Mathematics

Problem

The function

f:\mathbb R\to\mathbb R

is given by

f(x)=x^{2}

. A relation

g

is formed by interchanging

x

and

y

y=f(x)

, yielding

g(x)=\sqrt{x}

. Which statement correctly explains why

f\circ g

does NOT equal the identity on

\mathbb R

? a) Because

f

is not one–one on

\mathbb R

, its inverse is not a function, so

f\circ g

cannot be the identity. b) Because

g

is defined only for

x\ge 0

, the composition

f\circ g

is defined only for

x\ge 0

and thus is not the identity on all of

\mathbb R

. c) Because taking the square root gives two possible values

\pm\sqrt{x}

g

is not single-valued, so the composition cannot recover

x

. d) Because

f(g(x))=(\sqrt{x})^{2}=|x|

, the composition equals the absolute value of

x

, not

x

itself.

Intuitive explanation

When you compose a function with its "inverse," you expect to get back what you started with—but only if that inverse is actually a valid function on the same domain. Here, $g(x)=\sqrt{x}$ is only defined for $x \geq 0$ , so $f \circ g$ can only be computed there. More fundamentally, $f$ isn't one-to-one on all of $\mathbb{R}$ (both $2$ and $-2$ map to $4$ ), so swapping $x$ and $y$ doesn't produce a true function inverse. The composition works on $[0,\infty)$ , but fails on negative inputs where $g$ doesn't even exist.

Intuitive explanation

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