MAT 1320 A: Calculus I - University of Ottawa 16th.pdfMAT 1320 A: Calculus I Limits and continuity...
Transcript of MAT 1320 A: Calculus I - University of Ottawa 16th.pdfMAT 1320 A: Calculus I Limits and continuity...
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
MAT 1320 A: Calculus I
Paul-Eugene ParentDepartment of Mathematics and Statistics
University of Ottawa
September 16th, 2013
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Outline
1 Limits and continuity
2 Derivative
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Limits
Let f : D → R be a function, L ∈ R, and a ∈ R a point ofeither D or a boundary point of D.
Examples of acceptable “a”:
• If D = [1, 2] then a could be 1, 5;
• If D =]− 1, 3[ then a could also be 3 or −1 eventhough they are not in the domain of f .
DefinitionWe will write
limx→a
f (x) = L
if the value f (x) approaches the number L as x ∈ D goestoward a without being equal to a.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Limits
Let f : D → R be a function, L ∈ R, and a ∈ R a point ofeither D or a boundary point of D.
Examples of acceptable “a”:
• If D = [1, 2] then a could be 1, 5;
• If D =]− 1, 3[ then a could also be 3 or −1 eventhough they are not in the domain of f .
DefinitionWe will write
limx→a
f (x) = L
if the value f (x) approaches the number L as x ∈ D goestoward a without being equal to a.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Limits
Let f : D → R be a function, L ∈ R, and a ∈ R a point ofeither D or a boundary point of D.
Examples of acceptable “a”:
• If D = [1, 2] then a could be 1, 5;
• If D =]− 1, 3[ then a could also be 3 or −1 eventhough they are not in the domain of f .
DefinitionWe will write
limx→a
f (x) = L
if the value f (x) approaches the number L as x ∈ D goestoward a without being equal to a.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Limits
Let f : D → R be a function, L ∈ R, and a ∈ R a point ofeither D or a boundary point of D.
Examples of acceptable “a”:
• If D = [1, 2] then a could be 1, 5;
• If D =]− 1, 3[ then a could also be 3 or −1 eventhough they are not in the domain of f .
DefinitionWe will write
limx→a
f (x) = L
if the value f (x) approaches the number L as x ∈ D goestoward a without being equal to a.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative Remark: That value L must be independent of the way weapproach a.
Examples: 1) What is the value of limx→xo f (x) if it exists?
Answer: 5.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative Remark: That value L must be independent of the way weapproach a.
Examples: 1) What is the value of limx→xo f (x) if it exists?
Answer: 5.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative Remark: That value L must be independent of the way weapproach a.
Examples: 1) What is the value of limx→xo f (x) if it exists?
Answer: 5.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative2) What is the value of limx→xo f (x) if it exists?
Answer: 5.In other words what happens exactly at xo is not importantwhen computing the limit at that point.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative2) What is the value of limx→xo f (x) if it exists?
Answer: 5.In other words what happens exactly at xo is not importantwhen computing the limit at that point.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
y = 1x
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative We can expand the definition of limits and ask: “is f (x)approaching a definite value as |x | grows arbitrarily?” Wewrite in this case either
limx→∞
f (x) or limx→−∞
f (x).
In the case of y = 1x we have
limx→∞
f (x) = limx→−∞
f (x) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative We can expand the definition of limits and ask: “is f (x)approaching a definite value as |x | grows arbitrarily?” Wewrite in this case either
limx→∞
f (x) or limx→−∞
f (x).
In the case of y = 1x we have
limx→∞
f (x) = limx→−∞
f (x) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Directional limits
In some problems we might be interested in only a certaindirection of approach.
If we want to approach “a” from the right we write
limx→a+
f (x)
and from the leftlim
x→a−f (x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Directional limits
In some problems we might be interested in only a certaindirection of approach.
If we want to approach “a” from the right we write
limx→a+
f (x)
and from the leftlim
x→a−f (x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Directional limits
In some problems we might be interested in only a certaindirection of approach.
If we want to approach “a” from the right we write
limx→a+
f (x)
and from the leftlim
x→a−f (x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
TheoremThe limit limx→a f (x) exists if and only if both the right andthe left limit exist and are equal.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules when computing limits
Suppose both limx→a f (x) and limx→a g(x) are numbers andlet c ∈ R.
