MATH
Tic-Tac
G12 Advanced EOT2 Exam Coverage &
keyanswers
Finding Critical Points of a Function
Example Q5 Page: 258
Find all critical numbers and use the graph to determine whether the critical number represents a local
maximum, local minimum or neither for:
𝒇 𝒙 = −𝒙𝟑 + 𝟑𝒙𝟐 − 𝟑𝒙
Solution
𝒇 𝒙 = −𝒙𝟑 + 𝟑𝒙𝟐 − 𝟑𝒙
𝒇′ (𝒙) = −𝟑𝒙𝟐 + 𝟔𝒙 − 𝟑
From the graph:
Critical Numbers:
𝒂𝒕 𝒙 = 𝟏
𝒇′ 𝒙 = 𝟎 ⇒
−𝟑𝒙𝟐 + 𝟔𝒙 − 𝟑 = 𝟎
−𝟑(𝒙𝟐
− 𝟐𝒙 + 𝟏) = 𝟎
𝒇 𝟏 is neither local minimum nor
local maximum
−𝟑(𝒙 − 𝟏)(𝒙 − 𝟏) = 𝟎
𝒙=𝟏
When a cubic function has one critical
value it is neither local minimum nor
local maximum
Finding Critical Points of a Function
Exercise Q6 Page: 258
Find all critical numbers and use the graph to determine whether the critical number represents a local
maximum, local minimum or neither for:
𝒇 𝒙 = 𝒙𝟒 − 𝟐𝒙𝟐 + 𝟏
Solution
𝒇 𝒙 = 𝒙𝟒 − 𝟐𝒙𝟐 + 𝟏
𝒇′ 𝒙 = 𝟒𝒙𝟑 − 𝟒𝒙
Critical Numbers:
𝒇′ 𝒙 = 𝟎 ⇒
From the graph:
𝟒𝒙𝟑 − 𝟒𝒙 = 𝟎
⇒
𝟒𝒙(𝒙𝟐 − 𝟏) = 𝟎
⇒
𝟒𝒙(𝒙 − 𝟏)(𝒙 + 𝟏) = 𝟎
𝟒𝒙 = 𝟎 Or 𝒙 − 𝟏 = 𝟎 Or 𝒙 + 𝟏 = 𝟎
𝒙=𝟎
𝒙=𝟏
𝒙 = −𝟏
𝒇 𝟎 =𝟏
Local Maximum
𝒇 −𝟏 = 𝟎
Local Minimum
𝒇 𝟏 =𝟎
Local Minimum
Absolute Minima as well
Next
topic!
Finding Critical Values of a Function
Exercise Q20 Page: 258
Find all critical numbers and use the graph to determine whether the critical number
𝒙
represents a local maximum, local minimum or neither for: 𝒇 𝒙 = 𝟐
𝒙 +𝟏
Solution
𝒇 𝒙 =
𝒙
Using Quotient Rule
𝒙𝟐 + 𝟏
𝒅
𝒅
𝒙 ∙ 𝒙𝟐 + 𝟏 − 𝒙 ∙
𝒅𝒙
𝒇′ 𝒙 = 𝒅𝒙
( 𝒙𝟐 + 𝟏)𝟐
(𝟏) ∙
𝒇′
𝒙 =
𝒙𝟐
+𝟏−𝒙∙
𝒙𝟐 + 𝟏
𝒇′(𝒙) ≠ 𝟎
𝟐𝒙
𝟐 𝒙𝟐 + 𝟏
𝟑
𝟐
𝒇′ 𝒙 𝒊𝒔 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 ⇒ ( 𝒙 + 𝟏) = 𝟎
( 𝒙𝟐 + 𝟏)𝟐
⇒
𝒙𝟐 + 𝟏 − 𝒙𝟐
𝒇′
𝒙 =
𝒙𝟐 + 𝟏
( 𝒙𝟐 + 𝟏)𝟐
=
𝒙𝟐 + 𝟏 = 𝟎
𝟏
( 𝒙𝟐 + 𝟏)𝟑
𝟐
⇒ 𝒙 = −𝟏
No Critical Numbers
From the graph No local extrema
Finding Critical Values of a Function
Exercise Q21 Page: 258
Find all critical numbers and use the graph to determine whether the critical number
represents a local maximum, local minimum or neither for: 𝒇 𝒙 = |𝒙𝟐 − 𝟏|
Solution
𝒙𝟐 − 𝟏
𝟐
𝒇 𝒙 = | 𝒙 − 𝟏|
𝒙𝟐 − 𝟏 = 𝟎 ⇒
+++ + 𝟎 −− − 𝟎 +++ +
−𝟏
𝒙 = ±𝟏
𝒙𝟐
− 𝟏,
𝒙 ≤ −𝟏
𝒇 𝒙 = ൞𝟏 − 𝒙𝟐 , −𝟏 < 𝒙 ≤ 𝟏
𝒙𝟐 − 𝟏,
𝒙>𝟏
𝟐𝒙,
𝒙 < −𝟏
𝒇′ 𝒙 = ቐ−𝟐𝒙, −𝟏 < 𝒙 < 𝟏
𝟐𝒙,
𝒙>𝟏
Critical Points: 𝒙 = −𝟏
𝟏
Critical Numbers:
𝒂𝒕 𝒙 = 𝟏, 𝒙 = −𝟏
The function changes its sign
The function is not differentiable
−𝟐𝒙 = 𝟎 ⇒
𝒙=𝟎
From the graph:
𝒙 = 𝟎𝝐 (−𝟏, 𝟏)
𝒇 𝟎 =𝟏
Local Maximum
𝒇 −𝟏 = 𝟎
Local Minimum
𝒇 𝟏 =𝟎
Local Minimum
Absolute Minima as well
𝒙=𝟏
Next
topic!
