Download Making Functions Continuous Day 4

Making Functions Continuous
Consider: f ( x) 
sin x
x
Domain:  ,0u0, 
Let’s rewrite it as a piecewise function
 sin x , x  0

g ( x)   x
 a x  0
What should the value of a be such that g (x ) is continuous for all real numbers? Well, we are going to
discover the calculus trick.
To be continuous at x  0 .
g (0)  lim g ( x)
x0
Let’s try some!
1)
 x2  9 , x  3

f ( x)   x
x3

 b
2)
x 2  5x  3 , x  2
g ( x)  
 ax  1 x  2
 k

, x  2
9  x 2 x  2
3) h( x)   x 2
Let’s try some that are slightly harder:
4)
 x 2  5 x  3 , x  1

f ( x)   ax  b ,1  x  4
 11  3x
x4

5)
 cx  1 x  3
g ( x)   2
cx  1 x  3
Slightly Different
 x2  a2 , x  a

6) h( x)   x  a
xa
 2
Intermediate Value Theorem
This is an existence theorem and will not provide a solution. It just tells us of the existence of a
solution.
If f is continuous on the closed interval [a, b] and k is a number such that f (a )  k  f (b), then
there is at least one number c, such that a  c  b , such that f (c)  k .
In calculus: Let f ( x)  x 3  x  1 . Use the IVT to show that there is at least one root [zero] on [1, 2]
We know that since f (x ) is a polynomial, it is continuous on the interval. [Actually, it is continuous
on the interval. [Actually, it is continuous everywhere!] We have found that f (1)  1 and f (2)  5
.
Our Calculus-based proof is: By the IVT , there must exist some c, 1  c  2 , such that
f (1)  f (c)  f (2) . Since f (1)  1 and f (2)  5 , then  1  f (c)  5 so there must exist an
f (c)  0 for c.
Construct a convincing argument using the IVT to show that if g ( x)  x 3  5x 2  8x  9 , then there
exists some c such that g (c )  27
Here is an example of an IVT Question:
**Note this is a multiple choice question
x
0
1
2
F(x)
1
k
2
The function f is continuous on [0, 2] and has values given in the table above. The equation
f ( x) 
1
has at least two solutions in the interval [0, 2] if k=
2
a) 0
b) ½
c) 1
d) 2
e) 3
Let f be a continuous function. Selected values of f are given in the table below.
x
F(x)
1
-2
2
3
3
1
5
-5
8
7
What is the least number of solutions does the equation f(x)=1/2 have on the closed interval
1  x  8 [Hint: Draw a picture first]
Homework: Finish the worksheet for 1-4……Check out my work as well.