Finding Critical Points Part II

Hckyplayer8

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Jun 9, 2019
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94
A) Find the critical points for f(x) = x2 - x4
B) Determine the intervals where f is increasing and decreasing
C) Classify each point as a local max,min or neither
D) Determine the intervals where f is concave up or down
E) Determine the inflection points

A) f ' (x) = 2x-4x3 Setting that to zero will yield an x of 0. Factoring that yields -2x(2x2-1) which further factors to -2x(square root 2 x +1) (square root 2 x -1) So x will also equal square root 2 over 2 and negative square root 2 over 2.

B) To determine the intervals where f is increasing and decreasing, I test f ' at different x values.

Does all look well thus far?
 

Subhotosh Khan

Super Moderator
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Jun 18, 2007
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18,777
A) Find the critical points for f(x) = x2 - x4
B) Determine the intervals where f is increasing and decreasing
C) Classify each point as a local max,min or neither
D) Determine the intervals where f is concave up or down
E) Determine the inflection points

A) f ' (x) = 2x-4x3 Setting that to zero will yield an x of 0. Factoring that yields -2x(2x2-1) which further factors to -2x(square root 2 x +1) (square root 2 x -1) So x will also equal square root 2 over 2 and negative square root 2 over 2.

B) To determine the intervals where f is increasing and decreasing, I test f ' at different x values.

Does all look well thus far?
As far as I can see - it's good. Continue....
 

Otis

Senior Member
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Apr 22, 2015
Messages
1,861
… 2x-4x3 … yields -2x([square root 2]x +1) ([square root 2]x -1) So [x=0 or] square root 2 over 2 and negative square root 2 over 2 …
Hi Hckyplayer8. A space is not clear enough, to show that only 2 is inside the radical sign. You need to use grouping symbols, as inserted above in red. Also, the word 'and' (in blue) ought to be 'or' because x cannot represent two different values simultaneously.

💡 In the forum guidelines, there's a link to pages that show how to text math. One way (for example):

2x - 4x^3 = [-2x]*[sqrt(2)x + 1]*[sqrt(2)x - 1]

x = 0 or
x = sqrt(2)/2 or
x = -sqrt(2)/2


Cheers
 

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
94
Hi Hckyplayer8. A space is not clear enough, to show that only 2 is inside the radical sign. You need to use grouping symbols, as inserted above in red. Also, the word 'and' (in blue) ought to be 'or' because x cannot represent two different values simultaneously.

💡 In the forum guidelines, there's a link to pages that show how to text math. One way (for example):

2x - 4x^3 = [-2x]*[sqrt(2)x + 1]*[sqrt(2)x - 1]

x = 0 or
x = sqrt(2)/2 or
x = -sqrt(2)/2


Cheers
Thank you Otis.
 

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
94
A) Find the critical points for f(x) = x2 - x4
B) Determine the intervals where f is increasing and decreasing
C) Classify each point as a local max,min or neither
D) Determine the intervals where f is concave up or down
E) Determine the inflection points

A) f ' (x) = 2x-4x3 Setting that to zero will yield an x of 0. Factoring that yields -2x(2x2-1) which further factors to -2x(square root 2 x +1) (square root 2 x -1) So x will also equal square root 2 over 2 and negative square root 2 over 2.

B) To determine the intervals where f is increasing and decreasing, I test f ' at different x values.

Does all look well thus far?
Moving on since sqrt(2)/2 and -sqrt(2)/2 is approximately equal to .707 and -.707 I tested f ' (1), f ' (-1).
f ' (1) = -2
f ' (-1) = 2

So I know the function is increasing from [neg infinity,-sqrt(2)) and is decreasing from (sqrt(2),pos infinity].

Because [0,0] is a critical point, I have to test on either side of that as well. To keep it simple, I tested .1 and -.1

f ' (.1) = -.2
f ' (-.1) = .2

Still looking good?
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,765
Moving on since sqrt(2)/2 and -sqrt(2)/2 is approximately equal to .707 and -.707 I tested f ' (1), f ' (-1).
f ' (1) = -2
f ' (-1) = 2
Lets just consider sqrt(2)/2. Shouldn't you be inspectiong f ' (x) on either side of x=sqrt(2)/2? And then the same for x=-sqrt(2)/2?
 

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
94
Lets just consider sqrt(2)/2. Shouldn't you be inspectiong f ' (x) on either side of x=sqrt(2)/2? And then the same for x=-sqrt(2)/2?
Alright. So testing either side of sqrt(2)/2. The right side was f ' (1) = -2. I chose .5 for my left side. f ' (.5) = .5

This means for -sqrt(2)/2

f ' (-1) = 2
f ' (-.5) = -.5
 

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
94
Putting it all together...

x=0 is a local min because of a negative gradient before and a positive gradient after the critical point
x=sqrt(2)/2 and -sqrt(2)/2 are local maxes because of a positive gradient before and a negative gradient after

Next is testing for concavity.
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
5,178
For (b), since you already have \(\displaystyle f'= -2x(\sqrt{2}-x)(\sqrt{2}+x)= 2x(x- \sqrt{2})(x+ \sqrt{2})\), rather than evaluating at individual points (although that is valid), I would argue that
1) If \(\displaystyle x< -\sqrt{x}\) all three factors are negative so f' is negative and the function is decreasing.
2) If \(\displaystyle -\sqrt{x}< x< 0\) two factors are negative and one positive so f' is positive and the function is increasing.
3) If \(\displaystyle 0< x< \sqrt{x}\) one factor is negative and two positive so f' is negative and the function is decreasing.
4) If \(\displaystyle \sqrt{2}< x\) all factors are positive so f' is positive and the function is increasing.
 
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