Let f(x) = ax + b and g(x) = cx + d,where a n | vedclass

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(B) Given $f(x) = ax + b$ and $g(x) = cx + d$. We are given $a = 1$ and $b = 2$,so $f(x) = x + 2$. Now,calculate the composition $(fog)(x) = f(g(x)) =

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$c = 1$,$d$ is arbitrary

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(B) Given $f(x) = ax + b$ and $g(x) = cx + d$. We are given $a = 1$ and $b = 2$,so $f(x) = x + 2$. Now,calculate the composition $(fog)(x) = f(g(x)) = f(cx + d) = (cx + d) + 2 = cx + d + 2$. Next,calculate the composition $(gof)(x) = g(f(x)) = g(x + 2) = c(x + 2) + d = cx + 2c + d$. Since $(fog)(x) = (gof)(x)$ for all $x$,we equate the two expressions: $cx + d + 2 = cx + 2c + d$. Subtracting $cx$ and $d$ from both sides,we get $2 = 2c$,which implies $c = 1$. Since $d$ cancels out from both sides,$d$ can be any real number (arbitrary). Thus,$c = 1$ and $d$ is arbitrary.

Let $f(x) = ax + b$ and $g(x) = cx + d$,where $a \ne 0$ and $c \ne 0$. Assume $a = 1$ and $b = 2$. If $(fog)(x) = (gof)(x)$ for all $x$,what can you say about $c$ and $d$?

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