Exercises — Composition of functions — f(g(x)), g(f(x))

What "compose at a point" means: substitute a number, resolve the inner machine, then the outer.

$(f\circ g)(3)=f(g(3))$. Inner first: $g(3)=2\cdot 3=6$. Now the outer: $f(6)=6+4=10$. $\boxed{(f\circ g)(3)=10}$

$(g\circ f)(3)=g(f(3))$. Inner first: $f(3)=3+4=7$. Outer: $g(7)=2\cdot 7=14$. $\boxed{(g\circ f)(3)=14}$

Different answers ($10\ne 14$) — that is non-commutativity happening right in front of you.

$(g\circ f)(x)=g(f(x))$. Inner machine: $f(x)=x^2$. Feed it to $g$: $g(x^2)=x^2-5$. $\boxed{(g\circ f)(x)=x^2-5}$ Why it is NOT $(x-5{)}^2$: that would be $f(g(x))$, the other order. Reading right-to-left, in $g\circ f$ the function $f$ sits on the right, so $f$ runs first (squaring), then $g$ subtracts 5 — giving $x^2-5$, not the square of $x-5$.

$(f\circ g)(1)=f(g(1))$. From the table $g(1)=2$. Now $f(2)$: the table's $f$-column at $x=2$ gives $9$. $\boxed{(f\circ g)(1)=9}$ The trap avoided: you must look up $f$ at the value $g$ produced ($2$), not at the original input ($1$).

$(f\circ g)(x)=f(g(x))$: replace every $x$ in $f$'s rule with the whole expression $g(x)=x^2+2$. $f(x^2+2)=3(x^2+2)-1=3x^2+6-1=3x^2+5.$ $\boxed{(f\circ g)(x)=3x^2+5}$ $(g\circ f)(x)=g(f(x))$: replace every $x$ in $g$'s rule with $f(x)=3x-1$. $g(3x-1)=(3x-1{)}^2+2=9x^2-6x+1+2=9x^2-6x+3.$ $\boxed{(g\circ f)(x)=9x^2-6x+3}$

$(f\circ g)(x)=f(g(x))=f(x+1)=\frac{1}{x+1}.$ $\boxed{(f\circ g)(x)=\frac{1}{x+1}}$ Forbidden input: division by zero is undefined, so we need $x+1\ne 0$, i.e. $x\ne -1$. See Domain and range of function.

$(f\circ g)(x)=f(g(x))=2^{x+3}$. Using the exponent law $2^{a+b}=2^a{2}^b$: $2^{x+3}=2^x\cdot 2^3=8\cdot 2^x.$ $\boxed{(f\circ g)(x)=8\cdot 2^x}✓$ Why a shift inside becomes a scale outside: adding 3 to the exponent multiplies by the fixed factor $2^3=8$. This is a preview of Function transformations.

Step 1 — the formula: $(f\circ g)(x)=f(g(x))=\sqrt{6-x}$. Step 2 — where does $g$ accept the input? $g(x)=6-x$ works for all real $x$, so no restriction from $g$ itself. Step 3 — where can $f$ eat $g$'s output? $f$ needs a non-negative input, so we require $g(x)\ge 0$: $6-x\ge 0⟹x\le 6.$ $\boxed{\text{Domain}=(-\infty ,6]}$ Key point: trace the whole chain, not just the outer machine. See Domain and range of function.

Formula: $(f\circ g)(x)=f\left(\frac{1}{x}\right)=\frac{1}{\frac{1}{x}-2}$. Multiply top and bottom by $x$: $\frac{1}{\frac{1}{x}-2}=\frac{x}{1-2x}.$ $\boxed{(f\circ g)(x)=\frac{x}{1-2x}}$ Two separate exclusions must both be honoured:

1. $g$ must be defined: $x\ne 0$ (the inner $\frac{1}{x}$).
2. $f$ must accept $g$'s output: $g(x)\ne 2$, i.e. $\frac{1}{x}\ne 2\Rightarrow x\ne \frac{1}{2}$.

Note the simplified form $\frac{x}{1-2x}$ blows up at $x=\frac{1}{2}$ (that survives), but it hides the hole at $x=0$. You must keep both exclusions: $\boxed{x\ne 0\text{ and }x\ne \frac{1}{2}}$

$(f\circ g)(x)=f(x+b)=(x+b)+a=x+a+b.$ $(g\circ f)(x)=g(x+a)=(x+a)+b=x+a+b.$ Both equal $x+a+b$, so $f\circ g=g\circ f$. $\boxed{f\circ g=g\circ f=x+a+b}$ Why special: both machines are pure shifts, and shifting by $a$ then $b$ lands you in the same place as $b$ then $a$ — total shift is $a+b$ either way (addition commutes). This is a rare pair; non-commutativity is the general rule, as L1.1 showed.