• limx→a
(f (x)± g(x)) = limx→a
f (x)± limx→a
g(x);
• limx→a
cf (x) = c limx→a
f (x);
• limx→a
f (x)g(x) =(
limx→a
f (x))(
limx→a
g(x))
;and
• limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x)if lim
x→ag(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules when computing limits
Suppose both limx→a f (x) and limx→a g(x) are numbers andlet c ∈ R.
• limx→a
(f (x)± g(x)) = limx→a
f (x)± limx→a
g(x);
• limx→a
cf (x) = c limx→a
f (x);
• limx→a
f (x)g(x) =(
limx→a
f (x))(
limx→a
g(x))
;and
• limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x)if lim
x→ag(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules when computing limits
Suppose both limx→a f (x) and limx→a g(x) are numbers andlet c ∈ R.
• limx→a
(f (x)± g(x)) = limx→a
f (x)± limx→a
g(x);
• limx→a
cf (x) = c limx→a
f (x);
• limx→a
f (x)g(x) =(
limx→a
f (x))(
limx→a
g(x))
;
and
• limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x)if lim
x→ag(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules when computing limits
Suppose both limx→a f (x) and limx→a g(x) are numbers andlet c ∈ R.
• limx→a
(f (x)± g(x)) = limx→a
f (x)± limx→a
g(x);
• limx→a
cf (x) = c limx→a
f (x);
• limx→a
f (x)g(x) =(
limx→a
f (x))(
limx→a
g(x))
;and
• limx→a
f (x)
g(x)=
limx→a
f (x)
limx→a
g(x)if lim
x→ag(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Continuity
DefinitionLet f : D → R be a function and a ∈ D. We say that f iscontinuous at “a” if
f (a) = limx→a
f (x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples
For the following function
while the limx→xo f (x) exists and is equal to 5
it is notcontinuous at xo since f (xo) = 7 6= 5.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples
For the following function
while the limx→xo f (x) exists and is equal to 5 it is notcontinuous at xo since f (xo) = 7 6= 5.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Unfortunately in this case even though f (1) = 5, i.e., it isdefined at x = 1, it is not continuous since
1 = limx→1−
f (x) 6= limx→1+
f (x) = 5,
i.e., limx→1
f (x) does NOT exists!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Unfortunately in this case even though f (1) = 5, i.e., it isdefined at x = 1, it is not continuous since
1 = limx→1−
f (x) 6= limx→1+
f (x) = 5,
i.e., limx→1
f (x) does NOT exists!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Unfortunately in this case even though f (1) = 5, i.e., it isdefined at x = 1, it is not continuous since
1 = limx→1−
f (x) 6= limx→1+
f (x) = 5,
i.e., limx→1
f (x) does NOT exists!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous? YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous? YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous? YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous? YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous? YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous?
YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Examples of continuous functions
• All polynomials, e.g., x , x2 − 2, x3 + x − 1...
• All trigonometric functions.
• The exponential and logarithm functions.
• All rational functions, i.e., quotients of polynomial.
WARNING: One checks continuity only on x belonging tothe domain of a function.
Question: Is y = 1x continuous? YES!
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Example
Compute limx→0
(3 + x)2 − 9
x.
Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression. So
limx→0
(3 + x)2 − 9
x= lim
x→0
9 + 6x + x2 − 9
x
= limx→0
x(6 + x)
x.
As we are interested in the limit when approaching 0 we arepurposely avoiding x = 0 hence
limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Example
Compute limx→0
(3 + x)2 − 9
x.
Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression.
So
limx→0
(3 + x)2 − 9
x= lim
x→0
9 + 6x + x2 − 9
x
= limx→0
x(6 + x)
x.
As we are interested in the limit when approaching 0 we arepurposely avoiding x = 0 hence
limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Example
Compute limx→0
(3 + x)2 − 9
x.
Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression. So
limx→0
(3 + x)2 − 9
x= lim
x→0
9 + 6x + x2 − 9
x
= limx→0
x(6 + x)
x.
As we are interested in the limit when approaching 0 we arepurposely avoiding x = 0 hence
limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Example
Compute limx→0
(3 + x)2 − 9
x.
Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression. So
limx→0
(3 + x)2 − 9
x= lim
x→0
9 + 6x + x2 − 9
x
= limx→0
x(6 + x)
x.
