Answer
f(x) is continuous on (-$\infty$,-1)U(-1,$\infty$), meaning it is discontinuous at x=-1. It is continuous from the right at x=-1.
Work Step by Step
1. Check Continuity of $x^{2}$:
Polynomials are continuous everywhere, so our function is continuous on (-$\infty$,-1)
2. Check Continuity of x:
Continuous everywhere, so our function is continuous on (-1,1)
3. Check Continuity of $\frac{1}{x}$:
Rational functions are continuous where they are defined. This is undefined at 0. However, we are only looking at the interval (1,$\infty$) which it is continuous on.
So far, we know our function is continuous on (-$\infty$,-1)U(-1,1)U(1,$\infty$).
4. Check for continuity at x=-1
$lim_{x->-1}$f(x) does not exist because $lim_{x->-1^{-}}$f(x)$\ne$$lim_{x->-1^{+}}$f(x)
1$\ne$-1
It is discontinuous here.
However, $lim_{x->-1^{+}}$f(x)=f(-1)=-1 so it is continuous from the right.
5. Check for continuity at x=1
$lim_{x->1^{-}}$f(x)=$lim_{x->1^{+}}$f(x)=f(1)
1=1=1
So f(x) is continuous at 1.
6. Putting it all together:
f(x) is continuous on (-$\infty$,-1)U(-1,$\infty$), meaning it is discontinuous at x=-1. It is continuous from the right at x=-1.
