A. How to use the first derivative test? [Use a graphing calculator to see the graph.]
0  x  2
Example: f ( x)  cos x  2 sin x,
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Step 1: f is a continuous function.
Step 2: Find all critical numbers of f.
f ' ( x)  2 cos x( sin x)  2 cos x
 2 cos x(sin x  1)  0
cos x = 0 ==> x = π/2, 3π/2
sin x = -1 ==> x = 3π/2
Step 3: We have 3 intervals.
0 < x < π/2, π/2 < x < 3π/2, 3π/2 < x < 2π
0 < x < π/2 π/2 < x < 3π/2 3π/2 < x < 2π
Interval
−
'
Sign of f ( x)
f(x)
decreasing
+
−
increasing
decreasing
'
At x = π/2 : f changes from negative to positive, then f has a local minimum at π/2.
The local minimum value is f ( / 2)  cos ( / 2)  2 sin( / 2)  2
2
'
At x = 3π/2 : f changes from positive to negative, then f has a local maximum at 3π/2.
The local maximum value is f (3 / 2)  cos (3 / 2)  2 sin(3 / 2)  2
2
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B. How can we find intervals of concavity and inflection points?
f '' ( x)  2 sin x(1  sin x)  2 cos2 x
''
Step 1 : Find f ( x).
Step 2 : Solve for x, f ( x)  0 .
''
Using trig. identity, we get 2 sin x  sin x  1  0
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Factoring: (2 sin x  1)(sin x  1)  0
1. sin x = ½ , x = π/6, 5π/6.
2. sin x = -1 , x = 3π/2.
Interval
Sign of f
''
0 < x < π/6 π/6 < x < 5π/6 5π/6 < x < 3π/2 3π/2 < x < 2π
−
+
−
−
CD
Inflection points:
CU
CD
CD
 1
5  1
( , ), ( , )
6 4
6 4
1
,
4
(i)
x = π/6, f ( / 6)  cos ( / 6)  2 sin( / 6) 
(ii)
x = 5π/6, f (5 / 6)  cos (5 / 6)  2 sin( 5 / 6) 
2
2
1
.
4
C. How to use the second derivative test?
Step 1 : Find all critical numbers. x = π/2, 3π/2.
''
Step 2: Find f ( x).
f '' ( x)  2 sin x(1  sin x)  2 cos2 x  2(sin 2 x  cos2 x  sin x)
''
Step 3: Evaluate f ( x) at x = π/2, 3π/2.
f '' ( / 2)  4  0 , then f has a local minimum at x = π/2.
f '' (3 / 2)  0 , we don’t get any information from the second derivative test.
We need to use the first derivative test.
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