac{x}{\alpha}|}$ approaches 0. Therefore, $f(x) \to 0$ as $|x| \to \infty$. The graph must approach the x-axis asymptotically.
• Derivative Analysis:
  • For $x > 0$: $f(x) = \frac{1}{\alpha} e^{-\frac{x}{\alpha}}$. The derivative is $f'(x) = \frac{1}{\alpha} (-\frac{1}{\alpha}) e^{-\frac{x}{\alpha}} = -\frac{1}{\alpha^2} e^{-\frac{x}{\alpha}}$. This is negative, so the function decreases for $x > 0$.
  • For $x < 0$: $f(x) = \frac{1}{\alpha} e^{\frac{x}{\alpha}}$. The derivative is $f'(x) = \frac{1}{\alpha} (\frac{1}{\alpha}) e^{\frac{x}{\alpha}} = \frac{1}{\alpha^2} e^{\frac{x}{\alpha}}$. This is positive, so the function increases for $x < 0$.
  • At $x=0$: The derivative from the left ($x \to 0^-$) is $\frac{1}{\alpha^2}$, and the derivative from the right ($x \to 0^+$) is $-\frac{1}{\alpha^2}$. Since the left-hand and right-hand derivatives are not equal, the function has a sharp peak or cusp at $x=0$.

Comparing with Graph Options

Based on the analysis:

• The graph must be symmetric about the y-axis. This eliminates options 2 and 4.
• The graph must have a sharp peak (cusp) at $x=0$. Graph 1 shows a sharp peak at $x=0$, while Graph 3 shows a smooth peak.
• The function increases for $x<0$ and decreases for $x>0$, approaching 0 as $|x| \to \infty$. Graph 1 exhibits these characteristics.

Therefore, the graph that best represents the function $f(x) = \frac{1}{\alpha} e^{- |\frac{x}{\alpha}|}$ is the one with a sharp peak at the origin and symmetric exponential decay on both sides.

Conclusion: Option 1 matches these characteristics.