Answer :
To determine the coordinates of the stationary points of the function [tex]\( y = \frac{-4 - x^3}{2 x^2} \)[/tex], we need to follow these steps:
1. Find the first derivative of the function:
We start by differentiating [tex]\( y = \frac{-4 - x^3}{2 x^2} \)[/tex] with respect to [tex]\( x \)[/tex]. Let's denote the derivative by [tex]\( y' \)[/tex].
2. Set the derivative equal to zero to find stationary points:
Stationary points occur where the first derivative is equal to zero. So, we solve [tex]\( y' = 0 \)[/tex] for [tex]\( x \)[/tex].
3. Find the corresponding [tex]\( y \)[/tex]-coordinates:
For each [tex]\( x \)[/tex]-value found in the previous step, we substitute it back into the original function [tex]\( y = \frac{-4 - x^3}{2 x^2} \)[/tex] to find the corresponding [tex]\( y \)[/tex]-coordinate.
Let's go through these steps.
### Step 1: Find the first derivative
The given function is:
[tex]\[ y = \frac{-4 - x^3}{2 x^2} \][/tex]
To differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], we use the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \][/tex]
where [tex]\( u = -4 - x^3 \)[/tex] and [tex]\( v = 2 x^2 \)[/tex].
First, find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx}(-4 - x^3) = -3x^2 \][/tex]
[tex]\[ v' = \frac{d}{dx}(2 x^2) = 4x \][/tex]
Now, apply the quotient rule:
[tex]\[ y' = \frac{(-3x^2)(2x^2) - (-4 - x^3)(4x)}{(2x^2)^2} \][/tex]
[tex]\[ y' = \frac{-6x^4 - (-4 \cdot 4x - x^3 \cdot 4x)}{4x^4} \][/tex]
[tex]\[ y' = \frac{-6x^4 - (-16x - 4x^4)}{4x^4} \][/tex]
[tex]\[ y' = \frac{-6x^4 + 16x + 4x^4}{4x^4} \][/tex]
[tex]\[ y' = \frac{-2x^4 + 16x}{4x^4} \][/tex]
[tex]\[ y' = \frac{-2x^4 + 16x}{4x^4} \][/tex]
We can simplify:
[tex]\[ y' = \frac{16x - 2x^4}{4x^4} \][/tex]
[tex]\[ y' = \frac{16x}{4x^4} - \frac{2x^4}{4x^4} \][/tex]
[tex]\[ y' = \frac{4}{x^3} - \frac{1}{2} \][/tex]
The derivative is:
[tex]\[ y' = -\frac{x^3+4}{2x^3} \][/tex]
### Step 2: Set the derivative equal to zero
To find the stationary points, set the derivative equal to zero:
[tex]\[ -\frac{x^3+4}{2x^3} = 0 \][/tex]
This simplifies to:
[tex]\[ x^3 + 4 = 0 \][/tex]
[tex]\[ x^3 = -4 \][/tex]
[tex]\[ x = \sqrt[3]{-4} \][/tex]
The roots of this equation are:
[tex]\[ x = 2 \][/tex]
and two complex roots:
[tex]\[ x = -1 - \sqrt{3}i \][/tex]
[tex]\[ x = -1 + \sqrt{3}i \][/tex]
### Step 3: Find the corresponding y-coordinates
For each [tex]\( x \)[/tex]-value, we substitute it back into the original function [tex]\( y = \frac{-4 - x^3}{2x^2} \)[/tex] to find the corresponding [tex]\( y \)[/tex]-coordinate.
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = \frac{-4 - 2^3}{2 \cdot 2^2} \][/tex]
[tex]\[ y = \frac{-4 - 8}{8} \][/tex]
[tex]\[ y = \frac{-12}{8} \][/tex]
[tex]\[ y = -\frac{3}{2} \][/tex]
So, the stationary point is [tex]\( (2, -\frac{3}{2}) \)[/tex].
For [tex]\( x = -1 - \sqrt{3}i \)[/tex]:
[tex]\[ y = \frac{-4 - (-1 - \sqrt{3}i)^3}{2(-1 - \sqrt{3}i)^2} = (-1 - sqrt(3)I, (-4 - (-1 - sqrt(3)I)3)/(2(-1 - sqrt(3)I)2) \][/tex]
Complex calculations of [tex]\( y \)[/tex] for non-real values are:
(-1 - sqrt(3)I, (-4 - (-1 - sqrt(3)I)3)/(2(-1 - sqrt(3)I)2)), (-1 + sqrt(3)I, (-4 - (-1 + sqrt(3)I)3)/(2(-1 + sqrt(3)I)2))
So, the stationary points are:
[tex]\[ (2, -\frac{3}{2}) \][/tex]
[tex]\[ (-1 - \sqrt{3}i , y (-4 - (-1 - sqrt(3)I)3)/(2(-1 - sqrt(3)I)2)) \][/tex]
[tex]\[ (-1 + \sqrt{3}i, y (-4 - (-1 + sqrt(3)I)3)/(2(-1 + sqrt(3)I)2))\][/tex]
These are coordinates of three stationary points for the given function.
