Given:
[tex]f(x)=\begin{cases}4x+7ifx<0{} \\ 6x+4ifx\ge0{}\end{cases}[/tex]
Required:
To find the value of f(-8), f(0), f(4), and f(-100)+f(100).
Explanation:
f(-8) :
Clearly -8<0,
So
[tex]\begin{gathered} f(x)=4x+7 \\ f(-8)=4(-8)+7 \\ =-32+7 \\ =-25 \end{gathered}[/tex]
f(0) :
Clearly 0=0,
[tex]\begin{gathered} f(x)=6x+4 \\ =6(0)+4 \\ =4 \end{gathered}[/tex]
f(4) :
Clearly 4>0,
[tex]\begin{gathered} f(x)=6x+4 \\ f(4)=6(4)+4 \\ =24+4 \\ =28 \end{gathered}[/tex]
f(-100)+f(100) :
-100<0
[tex]\begin{gathered} f(x)=4x+7 \\ f(-100)=4(-100)+7 \\ =-400+7 \\ =-393 \end{gathered}[/tex]
100>0
[tex]\begin{gathered} f(x)=6x+4 \\ f(100)=6(100)+4 \\ =600+4 \\ =604 \end{gathered}[/tex][tex]\begin{gathered} f(-100)+f(100)=-393+604 \\ \\ =211 \end{gathered}[/tex]
Final Answer:
[tex]\begin{gathered} f(-8)=-25 \\ \\ f(0)=4 \\ \\ f(4)=28 \\ \\ f(-100)+f(100)=211 \end{gathered}[/tex]
