Question

заданы функции f(x)=3/2+x и g(x)=x+1/2.Найдите: 1)f(1)+g(3) 2)g(7)-f(7) 3)f(4)+g(4) 4)g(5)-f(10)

Answer 1

1) f(1)+g(3)=3/2 + 1 +3+1/2=4+2=6
2) g(7)-f(7)=7+1/2-3/2-7=-2/2=-1
3) f(4)+g(4)=3/2+4+4+1/2=8+2=10
4) g(5)-f(10)=5+1/2-3/2-10=-5-1=-6