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
Part C
f(3)
For x=3 ---> belong to the interval [0, infinite)
so
f(x)=x+7
substitute
f(3)=3+7
