Solution

A quadratic is not one-to-one on all $\mathbb{R}$, so we restrict the domain (given in each part).

Steps:

1. Let $y=f(x)$.
2. Interchange $x$ and $y$.
3. Solve for $y$.
4. Choose the correct branch using the given domain restriction.

(i) $f(x)=x^2-3,\; x\le 0$

Let $y=x^2-3$.

Interchange $x$ and $y$:

Since original domain is $x\le 0$, the inverse must output $\le 0$, so choose the negative branch:

Domain of $f$: $(-\infty,0]$.

Range of $f$: $[-3,\infty)$.

So:

(ii) $f(x)=x^2+6x+4,\; x<-3$

Complete square:

Let $y=(x+3)^2-5$.

Interchange $x$ and $y$:

Given $x<-3$ for the original function, we have $x+3<0$, so choose the negative branch:

Range of $f$ for $x<-3$ is $(-5,\infty)$ (minimum $-5$ not included).

So:

(iii) $f(x)=2x^2-8x+11,\; x\ge 2$

Complete square:

Let $y=2(x-2)^2+3$.

Interchange $x$ and $y$:

Since original domain is $x\ge 2$ and the vertex is at $x=2$, we take the increasing branch $y\ge 2$:

Range of $f$ is $[3,\infty)$, so:

(iv) $f(x)=3x^2-2x+6,\; x\ge 5$

Complete square:

Let $y=3\left(x-\frac{1}{3}\right)^2+\frac{17}{3}$.

Interchange $x$ and $y$:

Given original domain $x\ge 5$, we choose the branch that gives $y\ge 5$:

Range of $f$ for $x\ge 5$ starts at $f(5)=71$, so:

(v) $f(x)=2(x-3)^2+1,\; x\ge 3$

Let $y=2(x-3)^2+1$.

Interchange $x$ and $y$:

Given $x\ge 3$ for the original function, take $y\ge 3$:

Range of $f$ is $[1,\infty)$, so:

(vi) $f(x)=-3(x+4)^2-5,\; x<-4$

Let $y=-3(x+4)^2-5$.

Interchange $x$ and $y$:

Given original domain $x<-4$ so $(x+4)<0$, choose the negative branch:

Range of $f$ for $x<-4$ is $(-\infty,-5)$.

So: