Answer to Question #145764 in Algebra for liam donohue

Question

A function f is said to have a removable discontinuity at a if:

1. f is either not defined or not continuous at a.

2. f(a) could either be defined or redefined so that the new function is continuous at a.

Let f(x)=(2x2+5x−7)/(5x−5).

Show that f has a removable discontinuity at 5 and determine the value for f(5) that would make f continuous at 5.

Need to redefine f(5)=

Answer 1

Note : Most likely you have specified the function WRONGLY since

$$f(5)=\frac{2\cdot5^2+5\cdot5-7}{5\cdot5-5}=\frac{50+25-7}{25-5}=\frac{68}{20}=3.4$$

At the specified point $x_0=5$ , the specific function $y=f(x)$ has NO discontinuity.

But this function (if you don't change your condition) has a discontinuity (removable) at the point $x_0=1$ .

Therefore, first I will show how to eliminate the discontinuity of a function at a point $x_0=1$ , and then I will change the condition so that the discontinuity is at a point $x_0=5$ and try to investigate it.

1 case : the function does not change, a discontinuity under investigation changes $x_0=1$ .

By the definition, the function has a removable discontinuity of the first kind at the point $x=x_0$ if the condition

$$f(x_0-0)=f(x_0+0),\,\,\,\text{where}\\[0.3cm] f(x_0-0)=\lim\limits_{x\to x_0-0}f(x)\\[0.3cm] f(x_0+0)=\lim\limits_{x\to x_0+0}f(x)$$

( more information : https://en.wikipedia.org/wiki/One-sided_limit )

Then,

$$f(1-0)=\lim\limits_{x\to 1-0}f(x)=\lim\limits_{x\to 1-0}\frac{2x^2+5x-7}{5x-5}=\\[0.3cm] =\lim\limits_{x\to 1-0}\frac{2x^2-2x+7x-7}{5(x-1)}=\lim\limits_{x\to 1-0}\frac{2x(x-1)+7(x-1)}{5(x-1)}=\\[0.3cm] =\lim\limits_{x\to 1-0}\frac{(x-1)(2x+7)}{5(x-1)}=\lim\limits_{x\to 1-0}\frac{2x+7}{5}=\frac{9}{5}=1.4\\[0.3cm] f(1+0)=\lim\limits_{x\to 1+0}f(x)=\lim\limits_{x\to 1+0}\frac{2x^2+5x-7}{5x-5}=\\[0.3cm] =\lim\limits_{x\to 1+0}\frac{2x^2-2x+7x-7}{5(x-1)}=\lim\limits_{x\to 1+0}\frac{2x(x-1)+7(x-1)}{5(x-1)}=\\[0.3cm] =\lim\limits_{x\to 1+0}\frac{(x-1)(2x+7)}{5(x-1)}=\lim\limits_{x\to 1+0}\frac{2x+7}{5}=\frac{9}{5}=1.4\\[0.3cm] \boxed{f(1-0)=1.4=f(1+0)}$$

Conclusion,

If we define the function as follows, then the given function $y=f(x)$ will be continuous at the point $x_0=1$ :

$$f(x)=\left\{\begin{array}{l} \displaystyle\frac{2x^2+5x-7}{5x-5},\,\,\,x\neq1\\[0.3cm] \displaystyle\frac{9}{5},\,\,\,x=1 \end{array}\right.$$

2 case : the function changes, but the exploration point $x_0=5$ does not.

New function is

$$f_{new}=\frac{2x^2+5x-7}{5(x-5)}$$

Then,

$$f(5-0)=\lim\limits_{x\to 5-0}f(x)=\lim\limits_{x\to 5-0}\frac{2x^2+5x-7}{5(x-5)}=\\[0.3cm] =\lim\limits_{x\to 5-0}\frac{(x-1)(2x+7)}{5(x-5)}=-\infty\\[0.3cm] f(5+0)=\lim\limits_{x\to 5+0}f(x)=\lim\limits_{x\to 5+0}\frac{2x^2+5x-7}{5(x-5)}=\\[0.3cm] =\lim\limits_{x\to 5+0}\frac{(x-1)(2x+7)}{5(x-5)}=+\infty\\[0.3cm] \boxed{f(5-0)=-\infty\neq+\infty=f(5+0)}$$

Conclusion,

The function has an inremovable discontinuity, a discontinuity of the second kind at the point $x_0$ .

Note : show additional confirmation of our calculations in the form of a graph of the specified function