Solve for Positive Values of x in: y = -x² + 49 > 0

Given the function:

$y=-x^{2}+49$

Determine for which values of x the following is true: $f\left(x\right) > 0$

To determine for which values of $x$ the function $f(x) = -x^2 + 49$ is positive, we solve the inequality:

$-x^2 + 49 > 0$

We start by finding the roots of the associated quadratic equation:

$-x^2 + 49 = 0$

This can be rewritten as:

$x^2 = 49$

Solving for $x$, we have:

$x = \pm 7$

The roots are $x = 7$ and $x = -7$. These points are where the parabola intersects the x-axis. Since the parabola opens downwards (as the coefficient of $x^2$ is negative), the function is positive between the roots.

Therefore, the function $f(x) > 0$ over the interval:

$-7 < x < 7$

Thus, the correct choice is:

$-7 < x < 7$