Question

Please help.

Answer 1

a. 4
b. -4
c. 8
d. -8

Answer 2

@DanJS @Hero @PRAETORIAN.10 @HELP!!!! @HelpBlahBlahBlah @quickstudent @TheSmartOne

Answer 3

*x^2

Answer 4

are you in calculus, is that supposed to be f prime of -4?

Answer 5

\[f(x) = \frac{ x^2-16 }{ x+4 }\]

Answer 6

no f primes

Answer 7

s continuous at x = -4, find f(-4).

Answer 8

is

Answer 9

since they tell you it is continuous at -4, first factor the top

Answer 10

(x+4)(x-4)

Answer 11

\[(a^2 - b^2)^2 = (a+b)(a-b)\]

Answer 12

\[f(x) = \frac{ x^2-16 }{ x+4 } = \frac{ (x+4)(x-4) }{ x+4 } = x-4\]

Answer 13

right

Answer 14

but its telling to sub -4 in but that gives 0 and thts not an answer

Answer 15

@DanJS

Answer 16

f(x) = x - 4

f(-4) = -4 - 4 = - 8

Answer 17

ohhh

Answer 18

x=-4 would be undefined , extraneous, but they say it is continuous, there really is a hole in the graph at x=-4 for
f(x) = (x^2-16)/(x+4)

Answer 19

i forgot to cross it out...

Answer 20

If you just plugged in -4 you would get 0/0 zero over zero, that is nonsense, not a solution

Answer 21

Sorry

Answer 22

a number divided by 0 is undefined

Answer 23

Thanks. Can you check another problem for me; I'm a little confused

Answer 24

If f(x) is discontinuous, determine the reason.

a. f(x) is continuous for all real numbers
b. The limit as x approaches 1 does not exist
c. f(1) does not equal the limit as x approaches 1
d. f(1) is not defined

Answer 25

k thanks, let me find something for you , 1 second

Answer 26

Okay. :)

Answer 27

Here is a pretty straight forward thing i just found.
http://www.math.brown.edu/UTRA/discontinuities.html

Answer 28

I actually looked it up and I've seen that page, and it helped me cross out a and d but idk whether or not the answer is b or c

Answer 29

f(1) is defined, if x=1 f(1) = 1^2+4 = 5

Answer 30

what course are you taking?

Answer 31

calc

Answer 32

yes but the two seperate sides arent equivalent

Answer 33

so it dne?

Answer 34

ok
If you take the limit from both sides of a certain X value you get
\[\lim_{x \rightarrow 1^-}=5\]\[\lim_{x \rightarrow 1^+}=4\]

Answer 35

right i did that

Answer 36

since the Left and Right handed limits arent the same
\[\lim_{x \rightarrow 1}= does~ no t~exist\]

Answer 37

yes so b is the answer

Answer 38

The limit as x approaches 1 does not exist?

Answer 39

right, read the first 2 sections on removable and jump discontinuites here:
http://en.wikipedia.org/wiki/Classification_of_discontinuities

Answer 40

Read the sections, was helpful proved that the answer was indeed b!

Answer 41

yep

Answer 42

Thank you so much for all the help!

Answer 43

:)

Answer 44

there is a jump in the graph there, if you graph both those you can see