6. [-/4 Points] MY NOTES DETAILS SCALC9 3.3.013. ASK YOUR TEACHER Consider the following. (If an answer does not exist, enter DNE.) \[ f(x)=\frac{x^{2}-63}{x-8} \] (a) Find the interval(s) on which $f$ is increasing. (Enter your answer using interval notation.) (b) Find the interval(s) on which $f$ is decreasing. (Enter your answer using interval notation.) (c) Find the local minimum and maximum value of $f$. local minimum value local maximum value Need Help? Read It Submit Answer
Solution
Step 1 ：First, we find the derivative of the function \(f(x) = \frac{x^{2} - 63}{x - 8}\). The derivative is \(f'(x) = \frac{2x}{x - 8} - \frac{x^{2} - 63}{(x - 8)^{2}}\).
Step 2 ：Next, we find the critical points of the function. These are the points where the derivative is zero or undefined. The critical points are x = 7 and x = 9.
Step 3 ：We then determine the intervals on which the function is increasing or decreasing. The function is increasing where the derivative is positive and decreasing where the derivative is negative. The function is increasing on the intervals \((-\infty, 7)\) and \((9, \infty)\), and decreasing on the interval \((7, 9)\).
Step 4 ：Finally, we find the local minimum and maximum values of the function by substituting the critical points into the original function. The local minimum value of \(f\) is 14 and the local maximum value of \(f\) is 18.
Step 5 ：\(\boxed{\text{Final Answer:}}\) The function \(f\) is increasing on the intervals \((-\infty, 7) \cup (9, \infty)\), and decreasing on the interval \((7, 9)\). The local minimum value of \(f\) is \(\boxed{14}\) and the local maximum value of \(f\) is \(\boxed{18}\).
