TheEloCollector1’s AMA

It’s been a really long time since I’ve done an ama

Let f(x) = x^3 - 3x + 1. Find all real values of x such that f(f(x)) = 0.”

#2 …

#3 bro I ain’t got a degree yet

#4 YOU ANSWER THAT
#5whats amu bro
#6 I know that, but I have little proof
#7 you can’t
#8 as old as yo mama
#9 don’t visit me today

#10 why is everyone giving me complex problems

#10 if it greatly simplified I could have solved it prob

#11, #12, why tryna make me suffer

What is 0^0

Wait a sec, to #10


et f(x) = x^3 - 3x + 1. Find all real values of x such that f(f(x)) = 0.”
ChatGPT said:
We are given the function:

f
(
x
)
=
x
3
−
3
x
+
1
f(x)=x
3
−3x+1
and we are to find all real
x
x such that:

f
(
f
(
x
)
)
=
0
f(f(x))=0
Step 1: Let’s define the problem
We want:

f
(
f
(
x
)
)
=
0
f(f(x))=0
Let
y
=
f
(
x
)
y=f(x), then:

f
(
f
(
x
)
)
=
f
(
y
)
=
0
⇒
f
(
y
)
=
0
⇒
y
 is a root of 
f
⇒
f
(
x
)
∈
{
real roots of 
f
(
x
)
=
0
}
f(f(x))=f(y)=0⇒f(y)=0⇒y is a root of f⇒f(x)∈{real roots of f(x)=0}
Step 2: Find the real roots of
f
(
x
)
=
x
3
−
3
x
+
1
=
0
f(x)=x
3
−3x+1=0
This is a cubic, and we look for real roots.

Try Rational Root Theorem: test ±1

f
(
1
)
=
1
−
3
+
1
=
−
1
f(1)=1−3+1=−1
f
(
−
1
)
=
−
1
+
3
+
1
=
3
f(−1)=−1+3+1=3
So no rational roots. Let’s analyze its behavior or use trigonometric methods.

Step 3: Use depressed cubic trigonometric solution for:
x
3
−
3
x
+
1
=
0
x
3
−3x+1=0
This is a depressed cubic of the form:

x
3
+
p
x
+
q
=
0
 with 
p
=
−
3
,
q
=
1
x
3
+px+q=0 with p=−3,q=1
Since discriminant
Δ
=
(
q
2
)
2
+
(
p
3
)
3
=
(
1
2
)
2
+
(
−
1
)
3
=
1
4
−
1
=
−
3
4
<
0
Δ=(
2
q
​
)
2
+(
3
p
​
)
3
=(
2
1
​
)
2
+(−1)
3
=
4
1
​
−1=−
4
3
​
<0, the equation has three real roots.

Use trigonometric solution:

x
k
=
2
−
p
3
cos
⁡
(
θ
+
2
k
π
3
)
,
k
=
0
,
1
,
2
x
k
​
=2
−
3
p
​

​
cos(
3
θ+2kπ
​
),k=0,1,2
with
cos
⁡
θ
=
−
q
2
⋅
(
−
3
p
)
3
/
2
cosθ=−
2
q
​
⋅(−
p
3
​
)
3/2


But since
p
=
−
3
,
q
=
1
p=−3,q=1, we get:

x
k
=
2
cos
⁡
(
θ
+
2
k
π
3
)
x
k
​
=2cos(
3
θ+2kπ
​
)
with:

cos
⁡
θ
=
−
1
2
⇒
θ
=
2
π
3
cosθ=−
2
1
​
⇒θ=
3
2π
​

So roots are:

x
1
=
2
cos
⁡
(
2
π
9
)
x
2
=
2
cos
⁡
(
2
π
9
+
2
π
3
)
=
2
cos
⁡
(
8
π
9
)
x
3
=
2
cos
⁡
(
2
π
9
+
4
π
3
)
=
2
cos
⁡
(
14
π
9
)
x
1
​
=2cos(
9
2π
​
)x
2
​
=2cos(
9
2π
​
+
3
2π
​
)=2cos(
9
8π
​
)x
3
​
=2cos(
9
2π
​
+
3
4π
​
)=2cos(
9
14π
​
)
So the three real roots of
f
(
x
)
=
0
f(x)=0 are:

2
cos
⁡
(
2
π
9
)
,
  
2
cos
⁡
(
8
π
9
)
,
  
2
cos
⁡
(
14
π
9
)
2cos(
9
2π
​
),2cos(
9
8π
​
),2cos(
9
14π
​
)
​

Step 4: Now find all real
x
x such that
f
(
x
)
f(x) equals one of these roots
We are solving:

f
(
x
)
=
r
i
for each root 
r
i
f(x)=r
i
​
for each root r
i
​

i.e.,

x
3
−
3
x
+
1
=
r
i
⇒
x
3
−
3
x
+
(
1
−
r
i
)
=
0
x
3
−3x+1=r
i
​
⇒x
3
−3x+(1−r
i
​
)=0
Each is a new cubic equation. Since
f
f is a degree 3 polynomial, it will have 3 real roots for each
r
i
r
i
​
. Hence:

For each root
r
i
r
i
​
of
f
f, solve
f
(
x
)
=
r
i
f(x)=r
i
​

Each gives up to 3 real
x
x values
Since
f
(
x
)
=
x
3
−
3
x
+
1
f(x)=x
3
−3x+1 is a strictly increasing-decreasing-increasing function (has local extrema), and since each
r
i
r
i
​
is within its range, each equation will have exactly 3 real solutions.

So total real solutions of
f
(
f
(
x
)
)
=
0
f(f(x))=0 is:

3
×
3
=
9
 real solutions
3×3=9 real solutions
​

✅ Final Answer:
There are 9 real values of
x
x such that
f
(
f
(
x
)
)
=
0
f(f(x))=0.
Exact values are roots of:

x
3
−
3
x
+
1
=
2
cos
⁡
(
2
π
9
)
,
x
3
−
3
x
+
1
=
2
cos
⁡
(
8
π
9
)
,
x
3
−
3
x
+
1
=
2
cos
⁡
(
14
π
9
)
x
3
−3x+1=2cos(
9
2π
​
),x
3
−3x+1=2cos(
9
8π
​
),x
3
−3x+1=2cos(
9
14π
​
)

#20 0

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#25 idk #26 don’t get me locked it’s the middle of the night