Which values of x are point(s) of discontinuity for this function? Function x = –4 x = –2 x = 0 x = 2 x = 4

Mathematics

2 answers:

o-na [289]3 years ago

7 0

Answer:

x=0 and x=2

Step-by-step explanation:

We need to check at each point where the function changes definition

At x= -2

On the left side -4 on the right side = -( -2)^2 = -4 continuous

At x=0

The point is not defined since neither side has an equals sign

discontinuous

x =2

on the left side 2^2 =4 on the right side 2

It is discontinuous

poizon [28]3 years ago

3 0

Answer:

x = 0

x = 2

Step-by-step explanation:

Edge 2020

~theLocoCoco

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Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the quotient 4-x/x^2+5x-6 divided by x^2-11x+28/x^2+7x+6 in simplified form?

sashaice [31]

$\dfrac{4-x}{x^2+5x-6}:\dfrac{x^2-11x+28}{x^2+7x+6}=(*)$

$4-x=-(x-4)\\\\x^2+5x-6=x^2+6x-x-6=x(x+6)-1(x+6)=(x+6)(x-1)\\\\x^2-11x+28=x^2-7x-4x+28=x(x-7)-4(x-7)\\=(x-7)(x-4)\\\\x^2+7x+6=x^2+6x+x+6=x(x+6)+1(x+6)=(x+6)(x+1)\\\\(*)=\dfrac{-(x-4)}{(x+6)(x-1)}:\dfrac{(x-7)(x-4)}{(x+6)(x+1)}$
$=\dfrac{-(x-4)}{(x+6)(x-1)}\cdot\dfrac{(x+6)(x+1)}{(x-7)(x-4)}=\dfrac{-1}{x-1}\cdot\dfrac{x+1}{x-7}\\\\=-\dfrac{x+1}{x^2-7x-x+7}=-\dfrac{x+1}{x^2-8x+7}$