Section 5.5 The Intermediate Value Theorem Rolle’s Theorem The Mean Value Theorem 3.6.
4.2 Mean Value Theorem & Rolle’s Theorem
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4.2 Mean Value Theorem & Rolle’s Theorem
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4.2 Mean Value Theorem & Rolle’s Theorem. Mean Value Theorem. If f is a function that satisfies the following conditions: f is continuous on a closed interval [a, b] f is differentiable on the open interval (a, b) then there exists a number c in ( a, b) such that. - PowerPoint PPT Presentation
Transcript of 4.2 Mean Value Theorem & Rolle’s Theorem
4.2Mean Value Theorem &
Rolle’s Theorem
Mean Value TheoremIf f is a function that satisfies the
following conditions:◦f is continuous on a closed interval
[a, b]◦f is differentiable on the open
interval (a, b)
then there exists a number c in (a, b) such that ab
afbfcf
)()(
)('
In other words, there exists a point somewhere in (a, b) such that the instantaneous rate of change is equal to the average rate of change.
Mean Value Theorem
(a, f(a))
(b, f(b))
(c, f(c))
c
Mean Value TheoremExample:
The points P(1, 1) and Q(3, 27) are on the curve f(x) = x3. Using the Mean Value Theorem, find c in the interval (1, 3) such that f’(c) is equal to the slope of the secant of PQ.
13
127
PQm
13PQm
Since f(x) is defined for all real numbers, f is continuous on [1, 3]. Also, f’(x) = 3x2 is defined for all real numbers; thus, f’(x) is differentiable on (1, 3). Therefore, by MVT,
13)(' cf
133 2 c3
13c
(use only positive root since it is in the interval (1, 3))
Rolle’s TheoremIf f is a function that satisfies the
following three conditions:◦f is continuous on a closed interval
[a, b]◦f is differentiable on the open
interval (a, b)◦f(a) = f(b)
then there exists a number c in (a, b) such that f’(c) = 0.
Rolle’s Theorem
a b
(c, f(c))
c
f‘(c) = 0
Rolle’s TheoremExample:
If f(x) = x2 + 4x – 5, show that the hypotheses of Rolle’s Theorem are satisfied on the interval [–4, 0] and find all values of c that satisfy the conclusion of the theorem.
1.) f(x) = x2 + 4x – 5 is continuous everywhere since it is a polynomial.
2.) f’(x) = 2x + 4 is defined for all numbers and thus is differentiable on (–4, 0).
3.) f(0) = f(–4) = –5. Therefore, there exists a c in (–4, 0) such that f’(c) = 0
f‘(x) = 2x + 4 = 0 x = –
2
Increasing/Decreasing FunctionsA function is increasing on (x1, x2)
if:◦f(x1) < f(x2) OR
◦f‘(x) > 0 at each point of (x1, x2)
A function is decreasing on (x1, x2) if:◦f(x1) > f(x2) OR
◦f‘(x) < 0 at each point of (x1, x2)
AntiderivativesA function F(x) is an
antiderivative of a function f(x) if F’(x) = f(x) for all x in the domain of f.
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