Global minimum problem. PLEASE HELP

Question

Find the global minimum of the function f(x)= 2x^3 + 6x^2 -18x - 4 on the interval [0,3]

Answer 1

First find any local minimums on the interval by taking the derivative and setting it equal to zero. Then check the boundaries.

1) Find local extremes:

f '(x) = 6x² + 12x - 18

6x² + 12x - 18 = 0

x² + 2x - 3 = 0

(x + 3)(x - 1) = 0

x = -3, x = 1

Only x = 1 is on the interval in question

f(1) = 2(1)³ + 6(1)² - 18(1) - 4 = -14

2) Check the boundaries:

f(0) = -4

f(3) = 2(3)³ + 6(3)² - 18(3) - 4 = 50

So the global minimum occurs at x = 1 and is -14

Answer 2

Global mins (or maxs) are just where the first derivative equals zero, with checking a certain condition.

For a global min what we need is for the first derivative equal zero (First Order Condition) AND second derivative is greater than zero (Second Order Condition). Also, in my field I always check the end points (just to make sure).

HERE WE GO

f(x) = 2x³ + 6x² - 18x - 4 on [0,3]

f' = 6x² + 12x - 18

FOC: 6x² + 12x - 18 = 0 FOC is First Order Condition

simplify x² + 2x - 3 = 0 or (x-1) (x+3) = 0

solve that you get x = 1 OR x = -3 but -3 is NOT on the interval so possible answer is x = 1.

SOC check for x = 1

f" = 2x + 2 plug in x=1

f"(1) = 2 *(1) + 2 = 4 4 > 0 so SOC are good.

There you are global min is at x = 1