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amy0799
 one year ago
consider the following.
f(x) = (x)/(x^22)
Find the xvalues at which f is not continuous,
smaller value =
larger value =
Is the discontinuity at the smaller value removable?
Is the discontinuity at the larger value removable?
amy0799
 one year ago
consider the following. f(x) = (x)/(x^22) Find the xvalues at which f is not continuous, smaller value = larger value = Is the discontinuity at the smaller value removable? Is the discontinuity at the larger value removable?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0look for a value of x that makes the denominator zero, it won't be continuous there

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: for continuity, we can not divide by zero

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0i'm confused on what to do...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have tosolve this equation, first: \[\Large {x^2}  2 = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more precisely, we have two solutions: \[\Large x = \pm \sqrt 2 \]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0yea. so the smaller value is \[\sqrt{2}\] and the larger value is \[\sqrt{2} \]?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Correct! we have to exclude both values, since we can not divide by zero, so our function is not continuous at these points: \[\Large \begin{gathered} x = \sqrt 2 \hfill \\ x =  \sqrt 2 \hfill \\ \end{gathered} \]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0right. so how do you find out if it's removable?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that both discontinuities are not removable

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I explain with an example

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for example, the subsequent function: \[\Large f\left( x \right) = \frac{{{x^2}  1}}{{x  1}}\] is not continuous at x=1, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1nevertheless x=1 is a removable discontinuity, since we can write this: \[\Large f\left( x \right) = \frac{{{x^2}  1}}{{x  1}} = \frac{{\left( {x  1} \right)\left( {x + 1} \right)}}{{x  1}} = x + 1\]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0ooh ok. can you help me with another problem?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0consider the following. f(x) = {9x

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0f(x) = 9 if x less than or equal to 5 x^22x+2 if x greater than 5

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0find the xvalue at which f is not continuous and is the discontinuity removable

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1at x= 5, we have f(x)= 9 using your first part for f(x) and we have: \[{x^2}  2x + 2 = 25  2 \cdot 5 + 2 = 17 \ne 9\] so the discontinuity is at x=5

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that it is not removable

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1because it is a jumptype discontinuity

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks!. I have a few more I need help me, is that ok?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0g(x) = 5sin(x)/x if x less than 0 a4x if x greater than or less than 0 Find the constant a such that the function is continuous on the entire real line.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can write this: \[\Large \mathop {\lim }\limits_{x \to 0} \frac{{5\sin x}}{x} = 5\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1whereas at x=0, we have: \[\Large a  4x = a  4 \cdot 0 = a\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, for continuity, we have to request taht both values are equal each to other. So what do you get?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0Would the constant a be 5?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0g(x) = (x^2a^2)/(xa) if x is not equal to a 11 if x=a Find the constant a such that the function is continuous on the entire real line.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can write this: \[\Large \frac{{{x^2}  {a^2}}}{{x  a}} = \frac{{\left( {x  a} \right)\left( {x + a} \right)}}{{x  a}} = x + a\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and its value at x=a, is: \[\Large 2a\] Now, for continuity we have to request, as before, that those two values are equal each to other, so we have to solve this equation: \[\Large 2a = 11\]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand how you got 2a

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1after a simplification, I got this: \[\Large g\left( x \right) = x + a\] so, \[\Large g\left( a \right) = a + a = 2a\]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. so the constant a is 11/2?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0Describe the interval(s) on which the function below is continuous. \[f(x)=x \sqrt{x+1}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1your function is continuous inside all its domain, and its domain is the subset of all real numbers, such that the radicand is positive or equal to zero, namely: \[\Large x + 1 \geqslant 0\]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0[0,∞) is that how you would write it?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have: \[\Large x + 1 \geqslant 0 \Rightarrow x \geqslant  1\] so, the requested interval is: \[\Large [  1,\infty )\] am I right?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0yes. Thank you so much for the help and sorry if I took up your time.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no worries! :) :)
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