Given that f(t) = 1 - (3)/(t) and the range is \-1, 3, 7\, find the domain.

Part b)
To find the domain $D$ such that the range of $f(x) = \frac{1}{x-2}$ is $[-1, 0)$.

Step 1: Solve $y = \frac{1}{x-2}$ for $x$.

Step 2: $f'(x) = -\frac{1}{(x-2)^2} < 0$, so $f$ is strictly decreasing on $(-\infty, 2)$ and on $(2, \infty)$.
On $(-\infty, 2)$, $f(x) < 0$: as $x \to -\infty$, $f(x) \to 0^-$; as $x \to 2^-$, $f(x) \to -\infty$.

Step 3: Solve $f(x) = -1$.

$f(1) = -1$.

Step 4: Since $f$ is strictly decreasing on $(-\infty, 2)$, on $(-\infty, 1]$, $f$ decreases from $0^-$ to $-1$.
Thus, $f((-\infty, 1]) = [-1, 0)$.

Domain: $(-\infty, 1]$
(-\infty,1]

Part c)
$f(x) = x^2 - 2x$. Show $f(x) \ge -1$ for all real $x$.

Step 1: Complete the square.

Step 2: $(x-1)^2 \ge 0$ for all real $x$.

Equality at $x=1$: $f(1) = -1$.

Alternatively, $f'(x) = 2x - 2 = 0$ gives $x=1$, $f''(x)=2>0$ so minimum $f(1)=-1$.

Thus, $f(x) \ge -1$ for all $x$.
f(x)\ge-1

Part d)
Given $a^n = b^n$, show $\left( \frac{a}{b} \right)^n = 1$ (assume $b \neq 0$).

Step 1: Divide both sides by $b^n$.

If $b=0$, then $a^n=0$ so $a=0$, but $\frac{a}{b}$ undefined.

**\left( \frac{a{b} \right)^{n} = 1}}