Identify Negative Values for the Function y = x² + 16: An Inequality Challenge

We start by considering the function given: $y = x^2 + 16$. Our task is to find the values of $x$ making $f(x) < 0$, i.e., $x^2 + 16 < 0$.

Let's analyze the expression:

• The square of any real number $x$, denoted as $x^2$, is always non-negative. Thus, $x^2 \geq 0$ for all $x \in \mathbb{R}$.
• Adding 16 to a non-negative number (like $x^2$) results in a number that is at least 16, i.e., $x^2 + 16 \geq 16$.

Thus, the expression $x^2 + 16$ is always at least 16, and there are no $x$ values for which $y < 0$.

The solution is that there are No x values where $f(x) < 0$.

Hence, option 4 is correct: No x.