Find the relative min and relative max of h(x)=x^3-6x^2+12

Question

I'm not sure how to even start this. I also don't know how to use a graphing calculator so I'm really lost.

Answer 1

Draw a graph. You should be able to notice that the turning points look like they are at x=0 and x=4.

 

Plug it into the equation to get f(0) = 12 is the local max, and f(4) = 4³ - 6·4² + 12 = -20 is the local min.

 

Graphing calculators all have a utility that searches for minimums and maximums of any function. It asks you to enter a function, a guess where you think the min or max is and an interval to search in. It will give you the points (0,12) for max and (4,-20) for min.

 

There are also many methods without a calculator.

 

One is calculus which you will learn in a separate course.

 

The derivative (rate of change) of f(x) = xⁿ with respect to x is f'(x) = nx^n-1. This is called the power rule.

Derivatives are linear: (a f + b g)' = a f' + b g'

 

h'(x) = 3x² - 12x

 

Setting h'(x) = 0 gives

 

3x² - 12x = 0

x² - 4x = 0

x(x-4) = 0

 

x = 0 or x = 4

h(0) = 12 is max, h(4) = -20 is min.

 

 

The other is ordinary algebra. In a polynomial with leading coefficient 1, the coefficient if the 1-lower-degree term is negative the sum of roots.

 

Let C be the value at either a min or max and set h(x) = C

 

Thus we want h(x) - C = x³-6x²+12-C to have a double root.

 

The sum of the roots of h(x) - C is 6 so if the double root is R then the other is 6-2R

 

(x - R)²[x - (6 - 2R)] = (x² - 2Rx + R²)[x - (6 - 2R)]

 

= x³ - 6x² + [R² + 2R(6-2R)] + R²(2R-6) = x³ - 6x² - (3R² - 12R)x +R²(6-2R)

 

= x³ - 6x² - 3R(R-4)x +2R²(3-R)

 

This should equal x³ - 6x² + 12 so 3R(R-4)=0 making R=0 or 4.

 

Thus the critical points are at 0 and 4. Where h(0) = 12 is max and h(4) = -20 is min.