Animated Solution for Mathematics - Functions: The domain of the function $f(x)=\frac{{\cos }^{-1}\left(\frac{x^2-5x+6}{x^2-9}\right)}{{\log }_e(x^2-3x+2)}$ is

To find the domain of the function $f(x)=\frac{{\cos }^{-1}\left(\frac{x^2-5x+6}{x^2-9}\right)}{{\log }_e(x^2-3x+2)}$, we must satisfy three primary mathematical constraints:

1. The argument of the inverse cosine function must lie in the interval $[-1,1]$. 2. The argument of the logarithm must be strictly positive. 3. The denominator must not equal zero, meaning the logarithm cannot be zero.

The argument of the inverse cosine is $g(x)=\frac{x^2-5x+6}{x^2-9}$. Factoring the expression, we get:

We must immediately note that $x eq 3$ and $x eq -3$ to avoid division by zero. Simplifying the expression for $x eq 3$, we obtain $g(x)=\frac{x-2}{x+3}$.

We require $-1\le \frac{x-2}{x+3}\le 1$. Solving $\frac{x-2}{x+3}\le 1$:

This inequality holds when $x+3>0$, or $x>-3$.

Solving $\frac{x-2}{x+3}\ge -1$:

Using the Wavy Curve Method with critical points $x=-3$ and $x=-1/2$, the solution is $x\in (-\infty ,-3)\cup [-1/2,\infty )$. Intersecting these conditions, the numerator is valid for $x\in [-1/2,\infty )\setminus \{3\}$.

The logarithm ${\log }_e(x^2-3x+2)$ requires $x^2-3x+2>0$. Factoring gives $(x-1)(x-2)>0$, which implies $x\in (-\infty ,1)\cup (2,\infty )$.

Additionally, the denominator cannot be zero, so $\log_e(x^2-3x+2) eq 0$. This implies:

Using the quadratic formula, we find the roots to be $x=\frac{3\pm \sqrt{5}}{2}$. These values must be excluded from the domain.

We intersect the valid regions: 1. Numerator: $x\in [-1/2,\infty )\setminus \{3\}$ 2. Denominator: $x\in (-\infty ,1)\cup (2,\infty )$ 3. Exclusions: $x eq \frac{3 \pm \sqrt{5}}{2}$

The intersection of the numerator and denominator regions is $[-\frac{1}{2},1)\cup (2,3)\cup (3,\infty )$. Removing the points $\frac{3-\sqrt{5}}{2}\approx 0.38$ and $\frac{3+\sqrt{5}}{2}\approx 2.62$, we arrive at the final domain:

Domain $=