1. Find the first derivative of the function:
We start by differentiating [tex]\( y = \frac{-4 - x^3}{2 x^2} \)[/tex] with respect to [tex]\( x \)[/tex]. Let's denote the derivative by [tex]\( y' \)[/tex].
2. Set the derivative equal to zero to find stationary points:
Stationary points occur where the first derivative is equal to zero. So, we solve [tex]\( y' = 0 \)[/tex] for [tex]\( x \)[/tex].
3. Find the corresponding [tex]\( y \)[/tex]-coordinates:
For each [tex]\( x \)[/tex]-value found in the previous step, we substitute it back into the original function [tex]\( y = \frac{-4 - x^3}{2 x^2} \)[/tex] to find the corresponding [tex]\( y \)[/tex]-coordinate.
Let's go through these steps.
### Step 1: Find the first derivative
The given function is:
[tex]\[ y = \frac{-4 - x^3}{2 x^2} \][/tex]
To differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], we use the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \][/tex]
where [tex]\( u = -4 - x^3 \)[/tex] and [tex]\( v = 2 x^2 \)[/tex].
First, find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx}(-4 - x^3) = -3x^2 \][/tex]
[tex]\[ v' = \frac{d}{dx}(2 x^2) = 4x \][/tex]
Now, apply the quotient rule:
[tex]\[ y' = \frac{(-3x^2)(2x^2) - (-4 - x^3)(4x)}{(2x^2)^2} \][/tex]
[tex]\[ y' = \frac{-6x^4 - (-4 \cdot 4x - x^3 \cdot 4x)}{4x^4} \][/tex]
[tex]\[ y' = \frac{-6x^4 - (-16x - 4x^4)}{4x^4} \][/tex]
[tex]\[ y' = \frac{-6x^4 + 16x + 4x^4}{4x^4} \][/tex]
[tex]\[ y' = \frac{-2x^4 + 16x}{4x^4} \][/tex]
[tex]\[ y' = \frac{-2x^4 + 16x}{4x^4} \][/tex]
We can simplify:
[tex]\[ y' = \frac{16x - 2x^4}{4x^4} \][/tex]
[tex]\[ y' = \frac{16x}{4x^4} - \frac{2x^4}{4x^4} \][/tex]
[tex]\[ y' = \frac{4}{x^3} - \frac{1}{2} \][/tex]
The derivative is:
[tex]\[ y' = -\frac{x^3+4}{2x^3} \][/tex]
### Step 2: Set the derivative equal to zero
To find the stationary points, set the derivative equal to zero:
[tex]\[ -\frac{x^3+4}{2x^3} = 0 \][/tex]
This simplifies to:
[tex]\[ x^3 + 4 = 0 \][/tex]
[tex]\[ x^3 = -4 \][/tex]
[tex]\[ x = \sqrt[3]{-4} \][/tex]
The roots of this equation are:
[tex]\[ x = 2 \][/tex]
and two complex roots:
[tex]\[ x = -1 - \sqrt{3}i \][/tex]
[tex]\[ x = -1 + \sqrt{3}i \][/tex]
### Step 3: Find the corresponding y-coordinates
For each [tex]\( x \)[/tex]-value, we substitute it back into the original function [tex]\( y = \frac{-4 - x^3}{2x^2} \)[/tex] to find the corresponding [tex]\( y \)[/tex]-coordinate.
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = \frac{-4 - 2^3}{2 \cdot 2^2} \][/tex]
[tex]\[ y = \frac{-4 - 8}{8} \][/tex]
[tex]\[ y = \frac{-12}{8} \][/tex]
[tex]\[ y = -\frac{3}{2} \][/tex]
So, the stationary point is [tex]\( (2, -\frac{3}{2}) \)[/tex].
For [tex]\( x = -1 - \sqrt{3}i \)[/tex]:
[tex]\[ y = \frac{-4 - (-1 - \sqrt{3}i)^3}{2(-1 - \sqrt{3}i)^2} = (-1 - sqrt(3)I, (-4 - (-1 - sqrt(3)I)3)/(2(-1 - sqrt(3)I)2) \][/tex]
Complex calculations of [tex]\( y \)[/tex] for non-real values are:
(-1 - sqrt(3)I, (-4 - (-1 - sqrt(3)I)3)/(2(-1 - sqrt(3)I)2)), (-1 + sqrt(3)I, (-4 - (-1 + sqrt(3)I)3)/(2(-1 + sqrt(3)I)2))
So, the stationary points are:
[tex]\[ (2, -\frac{3}{2}) \][/tex]
[tex]\[ (-1 - \sqrt{3}i , y (-4 - (-1 - sqrt(3)I)3)/(2(-1 - sqrt(3)I)2)) \][/tex]
[tex]\[ (-1 + \sqrt{3}i, y (-4 - (-1 + sqrt(3)I)3)/(2(-1 + sqrt(3)I)2))\][/tex]
These are coordinates of three stationary points for the given function.
