Lesson 63 - Intermediate Value Theorem

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Presentation transcript:

Lesson 63 - Intermediate Value Theorem HL Math – Calculus

Intermediate Value Theorem Intermediate Value Theorem. Let f be a function which is continuous on the closed interval [a, b]. Suppose that d is a real number between f(a) and f(b); then there exists c in [a, b] such that f(c) = d.

Examples - Context 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.

Examples (1) Show that a “c” exists such that f(c)=2 for f(c)=x2 +2x-3 in the interval [0, 2] (2) Determine if f(x) has any real roots, where

Example 1 Show that a “c” exists such that f(c)=2 for f(c)=x2 +2x-3 in the interval [0, 2] First, f(x) is continuous Second f(0) = -3 Third, f(2) = 5 Thus, by the IVT, there exists a value, x = c, such that f(c) = 2 since -3 < 2 < 5

Example 2 Determine if f(x) has any real roots, where (1) f(x) is continuous on the interval [1,2] (2) f(1) = -ve # (3) f(2) = +ve #

Examples (3) Is any real number exactly one less than its cube? (4) Why does the IVT fail to hold for f(x) on [-1, 1]?

Examples (5) Show why a root exists in the given interval

Is any real number exactly one less than its cube? (Note that this doesn’t ask what the number is, only if it exists.) 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.

(4) Why does the IVT fail to hold for f(x) on [-1, 1]? Not Continuous in interval! Point of discontinuity at x = 0

(5) Show why a root exists in the given interval Continuous in interval f(-2)= -10 f(-1)= 1

(6) Show why a root exists in the given interval Continuous in interval f(1)= 1

Example 2 Consider the equation sin x = x – 2 . Use the intermediate Value Theorem to explain why there must be a solution between π/2 and π.

Example 3 Consider the function , Calculate f(6), f(-5.5), f(0) Can you conclude that there must be a zero between f(6) and f(-5.5)?