Review the response and rate each piece. Only mark steps incorrect if they contain a mathematical error. The model may add preamble or summary steps which seem superfluous, but do not mark them incorrect unless they contain a math error.
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Prompt: Consider the equation $x^2 + \frac{9}{x^2} = 10$. Solve for $x$ such that the inequality $x^3 - 3x > 0$ is also satisfied. Additionally, show that $y = x^3 - 3x$ is increasing at $x = 1$.

Now, we plug in 
x
=
1
x=1 into the inequality: 
1
3
−
3
(
1
)
=
1
−
3
=
−
2
<
0
1 
3
 −3(1)=1−3=−2<0. So, 
x
=
1
x=1 does not satisfy the inequality.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Finally, we plug in 
x
=
−
1
x=−1 into the inequality: 
(
−
1
)
3
−
3
(
−
1
)
=
−
1
+
3
=
2
>
0
(−1) 
3
 −3(−1)=−1+3=2>0. So, 
x
=
−
1
x=−1 satisfies the inequality.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Therefore, the solutions to the equation 
x
2
+
9
x
2
=
10
x 
2
 + 
x 
2
 
9
​
 =10 that also satisfy the inequality 
x
3
−
3
x
>
0
x 
3
 −3x>0 are 
x
=
3
x=3 and 
x
=
−
1
x=−1.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Now, we need to show that 
y
=
x
3
−
3
x
y=x 
3
 −3x is increasing at 
x
=
1
x=1. To do this, we can find the derivative of 
y
y with respect to 
x
x and evaluate it at 
x
=
1
x=1.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

The derivative of 
y
y with respect to 
x
x is given by 
d
y
d
x
=
3
x
2
−
3
dx
dy
​
 =3x 
2
 −3. Evaluating this at 
x
=
1
x=1, we get 
d
y
d
x
=
3
(
1
)
2
−
3
=
3
−
3
=
0
dx
dy
​
 =3(1) 
2
 −3=3−3=0.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

However, we need to check if the derivative is positive or negative in the neighborhood of 
x
=
1
x=1. Let's evaluate the derivative at 
x
=
1
+
ϵ
x=1+ϵ and 
x
=
1
−
ϵ
x=1−ϵ, where 
ϵ
ϵ is a small positive number.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Evaluating the derivative at 
x
=
1
+
ϵ
x=1+ϵ, we get 
d
y
d
x
=
3
(
1
+
ϵ
)
2
−
3
=
3
(
1
+
2
ϵ
+
ϵ
2
)
−
3
=
6
ϵ
+
3
ϵ
2
dx
dy
​
 =3(1+ϵ) 
2
 −3=3(1+2ϵ+ϵ 
2
 )−3=6ϵ+3ϵ 
2
 . Since 
ϵ
ϵ is positive, this expression is positive.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Evaluating the derivative at 
x
=
1
−
ϵ
x=1−ϵ, we get 
d
y
d
x
=
3
(
1
−
ϵ
)
2
−
3
=
3
(
1
−
2
ϵ
+
ϵ
2
)
−
3
=
−
6
ϵ
+
3
ϵ
2
dx
dy
​
 =3(1−ϵ) 
2
 −3=3(1−2ϵ+ϵ 
2
 )−3=−6ϵ+3ϵ 
2
 . Since 
ϵ
ϵ is positive, this expression is negative.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate

Therefore, the derivative of 
y
y with respect to 
x
x is increasing at 
x
=
1
x=1, since it changes from negative to positive as 
x
x increases through 
1
1.



Incorrect - LLM is appropriate


Correct - LLM is appropriate


Incorrect - Python is appropriate


Correct - Python is appropriate