MATH241 Lecture Notes - Lecture 7: Intermediate Value Theorem, Asymptote, Polynomial
9 May 2018
School
Department
Course
Professor
MATH241 - Lecture 7 - Continuity and Limits at Infinity
The Intermediate Value Theorem: suppose that is continuous on the closed interval f a,b][
and let be any number between and where , then there exist an(a)f(b)f(a) = (b)f/f
number in such that L a,b)( (c)f=n
Example
:
Show that on (2,3)n(x)l=x−√x
Solution:
n(x)l=x−√x
→ n(x)x−√x−l= 0
Let (x)n(x)f=x−√x−l
is continuous on (0, )(x)f∞
(2) n(2)f= 2 − √2 − l(3) n(3)f= 3 − √3 − l
= .4 .72 − 1 − 0 = .7 .13 − 1 − 1
= -0.1 < 0 = 0.2 > 0
Since by the IVT there exists a in (2,3) such that or (2) (3)f< 0 < f c (c)f= 0 n(c)l=c−√c
Example
:
For what values of the constant is the function continuous on c− , )( ∞ ∞
(x)f=x2+x+ 2 x< 1
x n(x)x3−c+l x ≥ 1
Solution:
(x)xlim
x → 1
−f= lim
x → 1
−c2+x+ 2 = c+ 1 + 2 = c+ 3
(x)x n(x)n(1)lim
x → 1
+f= lim
x → 1
+x3−c+l= 1 − c+l= 1 − c
Set (x) (x)lim
x → 1
−f= lim
x → 1
+f
→ 2c 1 2c c c+ 3 = 1 − c= + 3 = = = − 2 = = − 1
Thus (the limit exists) if (x)lim
x → 1 f= 2 −c= 1
*Note that , consequently and thus is continuous(1)f= 1 − c= 2 (x) (1)lim
x → 1 f=f(x)f
at by the limit definition of continuity if x= 1 −c= 1
Document Summary
Math241 - lecture 7 - continuity and limits at infinity. The intermediate value theorem : suppose that f f (b) and let and where in n be any number between (c) = x x a, b) n(x) f l f (a) = n is continuous on the closed interval a, b] f (a) = (b) Solution: l n(x) x x l. = x x l n(x) is continuous on (0, = 0. 2 > 0 by the ivt there exists a c in (2,3) such that f (c) For what values of the constant c is the function continuous on f (x) = x2 + x + 2 x3 c + l x x < 1 x 1 n(x) f. Solution: (x) lim x 1 lim x 1 + (x) f. = lim x 1 + x c 2 + x + 2 = c + 1 + 2 = c + 3 x3 c + l x n(x)
