Respuesta :

we have that

interval (-infinite,0) ------> f(x)=5x+4

interval [0, infinite) ------> f(x)=x+7

Part B

f(0)

For x=0 ------> belong to the interval [0, infinite)

so

f(x)=x+7

substitute

f(0)=0+7

f(0)=7

Part C

f(3)

For x=3 ---> belong to the interval [0, infinite)

so

f(x)=x+7

substitute

f(3)=3+7

f(3)=10

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