3.2 Rolle’s Theorem and the
Mean Value Theorem
We
Calculus!!!
Rolle’s Theorem
Let f be continuous on the closed interval [a, b]
and differentiable on the open interval (a, b).
If f(a) = f(b) then there is at least one number
c in (a, b) ' f’(c) = 0.
f’(c) means slope of tangent
line = 0. Where are the horiz.
tangent lines located?
f(a) = f(b)
a
c
c b
Ex. Find the two x-intercepts of f(x) = x2 – 3x + 2
and show that f’(x) = 0 at some point between the
two intercepts.
f(x) = x2 – 3x + 2
0 = (x – 2)(x – 1)
x-int. are 1 and 2
f’(x) = 2x - 3
0 = 2x - 3
x = 3/2
Rolles Theorem is satisfied as there is a point at
x = 3/2 where f’(x) = 0.
Let f(x) = x4 – 2x2 . Find all c in the interval (-2, 2)
such that f’(x) = 0.
Since f(-2) and f(2) = 8, we can use Rolle’s Theorem.
f’(x) = 4x3 – 4x = 0
4x(x2 – 1) = 0
x = -1, 0,
and 1
Thus, in the interval
(-2, 2), the derivative
is zero at each of these
three x-values.
8
The Mean Value Theorem
If f is continuous on the closed interval [a,b] and
differentiable on the open interval (a,b), then $
a number c in (a,b) '
f (b) - f (a )
f ' (c ) =
b-a
(b,f(b))
secant line
(a,f(a))
a
c
b
f (b) - f (a )
b-a
represents
slope of the
secant line.
Given f(x) = 5 – 4/x, find all c in the interval (1,4)
such that the slope of the secant line = the slope of
the tangent line.
f (4) - f (1) 4 - 1
f ' (c ) =
=
=1
4 -1
4 -1
?
4
f ' ( x) = ? = 2
x
4
2
4
=
x
=
1
x2
x = ±2
But in the interval of (1,4),
only 2 works, so c = 2.