Asked by tony
f(x)= |x−8| / x+8
Is the function is not continuous at x= -8,8 or just at -8.
Is the function is not continuous at x= -8,8 or just at -8.
Answers
Answered by
Jai
f(x)= |x−8| / (x+8)
The function is discontinuous only at x = 8 since it will yield a denominator equal to zero:
f(8) = |8-(-8)| /-8+8
f(8) = 16/0
x = 8 is an asymptotic discontinuity.
Hope this helps~ :3
The function is discontinuous only at x = 8 since it will yield a denominator equal to zero:
f(8) = |8-(-8)| /-8+8
f(8) = 16/0
x = 8 is an asymptotic discontinuity.
Hope this helps~ :3
Answered by
Jai
*oops sorry, I mean x = -8.
Answered by
tony
So not continuous at x=-8 and function is not defined right.
Answered by
Jai
Yep, not continuous at x = -8.
If you graph the function, it approaches (+) and (-) infinity at x = -8.
To prove, we get the lim as x -> (-8)+ (approaching -8 from the right side) and as x -> (-8)- (approaching -8 from the left side)
lim |x−8| / (x+8) as x -> (-8)+
What we do here is get a value close to -8 but greater than -8, for instance -7.99:
= |-7.99−8| / -7.99+8
= |-15.99| / 0.01
= 1599 (which is a large number and (+))
lim |x−8| / (x+8) as x -> (-8)-
What we do here is get a value close to -8 but less than -8, for instance -8.01:
= |-8.01−8| / -8.01+8
= |-16.01| / -0.01
= -1601 (which is numerically large but (-))
Hope this helps~ :3
If you graph the function, it approaches (+) and (-) infinity at x = -8.
To prove, we get the lim as x -> (-8)+ (approaching -8 from the right side) and as x -> (-8)- (approaching -8 from the left side)
lim |x−8| / (x+8) as x -> (-8)+
What we do here is get a value close to -8 but greater than -8, for instance -7.99:
= |-7.99−8| / -7.99+8
= |-15.99| / 0.01
= 1599 (which is a large number and (+))
lim |x−8| / (x+8) as x -> (-8)-
What we do here is get a value close to -8 but less than -8, for instance -8.01:
= |-8.01−8| / -8.01+8
= |-16.01| / -0.01
= -1601 (which is numerically large but (-))
Hope this helps~ :3
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