Find the global maximum and global minimum values of f(x) on the interval [-2,1]. Is there an interval for this function on which the maximum and minimum values occur at the endpoints? If so, find such an interval.
Let
f(x) = x^3 - 3x^2 + 2
Compute the derivative f'(x).
Find all critical points of f. Recall that a critical number is a value of x such that f'(x) = 0, and the corresponding critical point is (x, f(x)).
Mark the critical numbers on a number line and determine the sign of the first derivative on the intervals this creates. One way to do this is to test values: pick a number x in between two critical values and determine whether f'(x) is positive (+) or negative (-).
Use the first derivative test with your sign chart above to classify the critical points. For each critical point, is there a local maximum there, a local minimum, or neither? Remember: + to - through the critical point indicates a maximum, - to + indicates a maximum, (+ to + or - to - indicates neither).
Determine as many intercepts as you can for this function. Hint: the y-intercept and one of the x-intercepts are easy to find. (We don't need the other x-intercept to make a reasonable sketch below.)
Use the sign chart and your knowledge of the intercepts to make a sketch of the graph of f(x) without using a calculator. Your graph should be correct at the intercepts and perhaps a few other easy to determine values, and be increasing or decreasing on the correct intervals; it does not need to be perfect.
If time permits, repeat the steps from problem 2 for the function f(x) = x^4 - 4x^3 + 10.