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Please check if I got the extreme values correct
- Thread starter aruwin
- Start date
Hello, aruwin!
I can read about half of what you wrote.
Differentiate implicitly: .\(\displaystyle 3x^2 + 6x + 3y^2y' - 2y' \:=\:0 \quad\Rightarrow\quad 2y' - 3y^2y' \:=\:3x^2 + 6x \)
. . . . \(\displaystyle (2 - 3y^2)y' \:=\:3x)x+2) \quad\Rightarrow\quad y' \:=\:\dfrac{3x(x+2)}{2-3y^2}\)
\(\displaystyle \text{If }y' = 0,\text{ then: }\,3x(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:0,\,-2\)
If \(\displaystyle x = 0\), the equation becomes: .\(\displaystyle y^3-2y \:=\:0 \quad\Rightarrow\quad y(y^2-2) \:=\:0 \)
. . Hence: .\(\displaystyle y \:=\:0,\,\pm\sqrt{2}\)
We have three extreme points: .\(\displaystyle (0,\,0),\;(0,\sqrt{2}),\;(0,\text{-}\sqrt{2})\)
If \(\displaystyle x = -2\), the equation becomes: .\(\displaystyle -8 + 12 + y^3 - 2y \:=\:0 \quad\Rightarrow\quad y^3 - 2y + 4 \:=\:0\)
. . Hence: .\(\displaystyle (y + 2)(y^2-2y + 2) \:=\:0 \quad\Rightarrow\quad y \:=\:-2\)
And we have another extreme point: .\(\displaystyle (\text{-}2,\,\text{-}2)\)
I can read about half of what you wrote.
\(\displaystyle \text{Find the extreme values of the implicit function: }\:x^3 + 3x^2 + y^3 - 2y\:=\:0\)
Differentiate implicitly: .\(\displaystyle 3x^2 + 6x + 3y^2y' - 2y' \:=\:0 \quad\Rightarrow\quad 2y' - 3y^2y' \:=\:3x^2 + 6x \)
. . . . \(\displaystyle (2 - 3y^2)y' \:=\:3x)x+2) \quad\Rightarrow\quad y' \:=\:\dfrac{3x(x+2)}{2-3y^2}\)
\(\displaystyle \text{If }y' = 0,\text{ then: }\,3x(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:0,\,-2\)
If \(\displaystyle x = 0\), the equation becomes: .\(\displaystyle y^3-2y \:=\:0 \quad\Rightarrow\quad y(y^2-2) \:=\:0 \)
. . Hence: .\(\displaystyle y \:=\:0,\,\pm\sqrt{2}\)
We have three extreme points: .\(\displaystyle (0,\,0),\;(0,\sqrt{2}),\;(0,\text{-}\sqrt{2})\)
If \(\displaystyle x = -2\), the equation becomes: .\(\displaystyle -8 + 12 + y^3 - 2y \:=\:0 \quad\Rightarrow\quad y^3 - 2y + 4 \:=\:0\)
. . Hence: .\(\displaystyle (y + 2)(y^2-2y + 2) \:=\:0 \quad\Rightarrow\quad y \:=\:-2\)
And we have another extreme point: .\(\displaystyle (\text{-}2,\,\text{-}2)\)
Hello, aruwin!
I can read about half of what you wrote.
Differentiate implicitly: .\(\displaystyle 3x^2 + 6x + 3y^2y' - 2y' \:=\:0 \quad\Rightarrow\quad 2y' - 3y^2y' \:=\:3x^2 + 6x \)
. . . . \(\displaystyle (2 - 3y^2)y' \:=\:3x)x+2) \quad\Rightarrow\quad y' \:=\:\dfrac{3x(x+2)}{2-3y^2}\)
\(\displaystyle \text{If }y' = 0,\text{ then: }\,3x(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:0,\,-2\)
If \(\displaystyle x = 0\), the equation becomes: .\(\displaystyle y^3-2y \:=\:0 \quad\Rightarrow\quad y(y^2-2) \:=\:0 \)
. . Hence: .\(\displaystyle y \:=\:0,\,\pm\sqrt{2}\)
We have three extreme points: .\(\displaystyle (0,\,0),\;(0,\sqrt{2}),\;(0,\text{-}\sqrt{2})\)
If \(\displaystyle x = -2\), the equation becomes: .\(\displaystyle -8 + 12 + y^3 - 2y \:=\:0 \quad\Rightarrow\quad y^3 - 2y + 4 \:=\:0\)
. . Hence: .\(\displaystyle (y + 2)(y^2-2y + 2) \:=\:0 \quad\Rightarrow\quad y \:=\:-2\)
And we have another extreme point: .\(\displaystyle (\text{-}2,\,\text{-}2)\)
Yes, got those 4 extreme points but what are the maximum and minimum values??
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
You did not calculate the second derivative that I can see.
\(\displaystyle \frac{dy}{dx} = \frac{3x^2+6x}{2-3y^2}\)
So, the second derivative:
\(\displaystyle \frac{d^2y}{dx^2} = \frac{(2-3y^2)(6x+6)+(3x^2+6x)(6y\frac{dy}{dx})}{(2-3y^2)^2}\)
At \(\displaystyle (0,0)\), this is \(\displaystyle \frac{(2)(6)}{2^2} = 3\)
Local minimum.
At \(\displaystyle (0,\pm \sqrt{2})\), this is \(\displaystyle \frac{(2-3*2)(6)+(0)}{(2-3(2))^2} = -\frac{3}{2}\)
Both are local maximums.
At \(\displaystyle (-2,-2)\), this is \(\displaystyle \frac{(-10)(-6)+(0)}{(2-3(4))^2} = \frac{60}{100} = \frac{3}{5}\)
This is also a local minimum.
Based on the graph, it makes a circle around the point \(\displaystyle (0,\sqrt{\frac{2}{3}})\) of radius \(\displaystyle \sqrt{\frac{2}{3}}\) (or close to a circle). Then, the global maximum occurs when x approaches negative infinity, and the global minimum when x approaches infinity.
You should check out the graph at Wolfram Alpha
\(\displaystyle \frac{dy}{dx} = \frac{3x^2+6x}{2-3y^2}\)
So, the second derivative:
\(\displaystyle \frac{d^2y}{dx^2} = \frac{(2-3y^2)(6x+6)+(3x^2+6x)(6y\frac{dy}{dx})}{(2-3y^2)^2}\)
At \(\displaystyle (0,0)\), this is \(\displaystyle \frac{(2)(6)}{2^2} = 3\)
Local minimum.
At \(\displaystyle (0,\pm \sqrt{2})\), this is \(\displaystyle \frac{(2-3*2)(6)+(0)}{(2-3(2))^2} = -\frac{3}{2}\)
Both are local maximums.
At \(\displaystyle (-2,-2)\), this is \(\displaystyle \frac{(-10)(-6)+(0)}{(2-3(4))^2} = \frac{60}{100} = \frac{3}{5}\)
This is also a local minimum.
Based on the graph, it makes a circle around the point \(\displaystyle (0,\sqrt{\frac{2}{3}})\) of radius \(\displaystyle \sqrt{\frac{2}{3}}\) (or close to a circle). Then, the global maximum occurs when x approaches negative infinity, and the global minimum when x approaches infinity.
You should check out the graph at Wolfram Alpha
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