Given 3-x where $x \leq 2$ $x <= 2$ and $x^{2}$ $x^2$ where $x > 2$ $x >2$ , how do you find f(2)?

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Answer 1 · 2016-06-12T19:51:02

$f (2) = 1 .$ $f(2)=1.$

Given that $f (x) = 3 - x,$ $f(x)=3-x,$ x $\leq 2$ $<=2$
$f (x) = x^{2},$ $f(x)=x^2,$ x> $2 .$ $2.$

To find $f (2)$ $f(2)$ we examine the definition of $f (x)$ $f(x)$ in such a way that in it where $x = 2$ $x=2$ is allowed,

It is easy to see that when $x = 2, f (x) = 3 - x .$ $x=2, f(x)=3-x.$

$:. f(2)=3-2=1.$

Given 3-x where x <= 2x≤2x <= 2 and x^2x2x^2 where x >2x>2x >2, how do you find f(2)?