We are told the outer machine is $f(x)=\sqrt{x}$ — the last thing done is the square root. So whatever sits under the root must be the inner machine's output: $g(x)=5x-2.$ Check: $(f\circ g)(x)=f(5x-2)=\sqrt{5x-2}=h(x).$ ✓ $\boxed{g(x)=5x-2,f(x)=\sqrt{x}}$ Idea: to decompose, ask "what is the last operation performed?" — that is the outer function.

Rightmost runs first: order is $f$, then $g$, then $h$.

• $f(3)=3+1=4$
• $g(4)=4^2=16$
• $h(16)=2\cdot 16=32$ $\boxed{(h\circ g\circ f)(3)=32}$ General formula: $h(g(f(x)))=h(g(x+1))=h((x+1{)}^2)=2(x+1{)}^2.$ $\boxed{(h\circ g\circ f)(x)=2(x+1{)}^2}$ Check at $x=3$: $2(4{)}^2=32$ ✓. Associativity guarantees the bracketing doesn't matter.

We want $f(g(x))=x$, i.e. $3g(x)-5=x$. Solve for $g(x)$: $3g(x)=x+5⟹g(x)=\frac{x+5}{3}.$ $\boxed{g(x)=\frac{x+5}{3}}$ Check: $f(g(x))=3\cdot \frac{x+5}{3}-5=(x+5)-5=x.$ ✓ Because composing gives the identity $I(x)=x$, this $g$ is $f^{-1}$. Such an inverse exists here because $f$ is a bijection.

$(f\circ f)(x)=f\left(\frac{x}{x+1}\right)=\frac{\frac{x}{x+1}}{\frac{x}{x+1}+1}.$ Combine the denominator: $\frac{x}{x+1}+1=\frac{x+(x+1)}{x+1}=\frac{2x+1}{x+1}.$ So $(f\circ f)(x)=\frac{\frac{x}{x+1}}{\frac{2x+1}{x+1}}=\frac{x}{x+1}\cdot \frac{x+1}{2x+1}=\frac{x}{2x+1}.$ $\boxed{(f\circ f)(x)=\frac{x}{2x+1}}$ Valid where the chain is defined: $x\ne -1$ (inner) and $2x+1\ne 0\Rightarrow x\ne -\frac{1}{2}$.

$(f\circ g)(x)=f(\ln x)=e^{\ln x}.$ By definition, $\ln$ answers "which power of $e$ gives $x$?" and $e^{(\cdot )}$ raises $e$ to that power — they undo each other: $e^{\ln x}=x.$ $\boxed{(f\circ g)(x)=x,x>0}$ Domain caveat: although the result is $x$, we needed $g(x)=\ln x$ defined first, so $x>0$. The exp and log are inverses — but only on their proper domains.

Why the chain rule and not the power rule alone? $F$ is a composite: an inner machine $g$ feeds an outer machine $f$. The chain rule is the tool that measures how a small nudge in $x$ passes through both machines: $F^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x).$

• Outer derivative: $f(u)=u^5\Rightarrow f^{\prime }(u)=5u^4$, so $f^{\prime }(g(x))=5(x^2+1{)}^4$.
• Inner derivative: $g(x)=x^2+1\Rightarrow g^{\prime }(x)=2x$. Multiply: $F^{\prime }(x)=5(x^2+1{)}^4\cdot 2x=10x(x^2+1{)}^4.$ $\boxed{F^{\prime }(x)=10x(x^2+1{)}^4}$ Check a value: at $x=1$, $F^{\prime }(1)=10\cdot 1\cdot (2{)}^4=160$.

In $f\circ g$, which machine runs first? ::: $g$ (the rightmost / innermost) runs first. Why does simplifying a composite not restore excluded domain points? ::: Simplification hides the inner step that failed; the original chain was undefined there, so the exclusion stays. Which single extra factor does the chain rule add to the outer derivative? ::: The inner function's derivative $g^{\prime }(x)$. When is $f\circ g=g\circ f$ guaranteed for two shifts $x+a$, $x+b$? ::: Always — total shift $a+b$ is the same either way.