As we are interested in the limit when approaching 0 we arepurposely avoiding x = 0 hence
limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Example
Compute limx→0
(3 + x)2 − 9
x.
Solution: First of all we notice that it is a rational functionbut 0 is not in its domain. Hence we can’t simply substitute0 in the expression. So
limx→0
(3 + x)2 − 9
x= lim
x→0
9 + 6x + x2 − 9
x
= limx→0
x(6 + x)
x.
As we are interested in the limit when approaching 0 we arepurposely avoiding x = 0 hence
limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
And now we notice that the function 6 + x is a line (hencecontinuous) and that 0 is in its domain.
Conclusion: limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x) = 6 + 0 = 6.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
And now we notice that the function 6 + x is a line (hencecontinuous) and that 0 is in its domain.
Conclusion: limx→0
(3 + x)2 − 9
x= lim
x→0(6 + x) = 6 + 0 = 6.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Constructing continuous functions
Let f , g : D → R be two continuous functions (when we donot specify continuous at a particular point we meancontinuous everywhere on their domain).
• f ± g is continuous;
• f · g is continuous; and
•f
gis continuous whenever the quotient is defined, i.e., it
is continuous at all x ∈ D such that g(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Constructing continuous functions
Let f , g : D → R be two continuous functions (when we donot specify continuous at a particular point we meancontinuous everywhere on their domain).
• f ± g is continuous;
• f · g is continuous; and
•f
gis continuous whenever the quotient is defined, i.e., it
is continuous at all x ∈ D such that g(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Constructing continuous functions
Let f , g : D → R be two continuous functions (when we donot specify continuous at a particular point we meancontinuous everywhere on their domain).
• f ± g is continuous;
• f · g is continuous;
and
•f
gis continuous whenever the quotient is defined, i.e., it
is continuous at all x ∈ D such that g(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Constructing continuous functions
Let f , g : D → R be two continuous functions (when we donot specify continuous at a particular point we meancontinuous everywhere on their domain).
• f ± g is continuous;
• f · g is continuous; and
•f
gis continuous whenever the quotient is defined, i.e., it
is continuous at all x ∈ D such that g(x) 6= 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Intermediate value Theorem
Let f : D → R be a continuous function and suppose thatthere is an interval [a, b] ⊆ D.
TheoremFor each value “y” between f (a) and f (b) there isxo ∈ [a, b] such that
y = f (xo).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Intermediate value Theorem
Let f : D → R be a continuous function and suppose thatthere is an interval [a, b] ⊆ D.
TheoremFor each value “y” between f (a) and f (b) there isxo ∈ [a, b] such that
y = f (xo).
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
First application
Suppose we have a continuous function f : [0, 1]→ [0, 1].
Then one knows that there is xo ∈ [0, 1] such thatf (xo) = xo , i.e., f admits a fixed point.
Why?
Consider the new function
g : [0, 1] −→ Rx 7→ f (x)− x .
It is continuous as it is the sum of two continuous functions.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
First application
Suppose we have a continuous function f : [0, 1]→ [0, 1].
Then one knows that there is xo ∈ [0, 1] such thatf (xo) = xo , i.e., f admits a fixed point.
Why?
Consider the new function
g : [0, 1] −→ Rx 7→ f (x)− x .
It is continuous as it is the sum of two continuous functions.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
First application
Suppose we have a continuous function f : [0, 1]→ [0, 1].
Then one knows that there is xo ∈ [0, 1] such thatf (xo) = xo , i.e., f admits a fixed point.
Why?
Consider the new function
g : [0, 1] −→ Rx 7→ f (x)− x .
It is continuous as it is the sum of two continuous functions.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
First application
Suppose we have a continuous function f : [0, 1]→ [0, 1].
Then one knows that there is xo ∈ [0, 1] such thatf (xo) = xo , i.e., f admits a fixed point.
Why?
Consider the new function
g : [0, 1] −→ Rx 7→ f (x)− x .
It is continuous as it is the sum of two continuous functions.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
First application
Suppose we have a continuous function f : [0, 1]→ [0, 1].
Then one knows that there is xo ∈ [0, 1] such thatf (xo) = xo , i.e., f admits a fixed point.
Why?
Consider the new function
g : [0, 1] −→ Rx 7→ f (x)− x .
It is continuous as it is the sum of two continuous functions.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
On one hand if f (1) = 1, then we are done as we have justfound a fixed point.
