4. Suppose that f(2) = 16 and that f'(x) 2 – 3 for all x in the interval [2,7]. Determine the smallest possible value for f(7) by using the Mean Value Theorem.
4. Suppose that f(2) = 16 and that f'(x) 2 – 3 for all x in the interval [2,7]. Determine the smallest possible value for f(7) by using the Mean Value Theorem.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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(Please solve only using the given formula sheet) Thank you
(P4)
![4. Suppose that f(2) = 16 and that f'(x) 2 – 3 for all x in the interval [2,7]. Determine the smallest possible value for
f(7) by using the Mean Value Theorem.
10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3930b8f2-52cf-42d9-bbea-fa5b5c673abd%2F767365f6-f383-4539-bf93-2fd39626da5c%2Fgtj40bo_processed.png&w=3840&q=75)
Transcribed Image Text:4. Suppose that f(2) = 16 and that f'(x) 2 – 3 for all x in the interval [2,7]. Determine the smallest possible value for
f(7) by using the Mean Value Theorem.
10
![EXTREMA AND CRITICAL POINTS
c is a critical point of fif either f'(c) = 0 or f'(c) does not exist.
f(c) is an absolute maximum value of fiff (c) >f(x) for all x in the domain of f.
f(c) is an absolute minimum value of fif f(c) < f(x) for all x in the domain of f.
f(c) is a relative minimum value of fiff(c) 2 f (x) for all x near c.
f(c) is a relative maximum value of fiff(c) < f(x) for all x near c.
EXTREME VALUE THEOREM
FERMAT’S THEOREM
If fis continuous on a closed and bounded interval [a, b], then f
Iff(c) is a relative maximum or minimum value, then f'(c) = 0
attains a maximum and minimum on the interval.
ROLLE'S THEOREM
MEAN VALUE THEOREM
Let f be a function which satisfies the following:
Let f be a function which satisfies the following:
1. fis continuous on the closed interval [a, b].
1. fis continuous on the closed interval [a, b].
2. fis differentiable on the open interval (a, b).
2. fis differentiable on the open interval (a, b).
3. f(a) = f (b).
Then there is some c in the interval (a, b) where
Then there is some c in the interval (a, b) where f'(c) = 0.
f(a) – f(b)
f'(c) =
а —Ь
CALCULUS AND GRAPHS
If f'(x) > 0 on some interval (a,b), then f is increasing on (a, b).
If f'(x) < 0 on some interval (a,b), then f is decreasing on (a, b).
First Derivative Test:
Suppose c is a critical point of some function f.
a. Iff' changes from positive to negative at c, then f(c) is a relative maximum.
b. Iff' changes from negative to positive at c, then f(c) is a relative minimum.
c. If the sign of f' does not change while passing c, then f(c) is neither a maximum nor a minimum (it is a saddle point).
If f"(x) > 0 on some interval (a,b), then fis concave up on (a, b).
If f"(x) <0 on some interval (a, b), then f is concave down on (a, b).
Second Derivative Test:
Suppose c is a critical point of some function f.
a. If f"(c) < 0, then f(c) is a relative maximum.
b. If f"(c) > 0, then f(c) is a relative minimum.
c. If f"(c) = 0, then the test is inconclusive. (Could be a maximum, minimum, or saddle).
L’HOPTIAL’S RULE
AREA APPROXIMATION
ƒ(x)
takes either the
The area under a curve can be approximated using rectangles with
either right or left endpoints. You can subdivide an interval [a, b]
Iff and g are differentiable near a and lim =
x-a g(x)
00
or
then
b - a
for n rectangles by taking segments of length -
indeterminate form
00
f(x)
lim
f'(x)
= lim
x-a g'(x)
f(x)dx = lim
f(x*)Ax.
X-a g(x)
n00
i=1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3930b8f2-52cf-42d9-bbea-fa5b5c673abd%2F767365f6-f383-4539-bf93-2fd39626da5c%2F0rbrvg_processed.png&w=3840&q=75)
Transcribed Image Text:EXTREMA AND CRITICAL POINTS
c is a critical point of fif either f'(c) = 0 or f'(c) does not exist.
f(c) is an absolute maximum value of fiff (c) >f(x) for all x in the domain of f.
f(c) is an absolute minimum value of fif f(c) < f(x) for all x in the domain of f.
f(c) is a relative minimum value of fiff(c) 2 f (x) for all x near c.
f(c) is a relative maximum value of fiff(c) < f(x) for all x near c.
EXTREME VALUE THEOREM
FERMAT’S THEOREM
If fis continuous on a closed and bounded interval [a, b], then f
Iff(c) is a relative maximum or minimum value, then f'(c) = 0
attains a maximum and minimum on the interval.
ROLLE'S THEOREM
MEAN VALUE THEOREM
Let f be a function which satisfies the following:
Let f be a function which satisfies the following:
1. fis continuous on the closed interval [a, b].
1. fis continuous on the closed interval [a, b].
2. fis differentiable on the open interval (a, b).
2. fis differentiable on the open interval (a, b).
3. f(a) = f (b).
Then there is some c in the interval (a, b) where
Then there is some c in the interval (a, b) where f'(c) = 0.
f(a) – f(b)
f'(c) =
а —Ь
CALCULUS AND GRAPHS
If f'(x) > 0 on some interval (a,b), then f is increasing on (a, b).
If f'(x) < 0 on some interval (a,b), then f is decreasing on (a, b).
First Derivative Test:
Suppose c is a critical point of some function f.
a. Iff' changes from positive to negative at c, then f(c) is a relative maximum.
b. Iff' changes from negative to positive at c, then f(c) is a relative minimum.
c. If the sign of f' does not change while passing c, then f(c) is neither a maximum nor a minimum (it is a saddle point).
If f"(x) > 0 on some interval (a,b), then fis concave up on (a, b).
If f"(x) <0 on some interval (a, b), then f is concave down on (a, b).
Second Derivative Test:
Suppose c is a critical point of some function f.
a. If f"(c) < 0, then f(c) is a relative maximum.
b. If f"(c) > 0, then f(c) is a relative minimum.
c. If f"(c) = 0, then the test is inconclusive. (Could be a maximum, minimum, or saddle).
L’HOPTIAL’S RULE
AREA APPROXIMATION
ƒ(x)
takes either the
The area under a curve can be approximated using rectangles with
either right or left endpoints. You can subdivide an interval [a, b]
Iff and g are differentiable near a and lim =
x-a g(x)
00
or
then
b - a
for n rectangles by taking segments of length -
indeterminate form
00
f(x)
lim
f'(x)
= lim
x-a g'(x)
f(x)dx = lim
f(x*)Ax.
X-a g(x)
n00
i=1
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