The problem presented is an inequality:

$f(x) > g(x), \text{ where } f(x) = x^2 + 4 \text{ and } g(x) = 7.$

We need to solve this inequality:

$x^2 + 4 > 7.$

Step-by-Step Solution:

1. Start by solving the inequality $x^2 + 4 > 7$: $x^2 + 4 > 7$ Subtract 4 from both sides: $x^2 > 3.$

2. Now, solve $x^2 > 3$. This inequality means that $x$ cannot be between $-\sqrt{3}$ and $\sqrt{3}$. In interval notation, the solution is: $(-\infty, -\sqrt{3}) \cup (\sqrt{3}, +\infty).$

3. Approximating $\sqrt{3} \approx 1.732$, we can conclude that the inequality holds for $x \in (-\infty, -1.732) \cup (1.732, +\infty)$.

4. Now, let's match this solution to the provided options:

   • Option a: $(-\infty, 0)$
   • Option b: "kitas atsakymas" (another answer)
   • Option c: $(0, +\infty)$
   • Option d: $(1, +\infty)$

The correct solution is not covered precisely by any of the listed intervals, but the closest match in terms of the interval on the positive side is option d: $(1, +\infty)$, though this does not capture the full range of $x$. Thus, the correct response seems to be option b: "kitas atsakymas", as it refers to "another answer."

Would you like further clarification on any step or detail?

Related Questions:

1. How would the solution change if $g(x)$ was a different constant, say 6?
2. What is the graphical interpretation of the solution $f(x) > g(x)$?
3. How would you solve this inequality if $f(x) = x^2 - 2x + 3$?
4. Can we solve this inequality using factorization techniques?
5. How does the solution change if the inequality is $f(x) \geq g(x)$?

Tip:

Always visualize inequalities by plotting the functions $f(x)$ and $g(x)$ on a graph—it helps understand the solution intervals better!