NOTE: Topic 2 & 3 are the same.
Next
topic!
Functions for Which the Second Derivative Test is Inconclusive
Exercise Q9 Page: 276
Use the second derivative test to find the local extrema of:
𝒇 𝒙 = 𝒙𝟒 + 𝟒𝒙𝟑 − 𝟏
Studying the sign of the first derivative :
Solution
−− − 𝟎 ++ + 𝟎 ++ +
𝟒
𝟑
𝒇 𝒙 = 𝒙 + 𝟒𝒙 − 𝟏
𝒇′ 𝒙
𝒇′ 𝒙 = 𝟒𝒙𝟑 + 𝟏𝟐𝒙𝟐
Critical Number:
𝒇′ 𝒙 = 𝟎 ⇒ 𝟒𝒙𝟑 + 𝟏𝟐𝒙𝟐 = 𝟎
⇒
⇒ 4𝒙𝟐 = 𝟎
𝒙=𝟎
−𝟑
𝒙+𝟑 =𝟎
Or 𝒙 + 𝟑 = 𝟎
𝒙 = −𝟑
𝒇′′ 𝒙 = 𝟏𝟐𝒙𝟐 + 𝟐𝟒𝒙
𝒇′′ −𝟑 = 𝟏𝟐(−𝟑)𝟐 + 𝟐𝟒 −𝟑 = 𝟑𝟔 > 𝟎
⇒ local minimum
𝒇′′ 𝟎 = 𝟏𝟐(𝟎)𝟐 + 𝟐𝟒 𝟎 = 𝟎
⇒ We cannot determine if it is a
local extrema and its type
𝒇′ −𝟐 = 𝟒(−𝟐)𝟑 + 𝟏𝟐(−𝟐)𝟐 = 𝟏𝟔(+𝒗𝒆)
𝒇′ 𝟐 = 𝟒(𝟐)𝟑 + 𝟏𝟐(𝟐)𝟐 = 𝟖𝟎 (+𝒗𝒆)
𝟎
𝒇 𝒙
4𝒙𝟐
𝒇 −𝟑 = (−𝟑)𝟒 +𝟒(−𝟑)𝟑 −𝟏 = −𝟐𝟖
𝒇′ −𝟒 = 𝟒(−𝟒)𝟑 + 𝟏𝟐(−𝟒)𝟐 = −𝟔𝟒 (−𝒗𝒆)
decreasing−𝟑 increasing 𝟎 Increasing
𝒇′ 𝒙 > 𝟎 ⇒ 𝒇 𝒙 Is increasing on: −𝟑, 𝟎 ∪ 𝟎, ∞
𝒇′ 𝒙 < 𝟎⇒ 𝒇 𝒙 Is decreasing on: −∞, −𝟑
The derivative changes from negative
to positive 𝒍𝒐𝒄𝒂𝒍 𝑚𝑖𝑛𝑖𝑚𝒖𝒎 𝒂𝒕 𝒙 = −𝟑
𝟎, −𝟏
𝒇 −𝟑 = −𝟐𝟖 Is a local minimum
The derivative doesn′t change its
sign 𝑎𝑟𝑜𝑢𝑛𝑑 𝒙 = 𝟎
𝒇 𝟎 = (𝟎)𝟒 +𝟒(𝟎)𝟑 −𝟏 = −𝟏 Is not a
local extremum
−𝟑, −𝟐𝟖
Using Second Derivative Test to Find Extrema
Exercise Q10 Page: 276
Use the second derivative test to find the local extrema of:
𝒇 𝒙 = 𝒙𝟒 + 𝟒𝒙𝟐 + 𝟏
Solution
𝒇 𝒙 = 𝒙𝟒 + 𝟒𝒙𝟐 + 𝟏
𝒇′ 𝒙 = 𝟒𝒙𝟑 + 𝟖𝒙
Critical Number:
𝟒𝒙𝟑 + 𝟖𝒙 = 𝟎
