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  4. f: R arrow R defined by f(x)= f(x) = begincases 2x ; x> 3 x2 ; 1 < x le 3 3x ; x le 1 endcases then f(-2)+f(3)+f(4) is

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Q. $f: R \rightarrow R$ defined by $f(x)=$ $f(x) = \begin{cases} 2x ; x> 3 & \quad \\ x^2 ; 1 < x \le 3 & \quad \\ 3x ; x \le 1 & \quad \end{cases} $ then $f(-2)+f(3)+f(4)$ is

KCETKCET 2021Relations and Functions

Solution:

$f(-2)+f(3)+f(4)$
$-6+9+8$
$=11$