Question

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

Answer 1

Answer:    

x= -5

Step-by-step explanation:

we are given with the function:

$f(x)=\frac{x+5}{x^2+3x-10}$

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.

Answer 2

 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