Revised 11/13 16:30
Steps to determine the local and global extrema of a given function (See P 273) 1. Find all critical points (f0(c) = 0 or DNE)
2. Find the endpoints of the domain off
3. Use the first derivative test (P 263 and P275) 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
Interval of increasing: [1,∞); Interval of decreasing: (−∞,1]
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:
Sincef(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 off(x) on [−2,2], and [−2,0]? How do you know thatf(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
Interval of increasing: [1,2]; Interval of decreasing: [−2,1]
4. Local maxima at x=−2 and x= 2; local minimum atx= 1 (1st derivative test) 5. End points: f(−2) = 28; f(2) = 44; Critical points: f(−1) = 17; f(1) = 1
Global (abs) maximum atx= 2; global minimum at x= 1
1
• 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
Interval of increasing: [1,2] [1,2); Interval of decreasing: [0,1]
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 atx= 1
• Example 4: f(x) = xx2−42 , [−1,2)∪(2,∞)
1. Domain of the function = [−1,2)∪(2,∞); End points: −1 2. f0(x) = (x−2−8x4)2; Critical number: x= 0
3. Interval of increasing and decreasing: Note: first divide the domain into subintervals according to the location of critical numbers.
(−1,0) (0,2) (2,∞)
x −1/2 1 3
f0(x) + - -
f(x) Inc Dec Dec
Interval of increasing: [−1,0]; Interval of decreasing: [0,2)∪ (2,∞) 4. Local maximum at x= 0; local minimum: None at x=−1
5. Critical points: f(0) = 0; Endpoints: f(−1) =−1/3, limx→2− x2
x2−4 =−∞, limx→2+ x2x−24 =∞, limx→∞ x2 x2−4 = 1.
Global (abs) maximum: None ; global minimum: none.
6. graph of the function showing all significant features.
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0) x=−2
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2 x=2
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2 x=2
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2 x=2
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2 x=2
y=1
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2 x=2
y=1
x
y
−3 −2 −1 0 1 2 3
−10
−5 0 5 10 15
(0,0)
x=−2 x=2
y=1
x
y
2
