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# Intermediate Value Theorem - PowerPoint PPT Presentation

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 14 th and 15 th birthday, a boy went from 150 to 165 lbs. At some point he weighed 155lbs.

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## PowerPoint Slideshow about ' Intermediate Value Theorem' - hayden-dodson

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### Intermediate Value Theorem

2.4 cont.

• 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

Determine if f(x) has any real roots +2x-3 in the interval [0, 2]

f(x) is continuous on the interval

f(1)= - #

f(2)= + #

Since +2x-3 in the interval [0, 2]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.

Is any real number exactly one less than its cube?

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

Max-Min Theorem for Continuous Functions +2x-3 in the interval [0, 2]

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

Min at interior point and max at endpoint

Max and mins at endpoints

Why does the IVT fail to hold for f(x) on [-1, 1]? +2x-3 in the interval [0, 2]

Not Continuous in interval!

Point of discontinuity at x = 0

Show why a root exists in the given interval +2x-3 in the interval [0, 2]

Continuous in interval

f(-2)= -10

f(-1)= 1

Show why a root exists in the given interval +2x-3 in the interval [0, 2]

Continuous in interval

f(1)= 1