Animated Solution for Mathematics - Inverse Trigonometric Functions: The domain of the function $f(x)={\sin }^{-1}\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)+{\cos }^{-1}\left(\frac{x-1}{x+1}\right)$ is :

The function is defined as:

To find the domain, we must ensure that the arguments of both inverse trigonometric functions lie within the closed interval $[-1,1]$ and that the denominators are non-zero.

For the term ${\cos }^{-1}\left(\frac{x-1}{x+1}\right)$, the argument must satisfy $-1\le \frac{x-1}{x+1}\le 1$. We must also enforce the constraint $x eq -1$.

Solving the left inequality $\frac{x-1}{x+1}\ge -1$:

Using the Wavy Curve Method with critical points at $x=-1$ and $x=0$, we obtain $x\in (-\infty ,-1)\cup [0,\infty )$.

Solving the right inequality $\frac{x-1}{x+1}\le 1$:

This implies $x+1>0$, or $x>-1$. Intersecting these conditions, the domain for the ${\cos }^{-1}$ component is $x\in [0,\infty )$.

For the term ${\sin }^{-1}\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)$, we require $-1\le \frac{3x^2+x-1}{(x-1{)}^2}\le 1$, with $x eq 1$. Since $(x-1{)}^2>0$, we multiply by the denominator:

First, consider $-(x-1{)}^2\le 3x^2+x-1$:

This yields $x\in (-\infty ,0]\cup [\frac{1}{4},\infty )$.

Next, consider $3x^2+x-1\le (x-1{)}^2$:

This yields $x\in [-2,\frac{1}{2}]$. Intersecting these two results, the domain for the ${\sin }^{-1}$ component is $x\in [-2,0]\cup [\frac{1}{4},\frac{1}{2}]$.

To find the domain of the entire function, we intersect the domains obtained from both parts:

The intersection results in the set containing the point $0$ and the interval $[\frac{1}{4},\frac{1}{2}]$.

The final domain is: