If $f:R \to R$ be defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{2x:x > 3}\\{{x^2}:1 < x \le 3}\\{3x:x \le 1}\end{array}} \right.$.
Then, $f( - 1) + f(2) + f(4)$ is
If $f:R \to R$ be defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{2x:x > 3}\\{{x^2}:1 < x \le 3}\\{3x:x \le 1}\end{array}} \right.$.
Then, $f( - 1) + f(2) + f(4)$ is
Official Solution
It is given that,, $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{2x:x > 3}\\{{x^2}:1 < x \le 3}\\{3x:x \le 1}\end{array}} \right.$
$f( - 1) + f(2) + f(4) = 3( - 1) + {(2)^2} + 2 \times 4$
$= - 3 + 4 + 8 = 9$
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