Find the local and global extrema for the graph of ƒ(x) = x3 – 6x2.

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.  

it could be 1,3 and more others to create a cylinder.

step-by-step explanation:

so, converting firstly the mixed fractions to improper fractions and then "adding".