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Let \( f: R \rightarrow R \) be defined by \( f(x)=\left\{\begin{array}{cc}2 x ; & x>3 \\ x^{2} ; & 1 < x \leq 3 \\ 3 x ; & x \leq 1\end{array}\right. \)
Then \( f(-1)+f(2)+f(4) \) is
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Then \( f(-1)+f(2)+f(4) \) is
Solution:
1648 Upvotes
Verified Answer
The correct answer is:
\( 09 \)
Given function,
\( f(x)=\left\{\begin{array}{ll}2 x & ; x>3 \\ x^{2} & ; 1 < x \leq 3 \\ 3 x & ; x \leq 1\end{array}\right. \)
So,
\[
\begin{array}{l}
f(-1)+f(2)+f(4)=3(-1)+(2)^{2}+2(4) \\
=-3+4+8=9
\end{array}
\]
\( f(x)=\left\{\begin{array}{ll}2 x & ; x>3 \\ x^{2} & ; 1 < x \leq 3 \\ 3 x & ; x \leq 1\end{array}\right. \)
So,
\[
\begin{array}{l}
f(-1)+f(2)+f(4)=3(-1)+(2)^{2}+2(4) \\
=-3+4+8=9
\end{array}
\]
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