Chapter 9.3, Problem 67PPSE

Solution

To find: The points of discontinuity of the function, $y=\frac{x(2x-1)(x+1)}{(x+5)(x+1)}$ .

The points of discontinuity are $x=-5$ and $x=-1$ .

Given information:

The function, $y=\frac{x(2x-1)(x+1)}{(x+5)(x+1)}$ .

Calculation:

The function $y=\frac{x(2x-1)(x+1)}{(x+5)(x+1)}$ has a common factor $x+1$ in the numerator and denominator. So, the function has a hole at the point $x=-1$ . That means, the function is discontinuous at this point.

The function can be simplified to $y=\frac{x(2x-1)}{(x+5)}$ by cancelling the common factor.

Observe the function is undefined at points when the denominator is zero. That means, the function is defined when $x+5=0$ or $x=-5$ . So, the function is discontinuous at $x=-5$ as well.

There is no other restriction on the function. Therefore, the function is discontinuous at the points $x=-5$ and $x=-1$ .

Conclusion:

The points of discontinuity are $x=-5$ and $x=-1$ .