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?

- amy0799

- jamiebookeater

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- anonymous

look for a value of x that makes the denominator zero, it won't be continuous there

- Michele_Laino

hint:
for continuity, we can not divide by zero

- amy0799

i'm confused on what to do...

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## More answers

- Michele_Laino

we have tosolve this equation, first:
\[\Large {x^2} - 2 = 0\]

- amy0799

x=sqrt(2)

- Michele_Laino

more precisely, we have two solutions:
\[\Large x = \pm \sqrt 2 \]

- Michele_Laino

am I right?

- amy0799

yea. so the smaller value is \[-\sqrt{2}\] and the larger value is \[\sqrt{2} \]?

- 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

right. so how do you find out if it's removable?

- Michele_Laino

I think that both discontinuities are not removable

- amy0799

can you explain why?

- Michele_Laino

yes! I explain with an example

- 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

yes

- 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

ooh ok. can you help me with another problem?

- Michele_Laino

yes!

- amy0799

consider the following.
f(x) = {-9x

- Michele_Laino

ok!

- amy0799

f(x) = -9 if x less than or equal to 5
x^2-2x+2 if x greater than 5

- amy0799

find the x-value at which f is not continuous and is the discontinuity removable

- 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

is it not removable?

- Michele_Laino

I think that it is not removable

- Michele_Laino

because it is a jump-type discontinuity

- amy0799

ok thanks!. I have a few more I need help me, is that ok?

- Michele_Laino

ok!

- 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

we can write this:
\[\Large \mathop {\lim }\limits_{x \to 0} \frac{{5\sin x}}{x} = 5\]

- Michele_Laino

whereas at x=0, we have:
\[\Large a - 4x = a - 4 \cdot 0 = a\]

- Michele_Laino

now, for continuity, we have to request taht both values are equal each to other.
So what do you get?

- amy0799

Would the constant a be 5?

- Michele_Laino

that*

- Michele_Laino

correct! a=5

- 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

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

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

I don't understand how you got 2a

- 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

oh ok. so the constant a is 11/2?

- Michele_Laino

yes! correct!

- amy0799

Describe the interval(s) on which the function below is continuous.
\[f(x)=x \sqrt{x+1}\]

- 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

[0,∞)
is that how you would write it?

- 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

yes. Thank you so much for the help and sorry if I took up your time.

- Michele_Laino

no worries! :) :)

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