Respuesta :
After evaluation, the resultant value of the function f(2009) is 1.9021.
What are functions?
A function in mathematics from a set X to a set Y allocates exactly one element of Y to each element of X.
The sets X and Y are collectively referred to as the function's domain and codomain, respectively.
Any input for x results in a single output for y.
Considering that x is the input value, we would state that y is a function of x.
So, evaluate f(2009) as follows:
Given that f(x) is a function with all positive real numbers satisfying the requirement that f(x) > 0 and for all x > 0 and also:
[tex]\begin{aligned}&f(x-y)=\sqrt{f(x y)+1} \\&x > 0, y > 0 \\&\text { for, } x=1, \text { and } y=\frac{1}{2} \\&f\left(1-\frac{1}{2}\right)=\sqrt{f\left(1 \times \frac{1}{2}\right)+1} \\&f\left(\frac{1}{2}\right)^2=f\left(\frac{1}{2}\right)+1 \\&f\left(\frac{1}{2}\right)=\frac{1+\sqrt{5}}{2}\end{aligned}[/tex]
Now, consider: x = 2009 and y = 0
[tex]\begin{aligned}&f(2009-0)=\sqrt{f(2009 * 0)+1} \\&f(2009)=\sqrt{f(0)+1} \\&f(2009)=\sqrt{f\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)+1} \\&f(2009)=\sqrt{\sqrt{f\left(\frac{1}{\sqrt{2}} * \frac{1}{\sqrt{2}}\right)+1}+1} \\&f(2009)=\sqrt{\sqrt{f\left(\frac{1}{2}\right)+1}+1} \\&f(2009)=\sqrt{\sqrt{\frac{1+\sqrt{5}}{2}+1}+1} \\&f(2009)=\sqrt{\sqrt{\frac{3+\sqrt{5}}{2}}+1} \\&f(2009)=\sqrt{\sqrt{5.2362}+1} \\&=\sqrt{3.6180} \\&=1.9021\end{aligned}[/tex]
Therefore, after evaluation, the resultant value of the function f(2009) is 1.9021.
Know more about functions here:
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