Consider the function $f(x)=3 x^{2}-6 x+9,0 \leq x \leq 8$
The absolute maximum of $f(x)$ (on the given interval) is at $x=$
and the absolute minimum of $f(x)$ (on the given interval) is at $x=$
Answer
Final Answer: The absolute maximum of \(f(x)\) on the interval \([0, 8]\) is at \(x=\boxed{8}\) and the absolute minimum of \(f(x)\) on the interval \([0, 8]\) is at \(x=\boxed{1}\).
Step 1 ：Consider the function \(f(x)=3 x^{2}-6 x+9\), for \(0 \leq x \leq 8\)
Step 2 ：The absolute maximum of \(f(x)\) (on the given interval) is at \(x=\)
Step 3 ：And the absolute minimum of \(f(x)\) (on the given interval) is at \(x=\)
Step 4 ：To find the absolute maximum and minimum of the function \(f(x)\) on the interval \([0, 8]\), we need to find the critical points of the function and evaluate the function at these points and at the endpoints of the interval.
Step 5 ：The critical points of a function are the points where the derivative of the function is zero or undefined.
Step 6 ：In this case, the derivative of \(f(x)\) is \(f'(x) = 6x - 6\), which is never undefined.
Step 7 ：We can set \(f'(x) = 0\) to find the critical points.
Step 8 ：The critical point of the function is at \(x=1\).
Step 9 ：Now, we need to evaluate the function at this point and at the endpoints of the interval, \(x=0\) and \(x=8\), to find the absolute maximum and minimum.
Step 10 ：The absolute maximum of \(f(x)\) on the interval \([0, 8]\) is at \(x=8\) and the absolute minimum of \(f(x)\) on the interval \([0, 8]\) is at \(x=1\).
Step 11 ：Final Answer: The absolute maximum of \(f(x)\) on the interval \([0, 8]\) is at \(x=\boxed{8}\) and the absolute minimum of \(f(x)\) on the interval \([0, 8]\) is at \(x=\boxed{1}\).
