SL 2.5—Composite functions, identity, finding inverse

$(f\circ g)(x)=f(g(x))$, so we apply $g$ first, then $f$. Here $g(x)=x^3-2$ and $f(x)=x+5$.

Replace the input of $f$ with $g(x)=x^3-2$: $f(g(x))=f(x^3-2)=(x^3-2)+5$

$f(x^3-2)=x^3+3$

$(f\circ g)(x)=x^3+3$.

We need $(f\circ g)(x)=f(g(x))$, meaning $g$ acts first, then $f$ adds 5.

$(x+5{)}^3-2$ would result from $(g\circ f)(x)=g(f(x))$, not $f(g(x))$. This is the wrong order.

$x^3-7$ would arise if $f$ subtracted 5 instead of adding it. Since $f(x)=x+5$, we add 5, not subtract.

This expression contains an $x$ term which cannot arise from simply adding 5 to $x^3-2$. It is not a valid result of this composition.

Adding 5 to $x^3-2$ gives $x^3+3$, confirming the answer is $(f\circ g)(x)=x^3+3$.

$(g\circ f)(x)=g(f(x))$: apply $f$ first, then apply $g$ to the result.

$g(f(x))=g(4x+1)=(4x+1{)}^2-3$

$(4x+1{)}^2=16x^2+8x+1$

$16x^2+8x+1-3=16x^2+8x-2$

$(g\circ f)(x)=16x^2+8x-2$.

We need $g(f(x))$ where $f(x)=4x+1$ and $g(x)=x^2-3$. The result must be a degree-2 polynomial.

$4x^2-11$ would arise from squaring $2x$ and subtracting, which doesn't match $f(x)=4x+1$.

$16x^2+8x+1=(4x+1{)}^2$ — this is $(4x+1{)}^2$ without subtracting 3. The student forgot to subtract 3 from $g$.

$4x^2-3$ would result from $g(f(x))=(f(x){)}^2-3$ if $f(x)=2x$, not $4x+1$.

$(4x+1{)}^2-3=16x^2+8x+1-3=16x^2+8x-2$, confirming this option.