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

x = –5
x = –2
x = 2
x = 5

User Wsbaser
by
8.1k points

2 Answers

7 votes
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

User Adam Neal
by
7.2k points
3 votes

Answer:

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.


User Helper
by
7.6k points
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