Question bank — Composition of functions — f(g(x)), g(f(x))

False — composition is generally non-commutative; e.g. doubling then adding 3 gives $2x+3$, but adding 3 then doubling gives $2x+6$.

False — they may agree at isolated points yet differ everywhere else; equality as functions needs agreement for all $x$ in the domain.

False — that is the product of functions; composition means feeding $g(x)$ into $f$, i.e. $f(g(x))$, which is a nested substitution not a multiplication.

True — both sides unfold to $h(g(f(x)))$; the parentheses only regroup, they never change which function touches the input first.

True — $I$ returns its input untouched, so whether you apply it before or after $f$, nothing is added or removed; it behaves like $+0$ or $\times 1$.

True — every real $x$ is a legal input to $g$, and $g(x)$ is real so it is a legal input to $f$; nothing can break, so the chain is defined everywhere. (See Domain and range of function.)

False — e.g. $h(x)=(2x+1{)}^3$ splits as $g(x)=2x+1,f(x)=x^3$, but also trivially as $g(x)=x,f(x)=(2x+1{)}^3$; many splits exist, some just more useful.

True — feeding a larger input to $g$ gives a larger $g(x)$, and $f$ then turns that into a larger output; "bigger in ⇒ bigger out" survives the chain.

False — it is increasing; two order-reversals cancel, just like a negative times a negative is positive.

False — that only holds for special "self-inverse" functions like $f(x)=1/x$ or $f(x)=-x$; for $f(x)=x+1$ you get $f(f(x))=x+2$, nowhere near the identity.

The circle reads right-to-left: $(f\circ g)(x)=f(g(x))$ runs $g$ first. Reading it like English left-to-right is the classic trap.

Wrong constraint — the input to the square root is $x-5$, so we need $x-5\ge 0$, giving domain $[5,\infty )$, not $[0,\infty )$. The shift was ignored.

Incomplete — you must also require $x$ to lie in $g$'s domain and $g(x)$ to land in $f$'s domain; the whole chain must survive, not just the last link.

Backwards — you substitute $f(x)$ into $g$: take $g$'s formula and replace its variable with the whole expression $f(x)$. The inner function's output is what gets fed in.

You verify by re-composing: $f(g(x))$ must reproduce $h(x)$. If $g$ isn't exactly the inner piece $2x+1$, the rebuilt expression won't match — always substitute back to confirm.

Because $g$'s outputs become $f$'s inputs; if $g$ produces a value $f$ cannot accept (like a negative under a square root), the chain snaps at that value even though $g$ itself was happy.

It mirrors nested parentheses: in $f(g(x))$ you must resolve the innermost $g(x)$ before $f$ can act, so the physically-first operation is the rightmost symbol.

Every grouping evaluates the functions in the same fixed order (f, then g, then h); parentheses only say "which pair to name first", not "which runs first", so the final output is identical.

$I(x)=x$ returns exactly what it received, so inserting it into a chain neither transforms nor blocks the value — like an empty machine on the assembly line.

Because a composite scales change twice: $g$ magnifies the input's wiggle by $g^{\prime }$, then $f$ magnifies that by $f^{\prime }$ evaluated at $g(x)$; nested transformations multiply their sensitivities.

An inverse is built to undo $f$ so their chain returns the input; a general $f\circ f$ just applies the same transformation twice, which usually keeps moving the value further.

It is itself constant: $f(g(x))=f(7)$, a single fixed value for every $x$ (provided $7$ is in $f$'s domain), because $f$ always receives the same input.

Yes — the double filter (legal for $g$, and $g(x)$ legal for $f$) can shrink it below either original, e.g.  ​ of a shifted input trims off part of the real line.

The composite is undefined there even if $x$ was fine for $g$; that single output routes into a forbidden slot of $f$, so a hole appears in $f\circ g$'s domain.

Yes — the chain is still one-to-one (distinct inputs stay distinct through both stages) and onto (every target is reachable), so the whole assembly line is reversible.

$(f\circ g)(x)=e^{\ln x}=x$, but only for $x>0$, since $\ln x$ requires a positive input even though the simplified answer $x$ looks universal. (See Exponential and logarithmic functions.)

Still the identity $I(x)=x$ — doing "nothing" twice is doing nothing, so it stays the do-nothing element.

With $f=I$: $g\circ f=g(I(x))=g(x)$ trivially; and $f\circ g=I(g(x))=g(x)$ too — identity is invisible on either side, which is exactly why it earns the name.