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Example 2
1. Example 2 Let u(x)=x²+3x+2 and w(x)=1 x-1 a) Evaluate u(w(2)) b) Evaluate w(u(-3)) Method 1: Find u(w(x)) the sub into equation a) u(x) =x²+3x+2 u(w(x))=(1/(x-1))²+3(1/(x-1))+2 u(w(2))=(1/(2-1)²+3(1/(2-1))+2 =1²+3(1)+2 =6
2. Let u(x)=x²+3x+2 and w(x)=1 x-1 a) Evaluate u(w(2)) b) Evaluate w(u(-3)) Method 2: Find w(2) then sub into equation a) w(x)=1/(x-1) w(2)=1/(2-1) =1 u(w(2))=u(1) =1/(2-1) =6
3. Let u(x)=x²+3x+2 and w(x)=1 x-1 a) Evaluate u(w(2)) b) Evaluate w(u(-3)) Find u(-3) then sub into w(x) and evaluate b) u(x)=x²+3x+2 u(-3)=(-3)²+3(-3)+2 =2 w(u(-3))=w(2) =1/(2-1) =1
