91Ó°ÊÓ

First, we need to find the derivative of the function \(f(x) = 4 - x^2\). To do this, we differentiate the function with respect to \(x\) and get: $$f'(x) = -2x$$

Now, we find the critical points of the function. A critical point is a point at which \(f'(x) = 0\). To find the critical points, we set \(f'(x)\) equal to 0 and solve for x: $$-2x = 0$$ $$x = 0$$ So, the critical point is at x = 0.

To find the intervals where the function is increasing and decreasing, we can examine the sign of \(f'(x)\) around the critical point: For \(x < 0\), \(f'(x) = -2x > 0\). Thus, the function is increasing on the interval \((-\infty, 0)\). For \(x > 0\), \(f'(x) = -2x < 0\). Thus, the function is decreasing on the interval \((0, \infty)\).

Let's plot the graph of the original function \(f(x) = 4-x^2\) and its derivative \(f'(x) = -2x\). By observing the graphs, we can verify that the intervals where the function is increasing and decreasing match our findings. When you plot the graphs, you will see that \(f(x)\) is indeed increasing when \(x < 0\) and decreasing when \(x > 0\). The graph of the derivative, \(f'(x)\), agrees with this analysis since it is positive for \(x < 0\) (indicating an increasing f(x)) and negative for \(x > 0\) (indicating a decreasing f(x)). In conclusion, the function \(f(x) = 4 - x^2\) is increasing on the interval \((-\infty, 0)\) and decreasing on the interval \((0, \infty)\). Graphing the function \(f(x)\) and its derivative \(f'(x)\) confirms our conclusion.