explain why the function is discontinuous at the given point ( f(x) = { x^2-2x-8/ x-4 if x isnt equal to 4 and 3 if x=4 .. ( is it because there is a hole at 4 ?)
\[f(4)=3\] but \[\lim_{x\to 4}\frac{x^2-2x-8}{x-4}=6\]
satellite help me amistre could not answer my question
it isn't equal to anything at 4 because you will have \[4-4=0\] in the denominator
she has a piecewise function
satellite help me amistre could not answer my question
satellite help me amistre could not answer my question
it is true you can rewrite by factoring and canceling and gettin \[f(x)=x+2\] but that is only if \[x\neq4\]
oh i should learn to read hold on
yeah it is x\[\neq4\] lol
\[f(x) = \left\{\begin{array}{rcc} \frac{x^2-2x-8}{ x-4} & \text{if} & x \neq 4 \\ 3& \text{if} & x = 4 \end{array} \right. \]
yess
ooooh ok
so \[\frac{x^2-2x-8}{x-4}=\frac{(x-4)(x+2)}{x-4}=x+2\] which holds if \[x\neq 4\]
so this function, even though it looks like a rational function, is just the line \[y=x+2\] with a hole in it where (4,6) should be right?
satellite help me after this???
sure hold on
so i was right it was just the hole..
meaning that if you wanted to "remove" the discontinuity, you would just call it \[f(x)=x+2\] and be done
or if you want to spend a lot of ink you could write \[f(x) = \left\{\begin{array}{rcc} \frac{x^2-2x-8}{ x-4} & \text{if} & x \neq 4 \\ 6 & \text{if} & x = 4 \end{array} \right.\]
because the function really wants to be 6 when x = 4. but some idiot decided to call it 3 there, so it is not continuous
lol the idiot would be my math book -_-
and yes, you are right, it is a "hole" which is sometimes called a 'removable discontinuity" because if you define the function correctly you an "remove" it
yeah. ahh ok one last question my next problems involve using theorems (4,5,7, and 9) im not quite sure what that is.. for some functions.. x_x is it something my hw didnt specify?
come
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