Question

Can someone please help me find the removable discontinuities of d(x)=(x^2-12x+20)/3x?

Answer 1

there is none.

Answer 2

3x = 0
x = 0

a removable discontinuity is when the denominator and numerator is 0/0

Answer 3

oh okay, can you help me find the removable discontinuities of
1. z(x)=(x^2-7x-8)/x^3+64
2. f(x)=(2x^3-3x+1)/x^3-5x+7
3. y=(x^2+4x-5)/x^2+8x+15

? :) @shamil98

Answer 4

set the denominators to equal zero and solve for x.

Answer 5

or just use wolframalpha .lol

Answer 6

Well this is what I got,
1. -4, 2+2i√3, 2-2i√3
2. -2.74735
3. -5, -3

are those right?

Answer 7

on wolframalpha how do I find the removable discontinuities? I typed in the whole function z(x)=(x^2-7x-8)/x^3+64 and it graphed it, do i type in just the denominator?

Answer 8

x^3 = -64
x= -4
so the domain is all real numbers except -4.

Answer 9

so the removable discontinuities are only the real numbers and not the complex ones?

Answer 10

the first one has no removable discontinuity, only a point of discontinuity..
remember removable discontinuity = 0/0

Answer 11

i'm not entirely sure, i'm taking pre-calc as well lol. I covered this last chapter but i'm not entirely sure...

Answer 12

so with the answers I got by solving, since none of them were 0, none of them have a removable discontunity?

Answer 13

that was redundant sorry, i'm tired lol.

Answer 14

hahah its okay, i completely understand :p i've been trying to do this all day D:

Answer 15

removable discontinuity makes the numerator and denominator 0/0. If the numerator and denominator do not equal 0/0 then it doesn't have one i think..

Answer 16

thank you :)

Answer 17

factor out your numerator and denominator. If the numerator and denominator functions have a similar factor, then there is a removable discontinuity there.

For example, your 2nd equation is this:\[z(x)=\frac{ x^2-7x-8 }{ x^3+64}\] after we factor them out, we get this:\[z(x)=\frac{ (x-8)(x+1) }{ (x+4)(x^2-4x+16)}\]as we can see, there is no similar factor in the numerator or denominator, so there is no removable discontinuity there.

However, for your last equation:\[y(x)=\frac{x^2+4x-5}{ x^2+8x+15 }\]There is a removable discontinuity. Just factor it out and you'll see one.