Consider the functions $f(x)=3x-1$ and $g(x)=-x+2$. Eva was tasked with finding the equations of the composite functions $h_1=f\circ g$ and $h_2=g\circ f$, calculating their values for $x=\frac23 $, and comparing these values.
Eva’s procedure:
(1) She realized that $(f\circ g)(x)= f(g(x))$ and determined the equation for $h_1$ as follows: $$ h_1(x)=f(g(x))=3\cdot(-x+2)-1=-3x+6-1=-3x+5 $$
(2) She also realized that $(g\circ f)(x)= g(f(x))$ and determined the equation for $h_2$ as follows: $$ h_2(x)=g(f(x))=-(3x-1)+2=-3x+1+2=-3x+3 $$
(3) Then, she calculated the function value of $h_1$ for $x=\frac23 $: $$ h_1\left(\frac23 \right)=-3\cdot\frac23 +5=3 $$
(4) Next, she calculated the function value of $h_2$ for $x=\frac23 $: $$ h_2\left(\frac23 \right)=-3\cdot\frac23 +3=1 $$
(5) Finally, she concluded that the value of $h_1(\frac23)$ is $2$ greater than the value of $h_2(\frac23 )$.
Eva’s friends then commented on her procedure:
Susan: “You did great! Your procedure is completely correct!”
Elis: “You didn’t even need to calculate the last three steps. From the correctly determined equations of $h_1$ and $h_2$, it is clear that the values of $h_1$ will always be $2$ greater than the function values of $h_2$ at any point $x\in \mathbb{R}$, including $x=\frac23$.”
Alice: “Your procedure is not correct. You made the mistakes in steps $1$ and $2$ as you swapped the composite functions. It should have been $h_1(x)=-3x+3$ and $h_2(x)=-3x+5$. This means $h_1(\frac23 )$ is $2$ less than $h_2(\frac23 )$.”
Nora: “You made the mistake in step $1$. It should have been $h_1(x)=3\cdot(-x)+2-1=-3x+1$. However, you determined $h_2$ correctly. Therefore, $h_1(\frac23 )$ is $2$ less than $h_2(\frac23 )$.”
Which of Eva’s friends commented on her procedure correctly?
Susan and Elis
Alice
Nora
Only Susan
Only Elis
The composite function $h_1=f\circ g$ is defined by equality: $h_1(x)=f(g(x))$. This means that the function $g(x)$ is “inserted” into the function $f(x)$. In the equation for the function $f(x)$ every $x$ is replaced by $g(x)$. Therefore, Eva proceeded completely correctly. At the same time, also Elis was right. If the equations of two functions differ only by a constant, then their function values also differ by this constant at any point of their domain.