Animated Solution for Mathematics - Functions: The domain of the definition of the function $f(x)=\frac{1}{4-x^2}+{\log }_{10}(x^3-x)$ is :-

The function provided is $f(x)=\frac{1}{4-x^2}+{\log }_{10}(x^3-x)$. This function is a composition of a rational expression and a logarithmic term.

To determine the domain, we must identify the set of all $x$ values for which both components are simultaneously defined.

For the rational part, $f_1(x)=\frac{1}{4-x^2}$, we must ensure the denominator is non-zero to avoid division by zero.

This leads to the exclusion of two critical points: $x eq 2$ and $x eq -2$. These values represent the "no-go" zones where the function is undefined.

For the logarithmic part, $f_2(x)={\log }_{10}(x^3-x)$, the argument must be strictly positive. We set up the following inequality:

Factoring the expression, we obtain:

To solve the inequality $x(x-1)(x+1)>0$, we identify the critical points at $x=-1$, $x=0$, and $x=1$.

By applying the Wavy Curve Method, we test the intervals created by these points:

1. For $x>1$, the expression is positive. 2. For $0<x<1$, the expression is negative. 3. For $-1<x<0$, the expression is positive. 4. For $x<-1$, the expression is negative.

The solution to the inequality is the union of the intervals $(-1,0)\cup (1,\infty )$.

We must now intersect the logarithmic domain $(-1,0)\cup (1,\infty )$ with the rational constraints $x eq 2$ and $x eq -2$.

Note that $x=-2$ is already outside the logarithmic domain. However, $x=2$ lies within the interval $(1,\infty )$ and must be explicitly excluded.

By "punching a hole" at $x=2$, we arrive at the final domain:

Domain $=(-1,0)\cup (1,2)\cup (2,\infty )$