Else by construction
g(1) < 0 since f (1) < 1.
On the other hand if f (0) = 0, then again we are done. Elseby construction
g(0) > 0 since f (0) > 0.
Conclusion: By the intermediate value theorem, there isxo ∈ [0, 1] such that g(xo) = 0, i.e.,
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
On one hand if f (1) = 1, then we are done as we have justfound a fixed point. Else by construction
g(1) < 0 since f (1) < 1.
On the other hand if f (0) = 0, then again we are done. Elseby construction
g(0) > 0 since f (0) > 0.
Conclusion: By the intermediate value theorem, there isxo ∈ [0, 1] such that g(xo) = 0, i.e.,
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
On one hand if f (1) = 1, then we are done as we have justfound a fixed point. Else by construction
g(1) < 0 since f (1) < 1.
On the other hand if f (0) = 0, then again we are done.
Elseby construction
g(0) > 0 since f (0) > 0.
Conclusion: By the intermediate value theorem, there isxo ∈ [0, 1] such that g(xo) = 0, i.e.,
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
On one hand if f (1) = 1, then we are done as we have justfound a fixed point. Else by construction
g(1) < 0 since f (1) < 1.
On the other hand if f (0) = 0, then again we are done. Elseby construction
g(0) > 0 since f (0) > 0.
Conclusion: By the intermediate value theorem, there isxo ∈ [0, 1] such that g(xo) = 0, i.e.,
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
On one hand if f (1) = 1, then we are done as we have justfound a fixed point. Else by construction
g(1) < 0 since f (1) < 1.
On the other hand if f (0) = 0, then again we are done. Elseby construction
g(0) > 0 since f (0) > 0.
Conclusion: By the intermediate value theorem, there isxo ∈ [0, 1] such that g(xo) = 0, i.e.,
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
General culture
This result is true in higher dimensions!
It is known as Brouwer fixed point Theorem.
TheoremLet f : [0, 1]n → [0, 1]n be a continuous function forn = 1, 2, 3, ... . The there exists xo ∈ [0, 1]n such that
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
General culture
This result is true in higher dimensions!
It is known as Brouwer fixed point Theorem.
TheoremLet f : [0, 1]n → [0, 1]n be a continuous function forn = 1, 2, 3, ... . The there exists xo ∈ [0, 1]n such that
f (xo) = xo .
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Second application
Let p(x) = xn + an−1xn−1 + ... + a1x + a0 be a polynomialof odd degree greater or equal to 1, i.e., n can be any oddpositive integer greater than or equal to one.
TheoremOne automatically knows that there is at least one xo ∈ Rsuch that
p(xo) = 0.
Why?
Intuitively we know that
limx→±∞
p(x) = limx→±∞
xn.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Second application
Let p(x) = xn + an−1xn−1 + ... + a1x + a0 be a polynomialof odd degree greater or equal to 1, i.e., n can be any oddpositive integer greater than or equal to one.
TheoremOne automatically knows that there is at least one xo ∈ Rsuch that
p(xo) = 0.
Why?
Intuitively we know that
limx→±∞
p(x) = limx→±∞
xn.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Second application
Let p(x) = xn + an−1xn−1 + ... + a1x + a0 be a polynomialof odd degree greater or equal to 1, i.e., n can be any oddpositive integer greater than or equal to one.
TheoremOne automatically knows that there is at least one xo ∈ Rsuch that
p(xo) = 0.
Why?
Intuitively we know that
limx→±∞
p(x) = limx→±∞
xn.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Second application
Let p(x) = xn + an−1xn−1 + ... + a1x + a0 be a polynomialof odd degree greater or equal to 1, i.e., n can be any oddpositive integer greater than or equal to one.
TheoremOne automatically knows that there is at least one xo ∈ Rsuch that
p(xo) = 0.
Why?
Intuitively we know that
limx→±∞
p(x) = limx→±∞
xn.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Moreover, on one hand
limx→∞
xn =∞,
i.e., there is b ∈ R such that p(b) > 0.
On the other hand we also know since n is odd that
limx→−∞
xn = −∞,
i.e., there is a ∈ R such that p(a) < 0.
Conclusion: By the intermediate value theorem there mustexists xo ∈ [a, b] such that
p(xo) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Moreover, on one hand
limx→∞
xn =∞,
i.e., there is b ∈ R such that p(b) > 0.
