amy0799
  • amy0799
consider the following. f(x) = (x)/(x^2-2) Find the x-values 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
look for a value of x that makes the denominator zero, it won't be continuous there
Michele_Laino
  • Michele_Laino
hint: for continuity, we can not divide by zero
amy0799
  • amy0799
i'm confused on what to do...

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Michele_Laino
  • Michele_Laino
we have tosolve this equation, first: \[\Large {x^2} - 2 = 0\]
amy0799
  • amy0799
x=sqrt(2)
Michele_Laino
  • Michele_Laino
more precisely, we have two solutions: \[\Large x = \pm \sqrt 2 \]
Michele_Laino
  • Michele_Laino
am I right?
amy0799
  • amy0799
yea. so the smaller value is \[-\sqrt{2}\] and the larger value is \[\sqrt{2} \]?
Michele_Laino
  • Michele_Laino
Correct! 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
  • amy0799
right. so how do you find out if it's removable?
Michele_Laino
  • Michele_Laino
I think that both discontinuities are not removable
amy0799
  • amy0799
can you explain why?
Michele_Laino
  • Michele_Laino
yes! I explain with an example
Michele_Laino
  • Michele_Laino
for example, the subsequent function: \[\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}}\] is not continuous at x=1, right?
amy0799
  • amy0799
yes
Michele_Laino
  • Michele_Laino
nevertheless 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
  • amy0799
ooh ok. can you help me with another problem?
Michele_Laino
  • Michele_Laino
yes!
amy0799
  • amy0799
consider the following. f(x) = {-9x
Michele_Laino
  • Michele_Laino
ok!
amy0799
  • amy0799
f(x) = -9 if x less than or equal to 5 x^2-2x+2 if x greater than 5
amy0799
  • amy0799
find the x-value at which f is not continuous and is the discontinuity removable
Michele_Laino
  • Michele_Laino
at 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
amy0799
  • amy0799
is it not removable?
Michele_Laino
  • Michele_Laino
I think that it is not removable
Michele_Laino
  • Michele_Laino
because it is a jump-type discontinuity
amy0799
  • amy0799
ok thanks!. I have a few more I need help me, is that ok?
Michele_Laino
  • Michele_Laino
ok!
amy0799
  • amy0799
g(x) = 5sin(x)/x if x less than 0 a-4x 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
  • Michele_Laino
we can write this: \[\Large \mathop {\lim }\limits_{x \to 0} \frac{{5\sin x}}{x} = 5\]
Michele_Laino
  • Michele_Laino
whereas at x=0, we have: \[\Large a - 4x = a - 4 \cdot 0 = a\]
Michele_Laino
  • Michele_Laino
now, for continuity, we have to request taht both values are equal each to other. So what do you get?
amy0799
  • amy0799
Would the constant a be 5?
Michele_Laino
  • Michele_Laino
that*
Michele_Laino
  • Michele_Laino
correct! a=5
amy0799
  • amy0799
g(x) = (x^2-a^2)/(x-a) 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
  • Michele_Laino
we 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
  • Michele_Laino
and 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
  • amy0799
I don't understand how you got 2a
Michele_Laino
  • Michele_Laino
after a simplification, I got this: \[\Large g\left( x \right) = x + a\] so, \[\Large g\left( a \right) = a + a = 2a\]
amy0799
  • amy0799
oh ok. so the constant a is 11/2?
Michele_Laino
  • Michele_Laino
yes! correct!
amy0799
  • amy0799
Describe the interval(s) on which the function below is continuous. \[f(x)=x \sqrt{x+1}\]
Michele_Laino
  • Michele_Laino
your 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
  • amy0799
[0,∞) is that how you would write it?
Michele_Laino
  • Michele_Laino
we have: \[\Large x + 1 \geqslant 0 \Rightarrow x \geqslant - 1\] so, the requested interval is: \[\Large [ - 1,\infty )\] am I right?
amy0799
  • amy0799
yes. Thank you so much for the help and sorry if I took up your time.
Michele_Laino
  • Michele_Laino
no worries! :) :)

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