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

523- since2)("" IVTby cfca

Determine if f(x) has any real roots [1,2] intervalin 1)( 2 xxxf

f(x) is continuous on the interval

f(1)= - #

f(2)= + #

0)(2,1 IVTby cfc

Is any real number exactly one less than its cube?

(Note that this doesn’t ask what the number is, only if it exists.)

3 1x x

30 1x x

3 1f x x x

1 1f 2 5f

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.

Use your calculator to find an approximate solution.

1.32472

1)0( f

2

-2

-5 5

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]?

101

011)(

x

xxf

Not Continuous in interval!

Point of discontinuity at x = 0

1

-1

-2 2

Show why a root exists in the given interval

]1,2[22)( 23 onxxxf

Continuous in interval

f(-2)= -10

f(-1)= 1

0)(1,2 IVTby cfc

Show why a root exists in the given interval

1,211

1 4 onxx

Continuous in interval

f(1)= 1

0)(1,2

1 IVTby

cfc

16

1521

16

1)2

1(

f