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Example One
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  
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  

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Exampl

  • 2. 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  
  • 3. 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