The task in the image is to investigate the continuity of the given function at specific points ($x_1 = 0$ and $x_2 = 2$) and to plot the function. The function is:

$y = \frac{1}{2 - e^{x^2}}$

Steps to Analyze:

1. Continuity Check: A function $f(x)$ is continuous at a point $x = c$ if the following three conditions are met:

   • $f(c)$ is defined.

   • The limit $\lim_{x \to c} f(x)$ exists.

   • The value of the function at $c$ equals the limit: $f(c) = \lim_{x \to c} f(x)$.

   • Point $x = 0$: Substitute $x = 0$ into the denominator: $2 - e^{(0)^2} = 2 - e^0 = 2 - 1 = 1$ The denominator does not become zero, so the function is defined at $x = 0$. Next, check the limit as $x \to 0$: $\lim_{x \to 0} \frac{1}{2 - e^{x^2}} = \frac{1}{2 - e^0} = \frac{1}{1}.$ Since the limit exists and equals the value of the function, the function is continuous at $x = 0$.

   • Point $x = 2$: Substitute $x = 2$ into the denominator: $2 - e^{(2)^2} = 2 - e^4.$ Since $e^4$ is much larger than 2, the denominator becomes negative, but it does not equal zero. Thus, the function is defined at $x = 2$. Check the limit as $x \to 2$: $\lim_{x \to 2} \frac{1}{2 - e^{x^2}} = \frac{1}{2 - e^4}.$ As the limit exists and equals the function value, it is continuous at $x = 2$.

2. Graph of the Function:

   • The function is asymptotic to vertical lines where the denominator $2 - e^{x^2} = 0$, i.e., $x^2 = \ln(2)$. The approximate value of $\sqrt{\ln(2)}$ is $\pm 0.832$.
   • At these points, the function diverges to infinity.

Let me compute and generate the graph for this function to confirm observations. numpy as np import matplotlib.pyplot as plt

Define the function

def f(x): return 1 / (2 - np.exp(x**2))

Define x values avoiding discontinuity points (asymptotes)

x = np.linspace(-2, 2, 1000) y = f(x)

Identify points where the function is not defined

y[np.abs(2 - np.exp(x**2)) < 1e-6] = np.nan # Avoid division by zero

Plot the function

plt.figure(figsize=(10, 6)) plt.plot(x, y, label=r"$y = \frac{1}{2 - e^{x^2}}$", color='blue') plt.axhline(0, color='black', linewidth=0.5, linestyle='--') plt.axvline(0, color='black', linewidth=0.5, linestyle='--')

Highlighting the points x1=0 and x2=2

x1, x2 = 0, 2 plt.scatter([x1, x2], [f(x1), f(x2)], color='red', label=f'Points: x1={x1}, x2={x2}')

Title and Labels

plt.title("Graph of y = 1 / (2 - e^(x^2))") plt.xlabel("x") plt.ylabel("y") plt.ylim(-10, 10) # Limiting y for visualization purposes plt.legend() plt.grid()

Show the plot

plt.show()### Results:

1. The function $y = \frac{1}{2 - e^{x^2}}$ is continuous at $x_1 = 0$ and $x_2 = 2$, as the function is defined and the limits exist at both points.
2. The graph shows asymptotic behavior at $x = \pm \sqrt{\ln(2)}$, where the denominator $2 - e^{x^2} = 0$.

Would you like further explanation or calculations regarding the continuity, limits, or graph?

Related Questions:

1. How do you determine discontinuities in functions involving exponentials?
2. What are the implications of asymptotes for the behavior of functions?
3. Can we extend this analysis to study differentiability at $x_1 = 0$ and $x_2 = 2$?
4. How does the exponential term affect the continuity and range of this function?
5. What happens to the function as $x$ approaches infinity or negative infinity?

Tip:

Always check the denominator of a rational function to identify points of discontinuity or undefined behavior.