Question

Where is the removable discontinuity of f(x)= x+5/x^2+3x-10 located?

x = –5
x = –2
x = 2
x = 5

Answer 1

get the point of discontinuity we proceed as follows;
f(x)=x+5/x^2+3x-10
f(x)=4x+5x^2-10

this can be written is such a way that they have the same denominator, here we shall have:
f(x)=(4x^3-10x^2+5)/x^2
The denominator= x^2
The numerator=4x^3-10x^2+5

The discontinuity is at the point x=0

the removable discontinuity is the the point x=2

Answer 2

x= -5

Explanation:

we are given with the function:

`f(x)=(x+5)/(x^2+3x-10)`

We will factorize the denominator

`x^2+3x-10`

`x^2+5x-2x-10`

`x(x+5)-2(x+5)`

`(x-2)(x+5)`

Hence, We can see that (x+5) can be eliminated since, it can get cancelled with the numerator

Hence, the removable discontinuity is at (x+5) or x= -5

Removable discontinuity is that which can be eliminated from the function.