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Continuity
June 26th, 2012
Vocab
Continuous Discontinuous
Removable Discontinuity
Infinite Discontinuity
Jump Discontinuity Continuous from the left/right
Continuous on an interval
The intermediate value theorem
Concept Check Questions
1. Determine for whichx the function is continuous. Classify the
discontinuities that occur.
(a) f(x) = x
|x|
f(x) = x
|x|=
1 x >01 x 0Ifa 0 we have limxaf(x) = limxacos(x) = cos(a).Ifa
= 0 we have limx0+f(x) = limx0+cos(x) = 1.Ifa = 0 we have limx0f(x)
= limx0x
2 + 1 = 1.so f(x) is continuous for all real numbers.
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(e) f(x) =
x2 x 0ex x >0
Ifa 0 we have limxaf(x) = limxaex
=ea
.Ifa = 0 we have limx0+f(x) = limx0+x
2 = 0.Ifa = 0 we have limx0f(x) = limx0e
x = 1.so f(x) is continuous for x = 0 , at 0 f(x) has a jump
discontinuity.
(f) f(x) =
x2 x 01/x x >0
Ifa 0 we have limxaf(x) = limxa1/x= 1/a.
Ifa = 0 we have limx0+
f(x) = limx0+
1/x= .Ifa = 0 we have limx0f(x) = limx0x2 = 0.
so f(x) is continuous for x = 0 , at 0 f(x) has an infinite
discontinuity.
(g) f(x) = tan(x)
f(x) = sin(x)cos(x)
tan(x) has infinite discontinuities whenever cos(x) = 0 ie at
2
+n .
2. Show that there is an xthat solves the following
equations.
(a) sin(x) = 1Take x = /2. sin(/2) = 1, so such an xexists.
(b) x3 +x+ 1 = 0
The equation is already in the form function = value so we can
apply the intermediatevalue theorem.x3 +x+ 1 is a polynomail so it
is continuous on (, ). (1)3 + (1) + 1 = 1 and(0)3 + 0 + 1 = 1 so by
the intermediate value theorem x3 +x+ 1 has a root on in
theinterval (1, 0).
(c) cos(x) =x2
We rewrite the equation as in the form function = value so that
we can apply theintermediate value theorem.cos(x) x2 = 0cos(x) is a
trig function whose domain is all real numbers (and hence
continuous) and
x2 is a polynomail so their difference is continuous on (, ).
cos(0) 02 = 1 andcos(/2) (/2)2 = 2/4 so by the intermediate value
theorem cos(x) x2 has a rooton in the interval (0, /2).
(d) sin(x) =xsin(0) = 0, so x = 0 is a solution. Alternately, we
rewrite the equation as in the formfunction = value so that we can
apply the intermediate value theorem.sin(x) x= 0sin(x) is a trig
function whose domain is all real numbers (and hence continuous)
and xis continous everywhere so their difference is continuous on
(, ). sin(/2) /2 =
2
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1 /2 .5 and sin(/2) +/2 = 1 +/2 .5 so by the intermediate
valuetheorem sin(x) x has a root on in the interval (/2, /2).
(e) tan(x+ 2) = xtan(x) is continuous on (/2, /2) and similarly
on any interval of the form (/2 +n,/2 +n) so tan(x+ 2) is continous
on (/2 2, /2 2) and similarly on anyinterval of the form (/2 +n 2,
/2 +n 2), since x is continuous everywheretan(x+ 2) x is continuous
on the same intervals.Tangent takes the value of all real numbers
on each of these intervals so we expect eachto contain a solution.
Well consider the interval (/2 2, 3/2 2) (.4, 2.7).tan(+ 2 + 2) (2
+) = 2 0 so by the intermediate value theorem there is rooton (2 +
, 2.5) (note this isnt a question Id ask on an exam because it more
or lessrequires a calculator to finish.The other choice is to use
since limx3/22tan(x+ 2) x = there has to be somepoint where the
function is positive, which we can use to finish the intermediate
value
theorem arguement.
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