Solving y = -x² + 1: Finding Values Where Function is Negative

To solve this problem, we need to determine where the quadratic function $y = -x^2 + 1$ is negative.

• First, identify the roots of the equation by setting $y = 0$:
  $-x^2 + 1 = 0$ simplifies to $-x^2 = -1$ or $x^2 = 1$.
• Solving $x^2 = 1$ gives us the roots: $x = 1$ and $x = -1$.
• Now, examine the intervals defined by these roots: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.
• Test each interval:
  • For $x < -1$, choose $x = -2$:
    $y = -(-2)^2 + 1 = -4 + 1 = -3$, which is negative.
  • For $-1 < x < 1$, choose $x = 0$:
    $y = -(0)^2 + 1 = 1$, which is positive.
  • For $x > 1$, choose $x = 2$:
    $y = -(2)^2 + 1 = -4 + 1 = -3$, which is negative.
• Thus, the function is negative for $x < -1$ and $x > 1$.

Therefore, the values of $x$ for which $f(x) < 0$ are $x < -1$ or $x > 1$.