Respuesta :

CastleRook
 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:    

x= -5

Step-by-step explanation:

we are given with the function:

[tex]f(x)=\frac{x+5}{x^2+3x-10}[/tex]

We will factorize the denominator

[tex]x^2+3x-10[/tex]

[tex]x^2+5x-2x-10[/tex]

[tex]x(x+5)-2(x+5)[/tex]

[tex](x-2)(x+5)[/tex]

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.


Otras preguntas