On the other hand we also know since n is odd that
limx→−∞
xn = −∞,
i.e., there is a ∈ R such that p(a) < 0.
Conclusion: By the intermediate value theorem there mustexists xo ∈ [a, b] such that
p(xo) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Moreover, on one hand
limx→∞
xn =∞,
i.e., there is b ∈ R such that p(b) > 0.
On the other hand we also know since n is odd that
limx→−∞
xn = −∞,
i.e., there is a ∈ R such that p(a) < 0.
Conclusion: By the intermediate value theorem there mustexists xo ∈ [a, b] such that
p(xo) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Moreover, on one hand
limx→∞
xn =∞,
i.e., there is b ∈ R such that p(b) > 0.
On the other hand we also know since n is odd that
limx→−∞
xn = −∞,
i.e., there is a ∈ R such that p(a) < 0.
Conclusion: By the intermediate value theorem there mustexists xo ∈ [a, b] such that
p(xo) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Moreover, on one hand
limx→∞
xn =∞,
i.e., there is b ∈ R such that p(b) > 0.
On the other hand we also know since n is odd that
limx→−∞
xn = −∞,
i.e., there is a ∈ R such that p(a) < 0.
Conclusion: By the intermediate value theorem there mustexists xo ∈ [a, b] such that
p(xo) = 0.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
The derivative
Let f : D → R be a function and a ∈ D.
DefinitionWe say that f is differentiable at a ∈ D if the following limit
limh→0
f (h + a)− f (a)
h
is a number, i.e., it exists.
In that case we denote that number by f ′(a).
You will also see in the literaturedf
dx
∣∣∣∣x=a
.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
The derivative
Let f : D → R be a function and a ∈ D.
DefinitionWe say that f is differentiable at a ∈ D if the following limit
limh→0
f (h + a)− f (a)
h
is a number, i.e., it exists.
In that case we denote that number by f ′(a).
You will also see in the literaturedf
dx
∣∣∣∣x=a
.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
The derivative
Let f : D → R be a function and a ∈ D.
DefinitionWe say that f is differentiable at a ∈ D if the following limit
limh→0
f (h + a)− f (a)
h
is a number, i.e., it exists.
In that case we denote that number by f ′(a).
You will also see in the literaturedf
dx
∣∣∣∣x=a
.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
The derivative
Let f : D → R be a function and a ∈ D.
DefinitionWe say that f is differentiable at a ∈ D if the following limit
limh→0
f (h + a)− f (a)
h
is a number, i.e., it exists.
In that case we denote that number by f ′(a).
You will also see in the literaturedf
dx
∣∣∣∣x=a
.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules
Let f and g two differentiable functions at “a” and c ∈ R.
• (cf )′(a) = cf ′(a);
• (f ± g)′(a) = f ′(a)± g ′(a);
• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and
•(
f
g
)′(a) =
f ′(a) · g(a)− f (a) · g ′(a)
(g(a))2.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules
Let f and g two differentiable functions at “a” and c ∈ R.
• (cf )′(a) = cf ′(a);
• (f ± g)′(a) = f ′(a)± g ′(a);
• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and
•(
f
g
)′(a) =
f ′(a) · g(a)− f (a) · g ′(a)
(g(a))2.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules
Let f and g two differentiable functions at “a” and c ∈ R.
• (cf )′(a) = cf ′(a);
• (f ± g)′(a) = f ′(a)± g ′(a);
• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and
•(
f
g
)′(a) =
f ′(a) · g(a)− f (a) · g ′(a)
(g(a))2.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules
Let f and g two differentiable functions at “a” and c ∈ R.
• (cf )′(a) = cf ′(a);
• (f ± g)′(a) = f ′(a)± g ′(a);
• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a);
and
•(
f
g
)′(a) =
f ′(a) · g(a)− f (a) · g ′(a)
(g(a))2.
MAT 1320 A:Calculus I
Limits andcontinuity
Derivative
Rules
Let f and g two differentiable functions at “a” and c ∈ R.
• (cf )′(a) = cf ′(a);
• (f ± g)′(a) = f ′(a)± g ′(a);
• (f · g)′(a) = f ′(a) · g(a) + f (a) · g ′(a); and
•(
f
g
)′(a) =
f ′(a) · g(a)− f (a) · g ′(a)
(g(a))2.