1. Find the derivative of <span>P(x)=3x^3+2x^2-6x. It's P'(x)=9x^2 + 4x - 6. 2. Set this result equal to zero and solve for the critical values: </span> 9x^2 + 4x - 6 = 0 Using the quadratic formula, I got x = [-4 plus or minus sqrt(232)] / 18. Reducing this, x = [-4 plus or minus 2 sqrt(58)]; thus, there are two real, unequal roots and two real, unequal critical values. 3. One at a time, examine the two critical values: determine whether the derivative changes from neg to pos or from pos to neg at each of these values. Example: If the derivative is pos to the left of the first c. v. and neg to the right, we've got a local max.
4. Since there are only 2 critical values, you can have no more than 1 local max (corresponding to a change in the sign of the derivative from pos to neg) and one local min. (from neg to pos).
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