Alright, let's analyze the function [tex]\( G(x) = \frac{-4 \cdot x}{x + 1} \)[/tex] step by step.
### Step 1: Function Definition
The given function is:
[tex]\[ G(x) = \frac{-4x}{x+1} \][/tex]
### Step 2: Finding the Domain
The domain of the function includes all [tex]\( x \)[/tex] values for which the function is defined. The only restriction here is when the denominator is zero since division by zero is undefined.
Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
So, the domain of [tex]\( G(x) \)[/tex] is all real numbers except [tex]\( x = -1 \)[/tex].
### Step 3: Simplified Function Expression
We have the function already simplified as:
[tex]\[ G(x) = \frac{-4x}{x+1} \][/tex]
### Step 4: First Derivative of [tex]\( G(x) \)[/tex]
To find the first derivative [tex]\( G'(x) \)[/tex], we apply the quotient rule which states:
[tex]\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \][/tex]
Here, let [tex]\( u(x) = -4x \)[/tex] and [tex]\( v(x) = x+1 \)[/tex].
Calculate [tex]\( u' \)[/tex] and [tex]\( v' \)[/tex]:
[tex]\[ u'(x) = -4 \][/tex]
[tex]\[ v'(x) = 1 \][/tex]
Apply the quotient rule:
[tex]\[ G'(x) = \frac{(-4)(x+1) - (-4x)(1)}{(x+1)^2} \][/tex]
[tex]\[ G'(x) = \frac{-4x - 4 - (-4x)}{(x+1)^2} \][/tex]
[tex]\[ G'(x) = \frac{-4}{(x+1)^2} \][/tex]
So, the first derivative of [tex]\( G(x) \)[/tex] is:
[tex]\[ G'(x) = \frac{4x}{(x+1)^2} - \frac{4}{(x+1)} \][/tex]
### Step 5: Second Derivative of [tex]\( G(x) \)[/tex]
Next, we find the second derivative [tex]\( G''(x) \)[/tex]. We use the first derivative to apply the quotient rule again, or we simplify the first derivative and differentiate.
Given the first derivative [tex]\( G'(x) = \frac{4x}{(x+1)^2} - \frac{4}{(x+1)} \)[/tex]:
Find the second derivative by simplifying and differentiating term-by-term:
[tex]\[ G''(x) = \frac{d}{dx} \left( \frac{4x}{(x + 1)^2} - \frac{4}{x + 1} \right) \][/tex]
[tex]\[ G''(x) = \frac{(4 \cdot (x + 1)^2)' (4x) - 4x \cdot ((x + 1)^2)'}{(x + 1)^4} \][/tex]
[tex]\[ G''(x) = \frac{-8x}{(x + 1)^3} + \frac{8}{(x + 1)^2} \][/tex]
So, the second derivative of [tex]\( G(x) \)[/tex] is:
[tex]\[ G''(x) = \frac{-8x}{(x+1)^3} + \frac{8}{(x+1)^2} \][/tex]
### Summary of Results
Here is a concise summary of the information we derived:
[tex]\[ \text{Function, } G(x) = \frac{-4x}{x + 1} \][/tex]
[tex]\[ \text{Domain, } x \neq -1 \][/tex]
[tex]\[ \text{First derivative, } G'(x) = 4x/(x + 1)^2 - 4/(x + 1) \][/tex]
[tex]\[ \text{Second derivative, } G''(x) = \frac{-8x}{(x + 1)^3} + \frac{8}{(x + 1)^2} \][/tex]
