The image asks to select all correct statements related to the boundedness of certain functions over given intervals.

Let's go through each statement:

1. $y = e^{-x^2}$ is unbounded on $(-\infty, +\infty)$:

   • The function $e^{-x^2}$ is a Gaussian function, which approaches zero as $|x| \to \infty$ and has a maximum value at $x = 0$. Since it has a maximum and decays to zero at infinity, it is bounded on $(-\infty, +\infty)$. Therefore, this statement is false.

2. $y = \frac{1}{x^2}$ is bounded on $(1, +\infty)$:

   • For $x > 1$, $\frac{1}{x^2}$ decreases as $x$ increases, approaching zero as $x \to +\infty$. So, the function is bounded on this interval. This statement is true.

3. $y = \frac{1}{x^2 + 1}$ is bounded on $(-\infty, -1)$:

   • The function $\frac{1}{x^2 + 1}$ is bounded because the denominator always keeps the function value between 0 and 1. On any interval, including $(-\infty, -1)$, the function is bounded. This statement is true.

4. $y = \tan\left(x - \frac{\pi}{2}\right)$ is bounded on $(-1, 1)$:

   • The tangent function has vertical asymptotes at points where its argument is $\frac{\pi}{2} + k\pi$. Since $x - \frac{\pi}{2}$ will approach one of these asymptotes, the tangent function becomes unbounded on the interval $(-1, 1)$. Therefore, this statement is false.

5. $y = \frac{1}{x^2}$ is bounded below on $\mathbb{R}$:

   • The function $\frac{1}{x^2}$ has a vertical asymptote at $x = 0$ and grows without bound as $x \to 0$. However, for large $|x|$, the function approaches zero. While it is unbounded from above near zero, it is bounded below by 0, so this statement is true.

6. $y = \frac{1}{x^2}$ is bounded on $(-\infty, -1)$:

   • On $(-\infty, -1)$, the function $\frac{1}{x^2}$ decreases as $|x|$ increases, and it approaches zero. So, it is bounded on this interval. This statement is true.

Correct Statements:

• $y = \frac{1}{x^2}$ is bounded on $(1, +\infty)$
• $y = \frac{1}{x^2 + 1}$ is bounded on $(-\infty, -1)$
• $y = \frac{1}{x^2}$ is bounded below on $\mathbb{R}$
• $y = \frac{1}{x^2}$ is bounded on $(-\infty, -1)$

Would you like further details or clarifications?

Here are 5 related questions to expand your understanding:

1. How do you determine whether a function is bounded on a given interval?
2. What is the importance of asymptotes in analyzing boundedness?
3. How can limits help verify bounded behavior for functions like $\frac{1}{x^2}$?
4. Why does the function $e^{-x^2}$ have a maximum at $x = 0$?
5. How does the concept of boundedness relate to the continuity of a function?

Tip: When analyzing boundedness, always check for asymptotic behavior at extreme values and any potential discontinuities.