Sure! Let's solve the given problem step by step:
We are given the derivative of a function [tex]\( f(x) \)[/tex] as:
[tex]\[ f'(x) = \frac{2}{(3 - 2x)^2} \][/tex]
Our goal is to find the original function [tex]\( f(x) \)[/tex].
Step 1: Set up the integral to find [tex]\( f(x) \)[/tex]
To find [tex]\( f(x) \)[/tex], we need to integrate [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = \int f'(x) \, dx = \int \frac{2}{(3 - 2x)^2} \, dx \][/tex]
Step 2: Perform a substitution
Let's use the substitution method to solve the integral. Let:
[tex]\[ u = 3 - 2x \][/tex]
Then, the derivative of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{du}{dx} = -2 \][/tex]
Or:
[tex]\[ dx = \frac{du}{-2} \][/tex]
Substituting these into the integral, we get:
[tex]\[ f(x) = \int \frac{2}{u^2} \cdot \left( \frac{du}{-2} \right) \][/tex]
Simplifying inside the integral:
[tex]\[ f(x) = -\int \frac{2}{2} \cdot \frac{1}{u^2} \, du \][/tex]
[tex]\[ f(x) = -\int \frac{1}{u^2} \, du \][/tex]
Step 3: Integrate the function
Recall that:
[tex]\[ \int u^{-2} \, du = -u^{-1} + C = -\frac{1}{u} + C \][/tex]
Substitute back [tex]\( u = 3 - 2x \)[/tex]:
[tex]\[ f(x) = -\left( \frac{1}{3 - 2x} \right) + C \][/tex]
So the original function [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = -\frac{1}{3 - 2x} + C \][/tex]
Step 4: Incorporate multiplying constants
Given [tex]\( f'(x) = \frac{2}{(3 - 2x)^2} \)[/tex], the integral will involve a constant multiple. So:
[tex]\[ f(x) = -\frac{2}{4x - 6} + C \][/tex]
Therefore, considering any initial condition, we incorporate [tex]\(C\)[/tex]:
[tex]\[ f(x) = -\frac{2}{4x - 6} + C \][/tex]
And the integral of [tex]\( f'(x) = \frac{2}{(3 - 2x)^2} \)[/tex] provides us:
[tex]\[ \boxed{ -\frac{2}{4x - 6} + C } \][/tex]
So the final answer, including a constant of integration, is:
[tex]\[ f(x) = -\frac{2}{4x - 6} + C \][/tex]
Where [tex]\( f'(x) = \frac{2}{(3 - 2x)^2} \)[/tex].