𝒇′ 𝒙 = 𝟎 ⇒
⇒
𝟒𝒙(𝒙𝟐
+ 𝟐) = 𝟎
Or 𝒙𝟐 + 𝟐 = 𝟎
⇒ 𝟒𝒙 = 𝟎
𝒙=𝟎
𝒙𝟐 = −𝟐
𝒇′′ 𝒙 = 𝟏𝟐𝒙𝟐 + 𝟖
𝒇′′ 𝟎 = 𝟏𝟐(𝟎)𝟐 + 𝟖 = 𝟖
>𝟎
⇒
local minimum 𝒇 𝟎 = (𝟎)𝟒 +𝟒(𝟎)𝟐 +𝟏 = 𝟏
𝟎, 𝟏
Using Second Derivative Test to Find Extrema
Exercise Q11 Page: 276
Use the second derivative test to find the local extrema of:
𝒇 𝒙 = 𝒙𝒆−𝒙
Solution
𝒇 𝒙 = 𝒙𝒆−𝒙
𝟏,
𝒇′ 𝒙 = −𝒙𝒆−𝒙 + 𝒆−𝒙
Critical Number:
𝒇′ 𝒙 = 𝟎 ⇒ −𝒙𝒆−𝒙 + 𝒆−𝒙 = 𝟎
⇒
𝒆−𝒙 −𝒙 + 𝟏 = 𝟎
⇒ 𝒆−𝒙 = 𝟎
Or −𝒙 + 𝟏 = 𝟎
𝒙=𝟏
𝒆−𝒙 > 𝟎
𝒇′′ 𝒙 = 𝒙𝒆−𝒙 − 𝒆−𝒙 −𝒆−𝒙
𝒇′′ 𝒙 = 𝒙𝒆−𝒙 − 𝟐𝒆−𝒙
𝒇′′
𝟏 =
(𝟏)𝒆−(𝟏) −𝟐𝒆− 𝟏
𝟏
= − ≈ −𝟎. 𝟑𝟔𝟖
𝒆
𝟏
− 𝟏
= ≈ 𝟎. 𝟑𝟔𝟖
⇒ local maximum 𝒇 𝟏 = 𝟏 𝒆
𝒆
<𝟎
𝟏
𝒆
Using Second Derivative Test to Find Extrema
Exercise Q13 Page: 276
Use the second derivative test to find the local extrema of:
𝒙𝟐 − 𝟓𝒙 + 𝟒 Domain: ℝ/{𝟎}
𝒇 𝒙 =
Solution
𝒙
𝟐
𝒙 − 𝟓𝒙 + 𝟒
𝒇 𝒙 =
𝒙
𝒅 𝟐
𝒅
[𝒙 − 𝟓𝒙 + 𝟒] ∙ 𝒙 − (𝒙𝟐 − 𝟓𝒙 + 𝟒) ∙
[𝒙]
𝒅𝒙
𝒅𝒙
′
𝒇 𝒙 =
𝒙𝟐
𝟐 − 𝟓𝒙 − 𝒙𝟐 + 𝟓𝒙 − 𝟒
𝟐 − 𝟓𝒙 + 𝟒)(𝟏)
𝟐𝒙
(𝟐𝒙
−
𝟓)𝒙
−
(𝒙
𝒇′ 𝒙 =
=
𝟐
𝒙𝟐
𝒙
𝟐
𝒙 −𝟒
′
𝒇 𝒙 =
𝒙𝟐
Critical Number:
𝒙𝟐 − 𝟒 = 𝟎 ⇒ 𝒙 − 𝟐 𝒙 + 𝟐 = 𝟎
𝒇′ 𝒙 = 𝟎 ⇒
𝒙=𝟐
𝟖
𝒇 𝟐 =
=𝟏 >𝟎
𝟑
(𝟐)
′′
𝟐
⇒ local minimum 𝒇 𝟐 = (𝟐) −𝟓(𝟐)+𝟒 = −𝟏
(𝟐)
𝟖
𝒇′′ −𝟐 =
= −𝟏 < 𝟎
(−𝟐)𝟑
⇒ local maximum 𝒇 −𝟐 =
(−𝟐)𝟐 −𝟓(−𝟐)+𝟒
(−𝟐)
𝒙 = −𝟐
𝒅 𝟐
𝒅 𝟐
[𝒙 − 𝟒] ∙ (𝒙𝟐 ) − (𝒙𝟐 − 𝟒) ∙
[𝒙 ]
𝒅𝒙
𝒅𝒙
′′
𝒇 𝒙 =
𝒙𝟒
𝟑 − 𝟐𝒙𝟑 + 𝟖𝒙
𝟐 ) − (𝒙𝟐 − 𝟒)(𝟐𝒙)
𝟖
𝟖𝒙
𝟐𝒙
𝟐𝒙
(𝒙
′′
=
= 𝟒
=
𝒇 𝒙 =
𝒙𝟑
𝒙
𝒙𝟒
𝒙𝟒
−𝟐, −𝟗
𝟐, −𝟏
= −𝟗
Next
topic!