Intermediate Value Theorem
2.4 cont.
Examples

If between 7am and 2pm the temperature
went from 55 to 70.



At some time it reached 62.
Time is continuous
If between his 14th and 15th birthday, a boy
went from 150 to 165 lbs.


At some point he weighed 155lbs.
It may have occurred more than once.
Show that a “c” exists such that f(c)=2 for
f(c)=x^2 +2x-3 in the interval [0, 2]
f(x) is continuous on the interval
f(0)= -3
f(2)= 5
 by IVT  a " c"  f (c)  2 since - 3  2  5
Determine if f(x) has any real roots
f ( x)  x 2  x  1 in interval [1,2]
f(x) is continuous on the interval
f(1)= - #
f(2)= + #
by IVT  c  1,2  f (c)  0
Is any real number exactly one less than its cube?
(Note that this doesn’t ask what the number is, only if it exists.)
x  x3  1
f ( 0 )  1
0  x3  x  1
f 1  1
f  2  5
Since f is a continuous function, by the
intermediate value theorem it must
take on every value between -1 and 5.
Therefore there must be at least one
solution between 1 and 2.
f  x   x3  x  1
Use your calculator to find an approximate solution.
2
-5
5
-2
1.32472
Max-Min Theorem for Continuous
Functions
If f is continuous at every point of the closed
interval [a, b], then f takes on a minimum value m
and a maximum value M on [a, b]. That is, there
are numbers α and β in [a, b] such that f(α) = m,
f(β) = M, and m ≤ f(x) ≤ M at all points x in [a, b]
Max and mins at
interior points
Max and mins at
endpoints
Min at interior
point and max at
endpoint
Why does the IVT fail to hold for f(x) on
[-1, 1]?
 1  1  x  0
f ( x)  
0  x 1
1
Not Continuous in interval!
Point of discontinuity at x = 0
1
-2
2
-1
Show why a root exists in the given
interval
f ( x)  2 x  x  2
3
2
on [2,1]
Continuous in interval
f(-2)= -10
f(-1)= 1
by IVT  c   2,1  f (c)  0
Show why a root exists in the given
interval
1
 x 4  1 on
x
12 ,1
Continuous in interval
1
 15
f ( 1 )  1 2 
2 16
16
f(1)= 1
1 
 by IVT  c   ,1  f (c)  0
2 