Let f(x) = ax + b and g(x) = cx + d. The cond | vedclass

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(C) Step $1$: Compute $f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b$.Step $2$: Compute $g(f(x)) = g(ax + b) = c(ax + b) + d = cax + cb + d$.Step

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$f(d) = g(b)$

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(C) Step $1$: Compute $f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b$.Step $2$: Compute $g(f(x)) = g(ax + b) = c(ax + b) + d = cax + cb + d$.Step $3$: Equate the two expressions: $acx + ad + b = cax + cb + d$.Step $4$: Since this holds for all $x$, the constant terms must be equal: $ad + b = cb + d$.Step $5$: Rearrange the equation: $ad - d = cb - b$, which is $d(a - 1) = b(c - 1)$.Step $6$: Alternatively, observe that $f(d) = ad + b$ and $g(b) = cb + d$. From the equality $ad + b = cb + d$, we see that $f(d) = g(b)$.

Let $f(x) = ax + b$ and $g(x) = cx + d$. The condition $f(g(x)) = g(f(x))$ holds for all $x$ if and only if ...

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