Let's solve the expression for the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \sqrt{x^2 - 3x + 2} + \frac{1}{\sqrt{3 + 2x - x^2}} \][/tex]
1. Simplifying the Functions inside the Square Roots:
First, let's simplify the expressions inside the square roots.
For the first part:
[tex]\[ \sqrt{x^2 - 3x + 2} \][/tex]
We need to factorize [tex]\( x^2 - 3x + 2 \)[/tex].
[tex]\[
x^2 - 3x + 2 = (x-1)(x-2)
\][/tex]
So,
[tex]\[
\sqrt{x^2 - 3x + 2} = \sqrt{(x-1)(x-2)}
\][/tex]
2. Simplifying the Denominator inside the Second Square Root:
For the second part:
[tex]\[ \frac{1}{\sqrt{3 + 2x - x^2}} \][/tex]
We need to factorize [tex]\( 3 + 2x - x^2 \)[/tex].
[tex]\[
3 + 2x - x^2 = -(x^2 - 2x - 3)
\][/tex]
Next, we factorize [tex]\( x^2 - 2x - 3 \)[/tex].
[tex]\[
x^2 - 2x - 3 = (x-3)(x+1)
\][/tex]
Therefore,
[tex]\[
3 + 2x - x^2 = -(x-3)(x+1)
\][/tex]
So,
[tex]\[
\sqrt{3 + 2x - x^2} = \sqrt{-(x-3)(x+1)} = i \sqrt{(x-3)(x+1)}
\][/tex]
Thus, the second term becomes:
[tex]\[
\frac{1}{\sqrt{3 + 2x - x^2}} = \frac{1}{i \sqrt{(x-3)(x+1)}} = -\frac{i}{\sqrt{(x-3)(x+1)}}
\][/tex]
3. Combining the Two Parts:
Thus, our function [tex]\( f(x) \)[/tex] is:
[tex]\[
f(x) = \sqrt{(x-1)(x-2)} - \frac{i}{\sqrt{(x-3)(x+1)}}
\][/tex]
To conclude, the function [tex]\( f(x) \)[/tex] we have simplified to:
[tex]\[
f(x) = \sqrt{x^2 - 3x + 2} + \frac{1}{\sqrt{3 + 2x - x^2}}
\][/tex]
This is the detailed breakdown of the function.
