Analysis Of Behavior Of Functions Using Derivatives
Analyze the critical points of f (x) = x 3 - 27x - 20 .
Step 1: Find the derivative: f ¢(x) = 3x 2 - 27.
Step 2: Set the derivative equal to zero: f ¢(x) = 3x 2 - 27 = 0
Step 3: Solve for x: 3x - 27 = 0 .
x2 - 9 = 0
2
x =9
x = ±3
Step 4: Make a numberline and mark critical points only, then label algebraic sign of f ¢(x) for each
interval. Label the behavior of f, which increases when f ¢(x) is positive and decreases when f ¢(x) is
negative.
inc.
dec.
inc.
2
+
-3
-
+
3
Step 5: Identify local extrema. f has a local maximum point at x = -3 where f ¢(x) changes from
positive to negative. f has a local minimum at x = 3 point where f ¢(x) changes from negative to
positive.
Analyze the concavity of f (x) = x 3 - 27x - 20 .
Step 6: Find the second derivative: f ¢¢(x) = 6x . Label the algebraic sign of f ¢¢(x) for each interval.
Step 7: Label the behavior of f, which is concave up when f ¢¢(x) is positive and concave down when
f ¢¢(x) is negative.
concave down
−
concave up
0
+
Step 8: f has a point of inflection at x = 0 where f ¢¢(x) changes algebraic sign.
First Derivative Test:
Where f ¢(x) = 0 and changes sign from positive to negative, f (x) has a local maximum point.
Where f ¢(x) = 0 and changes sign from negative to positive, f (x) has a local minimum point.
Second Derivative Test:
Where f ¢(x) = 0 and f ¢¢(x) < 0 , f (x) has a local maximum point.
Where f ¢(x) = 0 and f ¢¢(x) > 0 , f (x) has a local minimum point.
Exercises:
Find intervals of increase and decrease of f (x) , as well as local extrema, showing derivatives and
numberlines, and using the First Derivative Test.
1. f (x) = -x 2 + 7x -17
f ¢(x) = -2x + 7 = 0
7
x=
2
7
f is increasing on æ -¥, ù
è
2 úû
7
f is decreasing on é ,¥ö
êë 2 ø
increasing
+
f ¢(x) = 0 and changes from positive to negative at x =
decreasing
7/2
−
7
, so f has a local maximum there.
2
2. f (x) = 3x 4 + 8x 3 - 6x 2 - 24x
x 2 + 3x + 2
3
2
x -1 x + 2x - x - 2
f ¢(x) = 12x + 24x - 12x - 24 = 0
3
2
= x + 2x - x - 2 = 0
= (x - 1)(x 2 + 3x + 2) = 0
= (x - 1)(x + 2)(x + 1) = 0
x = -2,-1,1
3
2
dec.
−
inc.
-2
+
dec.
-1
−
inc.
1
f is decreasing on (-¥, - 2] and [-1,1] .
+
f is increasing on [-2,-1] and [1, ¥) .
f ¢(x) = 0 and changes from positive to negative at x = -1 , so f has a local maximum there.
f ¢(x) = 0 and changes from negative to positive at x = -2 and at x = 1 , so f has a local minimum
there.
Find local extrema of f (x) , showing derivatives using the Second Derivative Test.
3. f (x) = x 3 -12x 2 + 45x
f ¢(x) = 3x - 24x + 45 = 0
2
= x - 8x + 15 = 0
= (x - 5)(x - 3) = 0
x = 3, 5
2
f ¢(x) = 3x - 24x + 45
f ¢¢(x) = 6x - 24
2
f ¢¢(3) = 6(3) - 24 < 0
f ¢¢(5) = 6(5) - 24 > 0
f ¢(x) = 0 and f ¢¢(x) > 0 at x = 5 , so f has a local minimum there.
f ¢(x) = 0 and f ¢¢(x) < 0 at x = 3 , so f has a local maximum there.
4. f (x) = x 4 - 8x 2 +1
f ¢(x) = 4x -16x = 0
2
4x(x - 4) = 4x(x - 2)(x + 2) = 0
x = 0, ± 2
3
f ¢(x) = 4x 3 -16x
2
f ¢¢(x) = 12x -16
f ¢¢(0) = 12(0)2 - 16 < 0
2
f ¢¢(-2) = 12(-2) - 16 > 0
f ¢¢(2) = 12(2)2 - 16 > 0
f ¢(x) = 0 and f ¢¢(x) > 0 at x = -2 , so f has a local minimum there.
f ¢(x) = 0 and f ¢¢(x) > 0 at x = 2 , so f has a local minimum there.
f ¢(x) = 0 and f ¢¢(x) < 0 at x = 0 , so f has a local maximum there.