MATH241 Lecture Notes - Lecture 24: Intermediate Value Theorem
9 May 2018
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Department
Course
Professor
MATH241 - Lecture 24 - Maximum and Minimum Values, and the Mean Value Theorem
4.1: Maximum and Minimum Values (conclusion)
Example
:
Find the extreme values of f
on x in xf (x) = − 2
1+s0, 2π
[ ]
Solution:
os xf′(x) = − 2
1+c
→ (find the critical numbers)os x−2
1+c= 0
→ os xc =2
1
→ OR x=3
πx=3
5π
in 0f(0) = 0 + s= 0
in f(3
π)= − 2
1(3
π)+s(3
π)
.5 .9 .4= − 6
π+2
√3 ≈ − 0 + 0 = 0
in f(3
5π )= − 2
1(3
5π )+s(3
5π )
.6 .5 = − 6
5π + −
(2
√3 )≈ − 2 + (− .9)0 = − 3
in 2π .1f(2π) = − 2
1(2π) + s= − π ≈ − 3
4.2: The Mean Value Theorem
Rolle’s Theorem: If is continuous on and differentiable on , and ,f a,b
[ ](a,b) f(a) = f(b)
then there exists a number in such that c(a,b) f′(c) = 0
Example
:
Show that the polynomial has at most one root in the interval xx3+ 4 + 1 − , 2
[2 ]
Solution:
Let xf (x) = x3+ 4 + 1
5f(− )2 = − 8 − 8 + 1 = − 1
7f(2) = 8 + 8 + 1 = 1
Document Summary
Math241 - lecture 24 - maximum and minimum values, and the mean value theorem. Find the extreme values of f f (x) = 2. Os x c x = 3 (find the critical numbers) os x. Or f (0) = 0 + s f ( 3. 3 0 + 0 = 0. Rolle"s theorem: if then there exists a number is continuous on (a, b) in c f a, b and differentiable on (a, b) , and f (a) = f (b) such that f (c) = 0. Show that the polynomial x3 + 4 + 1 x has at most one root in the interval. Let x f (x) = x3 + 4 + 1 f ( )2 = 8 8 + 1 = 1. By the intermediate value theorem (ivt) (section 2. 5) there exists a number f (a) = 0 that. Suppose there is another number b in ( , 2)
