(1 point) The derivative of $f(x)$ is given by $f^{\prime}(x)=(x-4)^{2}(x+7)^{2}$. Find the critical points and local extrema of $f$, and the open intervals on which $f$ is increasing and decreasing. Critical points of $f(x)$ are $4,-7$ (If there is more than one critical point, enter a comma-separated list. If there are no critical points, enter "none".) $f(x)$ is increasing on (-infinity, infinity) $f(x)$ is decreasing on (Enter your answers using interval notation.)

Step 1 ：The critical points of a function are the points where the derivative of the function is either zero or undefined. In this case, the derivative of the function \(f(x)\) is given by \(f^\prime(x)=(x-4)^{2}(x+7)^{2}\). This derivative is defined for all real numbers, so we only need to find where it is zero.

Step 2 ：Setting \(f^\prime(x)=0\), we get \((x-4)^{2}(x+7)^{2}=0\). This equation is satisfied when \(x=4\) or \(x=-7\). Therefore, the critical points of \(f(x)\) are \(4\) and \(-7\).

Step 3 ：To determine the local extrema and the intervals on which \(f\) is increasing or decreasing, we can use the first derivative test. This involves evaluating the derivative at points to the left and right of each critical point.

Step 4 ：For \(x<-7\), \(f^\prime(x)=(x-4)^{2}(x+7)^{2}\) is positive because a square of a real number is always nonnegative and we have two negative factors \((x-4)\) and \((x+7)\).

Step 5 ：For \(-7

Step 6 ：For \(x>4\), \(f^\prime(x)=(x-4)^{2}(x+7)^{2}\) is positive because a square of a real number is always nonnegative and we have no negative factors.

Step 7 ：Therefore, \(f(x)\) is increasing on \((-\infty, -7)\), \((-7, 4)\), and \((4, \infty)\), and there are no intervals on which \(f(x)\) is decreasing.

Step 8 ：Since \(f(x)\) changes from increasing to increasing at \(x=-7\) and \(x=4\), there are no local extrema.