Answer:
Global extrema: none. Local extrema: (0, 0) and (4, -32)
Step-by-step explanation:
ƒ(x) = x3 – 6x2 should be written as ƒ(x) = x^3 – 6x^2. Use " ^ " to denote exponentiation, please.
One way to answer this problem would be to make a careful graph of ƒ(x) = x^3 – 6x^2. Notice that this graph begins in Quadrant III and ends in Quadrant I; this is one outcome of its being an ODD function. The graph will increase, reach a peak (a local max), decrease, reach a valley (a local min) and then grow from then on.
Another way is to use calculus. You don't say what course you're in, so I can't be sure that calculus would make sense to you.
Find the first derivative of ƒ(x) = x^3 – 6x^2. It is f '(x) = 3x^2 - 12x. Set this derivative = to 0 and find the roots. Hint: find the roots of 3x^2(x - 4). They are x = 0 and x = 4. At x = , y = f(0) = 0. Thus, the local max is
(0, 0). At x = +4, y = f(4) = 64 - 6(16) = -32. Thus, the local min is at (4, -32 ).
This graph rises without bound as x goes to ∞, and decreases without bound as x goes to -∞. Thus, there is neither a global max nor a global min.
