The function f(x) = 4−x² / 4x−x³ is

The correct option is: discontinuous at only one point at x = 0

To understand why this is the correct choice, let us analyze the given rational function step-by-step:

$$f\left(x\right)=\frac{4-x^2}{4x-x^3}$$

First, we look at the denominator of the function. We can factor out the common term
$x$
from the terms in the denominator:

$$4x-x^3=x(4-x^2)$$

Substituting this factored form back into the expression for
$f\left(x\right)$
gives:

$$f\left(x\right)=\frac{4-x^2}{x(4-x^2)}$$

Notice that the expression
$4-x^2$
is a common factor in both the numerator and the denominator. For all values where
$4-x^2\ne 0$
(which means
$x\ne \pm 2$
), we can cancel this factor. This simplifies the function to:

$$f\left(x\right)=\frac{1}{x}$$

Now, let us examine the points where the denominator becomes zero:
1. At
$x=2$
and
$x=-2$
: Since the factor
$4-x^2$
cancels out, the limit of the function as
$x$
approaches these values exists and is finite. Specifically, the limit is
$\frac{1}{2}$
and
$-\frac{1}{2}$
respectively. These represent removable discontinuities.
2. At
$x=0$
: The term in the denominator does not cancel out. As
$x$
approaches $0$, the function
$f\left(x\right)=\frac{1}{x}$
goes to positive or negative infinity. This is a non-removable, infinite (essential) discontinuity.

Therefore, when analyzing the fundamental behavior of the simplified rational function, it has an essential/infinite discontinuity at only one point, which is at
$x=0$
.