Revised November 7, 2009 Steps to determine the local and global extrema of a given function 1. Find all critical points (f

Revised November 7, 2009

Steps to determine the local and global extrema of a given function 1. Find all critical points (f⁰(c) = 0 or DNE)

2. Find the endpoints of the domain of f

3. Use the first derivative test to identify the local extrema.

• Example 1: f(x) = 3x⁴+ 4x³− 6x²− 12x + 12

1. Domain of the function = (−∞.∞) (no endpoints) 2. f⁰(x) = 12x³+ 12x²− 12x − 12 = 12(x + 1)²(x− 1);

Critical numbers: x = 1 and x =−1.

3. Interval of increasing and decreasing: Note: first divide the domain into subintervals according to the location of critical numbers.

(−∞, −1) (−1, 1) (1, ∞)

x −2 0 2

f⁰(x) - - +

f (x) Dec Dec Inc

4. Local maximum: None; local minimum at x = 1 (1st derivative test)

5. Global (abs) maximum: None (limx→∞f (x) =∞); global minimum at x = 1:

Since f (x) is a polynomial, it is continuous on (−∞, ∞). Since the function is decreas- ing to the left of 1 and increasing to the right of 1.

6. Other questions: (Without sketching the graph) Can you find the abs max/min of f (x) on [−2, 2], and [−2, 0]? How do you know that f(x) has no x-intercept (or f(x) 6= 0)?

• Example 2: f(x) = 3x⁴+ 4x³− 6x²− 12x + 12, [−2, 2]

1. Domain of the function = [−2, 2]: Endpoints: −2, 2 2. f⁰(x) = 12x³+ 12x²− 12x − 12 = 12(x + 1)²(x− 1);

Critical numbers: x = 1 and x =−1 (All critical points are in the domain).

3. Interval of increasing and decreasing: Note: first divide the domain into subintervals according to the location of critical numbers.

(−2, −1) (−1, 1) (1, 2)

x −1.5 0 1.5

f⁰(x) - - +

f (x) Dec Dec Inc

4. Local maxima at x =−2 and x = 2; local minimum at x = 1 (1st derivative test) 5. End points: f (−2) = 28; f(2) = 44; Critical points: f(−1) = 17; f(1) = 1

Global (abs) maximum at x = 2; global minimum at x = 1

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• Example 3: f(x) = 3x⁴+ 4x³− 6x²− 12x + 12, [0, 2)

1. Domain of the function = [0, 2): Endpoints: 0. (Note 2 is not in the domain) 2. f⁰(x) = 12x³+ 12x²− 12x − 12 = 12(x + 1)²(x− 1);

Critical numbers: x = 1. (−1 is not in the domain).

3. Interval of increasing and decreasing: Note: first divide the domain into subintervals according to the location of critical numbers.

(0, 1) (1, 2)

x .5 1.5

f⁰(x) - +

f (x) Dec Inc

4. Local maxima at x = 0; local minimum at x = 1 (1st derivative test) 5. Critical points: f (1) = 1; Endpoints: f (0) = 12, limx→2f (x) = 44

Global (abs) maximum: None ; global minimum at x = 1

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