Find the absolute extrema on [-2,2]

Question

f(x)= 10–12x+6x²–x³

 

 

 

Answer 1

For f(x) equal to 10 - 12x + 6x² - x³:

take the first derivative f'(x) as 0 - 12 + 12x - 3x²;

take the second derivative f''(x) as 12 - 6x.

Solve the first derivative equated to 0 for critical 

values of x; rewrite 0 - 12 + 12x - 3x² = 0 as 

3x² - 12x + 12 = 0.

Simplify 3x² - 12x + 12 = 0 to x² - 4x + 4 = 0.

Express x² - 4x + 4 = 0 as (x - 2)(x - 2) = 0, which indicates

a double root of x = 2.

Apply x = 2 to the second derivative to obtain f''(2) or 

12 - 6(2) = 0. This zero outcome indicates that f(x) has neither

a maximum nor minimum at x = 2; 2 is instead the location 

of an inflection point where f(x) changes from upward to

downward concavity.

For -2 ≤ x ≤ 2, the graph of f(x) gives f(-2) = 66

and f(2) = 2 {with f(x) decreasing throughout [-2, 2]}

as the absolute extrema on the interval given.