2.8 Topics: Rolle’s Theorem and the Mean Value Theorem
SOLUTIONS
1. Find the two x-intercepts for the function f  x   x2  x  2 and show that f   x   0 at some point between the two x-intercepts.
f  x   x2  x  2
We find that the two x intercepts have x-values of 2 and  1 .
0  x2  x  2
Since f  2   f  1  0 we can expect to find an x value
f   x   2x 1
inbetween 2 and  1 where f   x   0. This is due to
0  2x 1
the fact that since we have a function that is cts on  1, 2
x
0   x  2  x  1
x  2, x  1
x int   2, 0  ( 1, 0)
The place where this occurs is
and diff'able on  1, 2  the result of Rolle's Theorem is gauranteed.
1
2
Note:
1
  1, 2 
2
Determine if Rolle’s Theorem can be applied to the given function on the closed interval  a, b . If Rolle’s Theorem can be applied,
find all values of c in  a, b  such that f   c   0 . If Rolle’s Theorem cannot be applied explain why not.
0,3
2. f  x    x2  3x
3. f  x   1  x  1
Yes, Rolle’s Theorem does apply
since the function is a polynomial
which is diff’able everywhere (and
consequently cts. everywhere)
f  x    x 2  3x
0, 2
No, Rolle’s Theorem can NOT
be applied to this function on
this specific interval BECAUSE
this function is not diff’able on
 0, 2 . In particular it is not
diff’able at x = 1.
0,3
f   x   2 x  3
0, 2 
Yes, Rolle’s Theorem does apply
since the sine function is
diff’able everywhere (and
consequently cts. everywhere)
f  x   sin x  0, 2 
f   x   cos x
0  cos x
0  2 x  3
x
4. f  x   sin x

 k   where k is any integer.
2
But the only solutions in the given
x
3
3
Note:   0,3
2
2
interval are

2
and
3
2
For the given function and closed interval  a, b , determine all values of c in the open interval  a, b  that satisfy the Mean Value
Theorem.
2,1
 2,1
5. f  x   x2
f  x   x2
f   x   2x
2c 
f c 
MVT
f 1  f  2 
1  2
Note: MVT applies as the function is a polynomial so it is diff’able everywhere and consequently cts. too.
 2c 
f  x   x3  2 x
 1,1
 1,1
f   x   3x 2  2
MVT
6. f  x   x3  2 x
f b   f  a 
ba
1 4
 2c  1 
3
c
1
  2,1
2
Note: MVT applies as the function is a polynomial so it is diff’able everywhere and consequently cts. too.
f c 
f b   f  a 
ba
3
 3
3c 2  2 
 3c 2  2 
 3c 2  2  3
1  1
2
1
1
1
3c 2  1  c 2 
 c
 c
  1,1
3
3
3
f 1  f  1
0,  Note:
0,  
7. f  x   sin x
f  x   sin x
f   x   cos x MVT
cos c 
c

2
f  c  
f    f  0 
MVT applies as the sine function is diff’able everywhere and consequently cts. too.
f b  f  a 
 cos c 
 0
ba
00

 cos c  0
 k   where k is any integer.But the only solution in the given
interval is

2
1
 1,1
x
The mean value theorem can NOT be applied to this function on this interval. This function is discontinuous at x = 0 which lies inside
the given interval!!
8. f  x  
9. A. Use the Intermediate Value Theorem and Rolle’s Theorem to show that the equation x5  x3  x  1  0 has exactly one real
solution.
f  x   x5  x3  x  1
since the function is a polynomial we know that it is diff'able everywhere
and consequently is cts.everywhere too.
f  1  2 and
f  0   1 so by the intermediate value theorem
the function must take on all y-values between  2 and 1 (note: the IVT
applies as the function is cts.). Since 0   2,1 there must be at least one
c value in  1, 0  such that f  c   0.
Now, how do we know there isn't more than one zero for this function?
Well, suppose that there are two c's such that f  c1   0 and f  c2   0.
If this were true then Rolle's Theorem would indicate that there must be
a place where the derivative equals zero. BUT....
f   x   5 x 4  3x 2  1 is always POSITIVE no matter what x you put into it.
Thus f  x   x 5  x 3  x  1 has exactly one real zero.
B. Could the one real solution be rational? Explain.
NO, the one real zero cannot be rational. The Rational Roots Theorem tells us that the only possible rational zeros that this particular
polynomial could have would have to come from the following list of p/q’s.
x5  x3  x  1  0
p : 1
q : 1
p
: 1
q
A quick check shows that neither of the two possible rational zeros are actually zeros.
C. Explain what you could do to approximate the one real solution.
You could use Newton’s Method to approximate the zero! If you have time you may want to review Newton’s Method and give it a
shot. You can check your answer by going to www.wolframalpha.com and typing in x^5+x^3+x+1.