Let f(x)=3x^4-7x^3+2x^2-bx+1. For what value of b is f(1)=1

First, you substitute 1 in for x, and you will get \(3\cdot1^4-7\cdot1^3+2\cdot1^2-b+1=1\). 

Simplify.\(3-7+2-b+1=1\)

\(-1-b=1\)

\(b=-2\)