JazarahC22060 JazarahC22060
  • 04-11-2022
  • Mathematics
contestada

I tried my best but I can't seem to get it

I tried my best but I cant seem to get it class=

Respuesta :

DontavisI674018 DontavisI674018
  • 04-11-2022

Given

[tex]f(x)=\begin{cases}2x+4;{x<0} \\ 2x+8;x\ge0{}\end{cases}[/tex]

Therefore,

[tex]\begin{gathered} -1<0 \\ \Rightarrow f(-1)=2(-1)+4=-2+4=2 \\ \Rightarrow f(-1)=2 \end{gathered}[/tex]

Similarly,

[tex]\begin{gathered} 0\ge0 \\ \Rightarrow f(0)=2(0)+8=8 \\ \Rightarrow f(0)=8 \\ 2\ge0 \\ \Rightarrow f(2)=2(2)+8=4+8=12 \\ \Rightarrow f(2)=12 \end{gathered}[/tex]

The answers are f(-1)=2, f(0)=8, f(2)=12

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