This document provides steps to evaluate a composite function u(w(2)). It defines u(x)=x^2+3x+2 and w(x)=1/x-1. It shows that to evaluate u(w(2)), you first find w(2)=1, then substitute 1 into u(x) to get u(1)=6. Therefore, the value of the composite function u(w(2)) is 6.

1. Example TwoEvaluate a Composite Function By: Madeline, Sam, and Carlie

2. 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1 Evaluate u(w(2))

3. 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1 Evaluate u(w(2)) Start with u(x)=𝑥2+3x + 2

4. 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1 Evaluate u(w(2)) Start with u(x)=𝑥2+3x + 2 2. LS Replace x with w(x). This is simple substitution. You’ll get u(w(x)). Do the same on RS to get: 𝑢𝑤𝑥=1𝑥−12+31𝑥−1+2

5. 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1 Evaluate u(w(2)) Start with u(x)=𝑥2+3x + 2 2. LS Replace x with w(x). This is simple substitution. You’ll get u(w(x)). Do the same on RS to get: 𝑢𝑤𝑥=1𝑥−12+31𝑥−1+2 3. Q-What do you do to go from u(w(x)) to u(w(2))?

6. 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1 Evaluate u(w(2)) Start with u(x)=𝑥2+3x + 2 2. LS Replace x with w(x). This is simple substitution. You’ll get u(w(x)). Do the same on RS to get: 𝑢𝑤𝑥=1𝑥−12+31𝑥−1+2 3. Q-What do you do to go from u(w(x)) to u(w(2))? A-Replace the x with 2: 𝑢𝑤2=12−12+312−1+2

7. 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1 Evaluate u(w(2)) Start with u(x)=𝑥2+3x + 2 2. LS Replace x with w(x). This is simple substitution. You’ll get u(w(x)). Do the same on RS to get: 𝑢𝑤𝑥=1𝑥−12+31𝑥−1+2 3. Q-What do you do to go from u(w(x)) to u(w(2))? A-Replace the x with 2: 𝑢𝑤2=12−12+312−1+2 4. Solve. 12+31+2=6

8. The same can be done in reverse: 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1

9. The same can be done in reverse: Previous Way: LS = u(x) = u(w(x)) = u(w(2)) 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1

10. The same can be done in reverse: Previous Way: LS = u(x) = u(w(x)) = u(w(2)) 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1

11. The same can be done in reverse: Reverse: 1) Determine w(2) 𝑤2=12−1 =1 2) Substitute into u(x) 𝑢𝑤2=12+31+2 =6 Previous Way: LS = u(x) = u(w(x)) = u(w(2)) 𝑢𝑥=𝑥2+3x + 2 w(x) = 1𝑥−1

12. Example TwoEvaluate a Composite Function By: Madeline, Sam, and Carlie