Increasing/Decreasing Functions
x
0.5x2
x
0.5x2
-2
2
0
0
-1
0.5
2
2
0
0
3
4.5
As x increases
y decreases
This function is
decreasing when x < 0
As x increases
y increases
This function is
increasing when x > 0
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Remember: First derivative gives a formula for the
slope of the tangent line on a curve.
slope is negative
graph is decreasing
slope is positive
graph is increasing
Test for Increasing or Decreasing Functions
If f ′(x) > 0 for all x in an interval I, then f is increasing on I.
If f ′(x) < 0 for all x in an interval I, then f is decreasing on I.
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EXAMPLE 1 Calculus Approach
Find the interval on which the function
f(x) = x2 + 2x – 3 is increasing and decreasing.
f ′ (x) = 2x + 2
Step 1: Find the derivative
Step 2: Solve the inequalities
2x + 2 < 0
x < –1
The function is decreasing
when x < –1
2x + 2 > 0
x > –1
The function is increasing
when x > –1
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EXAMPLE 1 Graphing Approach
Find the interval on which the function
f(x) = x2 + 2x – 3 is increasing and decreasing.
Critical Value
f ' (x) = 2x + 2
2x + 2 = 0
x = –1
f –1) = (–1)2 + 2(–1) – 3
f(–1) = = – 4
Turning point
The function is decreasing
when x < –1
(–1, –4)
The function is increasing
when x > –1
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EXAMPLE 2
f(x) = 2x3 – 6x2
Critical values of x occur where the tangent has a slope of 0.
f ′(x) = 6x2 – 12x
6x2 – 12x = 0
The function increases when 6x(x – 2) > 0
x < 0 or x > 2
6x(x – 2) = 0
The function decreases when 6x(x – 2) < 0
0<x<2
x = 0 or x = 2
–
+
𝟔 −𝟏 −𝟏 − 𝟐
𝟏𝟖
0
𝟔 𝟏 𝟏−𝟐
−𝟔
+
2
𝟔 𝟑 𝟑−𝟐
𝟏𝟖
5
f(x) = 2x3 – 6x2
𝒇 𝟎 = 𝟐 𝟎𝟑 − 𝟔 𝟎 𝟐 = 𝟎
𝒇 𝟐 = 𝟐 𝟐𝟑 − 𝟔 𝟐𝟐 = −𝟖
turning points
(𝟎, 𝟎)
decreasing
0<x<2
increasing
x>2
increasing
x<0
(𝟐, − 𝟖)
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EXAMPLE 3
Find the interval on which the function
𝟏
f(x) = x3 + 2x2 – 5x + 5 is increasing and decreasing.
𝟑
Step 1: Find the derivative
f ′ (x) = x2 + 4x – 5
Step 2: Factor the derivative f ′ (x) = x2 + 4x – 5
f ′ (x) = (x + 5)(x – 1)
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EXAMPLE 3 con’t
Step 3: Solve the inequalities
Decreasing
(x + 5 )(x – 1) < 0
Increasing
(x + 5)(x – 1) > 0
–
+
−𝟔 + 𝟓 −𝟔 − 𝟏
𝟕
–5
𝟎+𝟓 𝟎−𝟏
−𝟓
+
1
𝟐+𝟓 𝟐−𝟏
𝟕
The function is increasing when x < – 5 or x > 1
The function is decreasing when –5 < x < 1
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𝟏
𝟑
f(x) = x3 + 2x2 – 5x + 5
𝟏
𝟑
f(-5) = (-5)3 + 2(-5)2 – 5(-5) + 5 = 38.33…
𝟏
𝟑
f(-5) = (1)3 + 2(1)2 – 5(1) + 5 = 2.33…
turning point (−𝟓,
increasing
x < –5
decreasing
𝟑𝟖. 𝟑𝟑 … ) –5 < x < 1
increasing
x>1
turning point (𝟏, 𝟐. 𝟑𝟑 … )
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−𝒙
Rational Functions 𝒇 𝒙 =
𝒙−𝟔
𝒇 𝟔
𝑫𝑵𝑬
Remember Limits and the asymptotes on the graph?
−𝑥
lim
𝑥→6 𝑥 − 6
DNE
Vertical asymptote
x=6
−𝑥
lim
𝑥→∞ 𝑥 − 6
𝒙
−𝒙
𝐥𝐢𝐦
𝟔
𝒙→∞ 𝒙
−
𝒙 𝟔
−𝟏
𝟏−𝟎
Horizontal asymptote
y = –1
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Example 4
Rational Functions
−𝒙
𝒇 𝒙 =
𝒙−𝟔
Find the derivative
𝒙 − 𝟔 −𝟏 − (−𝒙)(𝟏)
′
𝒇 𝒙 =
𝒙−𝟔 𝟐
𝟔
=
𝒙−𝟔
𝟐
This function always increases because
𝟔
𝒇′(𝒙) =
𝒙−𝟔
𝟐
is always positive.
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x=6
y = –1
increasing
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EXAMPLE 5
−𝟐
𝒇 𝒙 = 𝟐
𝒙 −𝟏
𝒙 ≠ ±𝟏
𝟐 − 𝟏 𝟎 − (−𝟐)(𝟐𝒙)
𝒙
𝒇′ 𝒙 =
𝒙𝟐 − 𝟏 𝟐
𝟒𝒙
𝒇 𝒙 = 𝟐
𝒙 −𝟏
′
𝟐
always positive
Increasing when
Decreasing when
4x > 0
𝒙 > 𝟎, 𝒙 ≠ 𝟏
4x < 0
𝒙 < 𝟎, 𝒙 ≠ −𝟏
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EXAMPLE 5
decreasing
4x < 0
𝒙 < 𝟎, 𝒙 ≠ −𝟏
increasing
4x > 0
𝒙 > 𝟎, 𝒙 ≠ 𝟏
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