Calculus I, Columbia University, Fall 2014 Instructor: Paul Siegel November 4, 2014

Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Calculus I, Columbia University, Fall 2014
Instructor: Paul Siegel
November 4, 2014
Calculus I,
Columbia
University,
Fall 2014
The Derivative of a Function
Instructor:
Paul Siegel
Limits allow us to determine basic qualitative features of the
graph of a function, e.g. end behavior, asymptotes,
discontinuities.
More refined questions: where is a given function increasing /
decreasing and how quickly? How can we find ”bumps” in the
graph of a function?
These questions can be answered using the derivative of a
function.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
The tangent line to a function f at a point x0 is defined to be
the line which best approximates the graph of f near the point
P = (x0 , f (x0 )).
Sometimes a function can’t be well approximated by a line at
certain points:
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
(Aside: some continuous functions can’t be well approximated
by a line at *any* point!)
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
How do we find the tangent line to a function f at a point x0 ,
assuming it exists?
Idea: find the slope of the line joining (x0 , f (x0 )) and another
nearby point on the graph of f , and take a limit as the nearby
point approaches (x0 , f (x0 )).
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Point: (x0 , f (x0 ))
Nearby point: (x0 + h, f (x0 + h)) where h is small
Slope:
f (x0 + h) − f (x0 )
f (x0 + h) − f (x0 )
=
x0 + h − x0
h
Slope of tangent line:
f (x0 + h) − f (x0 )
h→0
h
lim
Example: Find the slope of the tangent line to the function
f (x) = x 2 at the point x = 1
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Notation: the slope of the tangent line to the function f at the
point x0 is called the derivative of f at x0 and denoted f 0 (x0 ).
f 0 (x0 ) = lim
h→0
f (x0 + h) − f (x0 )
h
If this limit exists, we say that f is differentiable at x0 .
Interpretation: f 0 (x0 ) represents the infinitesimal rate at which
f is changing near x0 ; if f 0 (x0 ) > 0 then f is increasing at x0 ,
while if f 0 (x0 ) < 0 then f 0 (x0 ) is decreasing at a.
Calculus I,
Columbia
University,
Fall 2014
Basic properties of derivatives
Instructor:
Paul Siegel
Let f and g be functions which are differentiable at a and let c
be a constant. Then:
• f and g are continuous at a (differentiability implies
continuity)
• f + g is differentiable at a, and (f + g )0 (a) = f 0 (a) + g 0 (a)
• cf is differentiable at a, and (cf )0 (a) = cf 0 (a)
Note that we did not give a rule for finding the derivative of
the product or quotient of two differentiable functions; such
rules exist, but they are more complicated than you think!
Calculus I,
Columbia
University,
Fall 2014
Derivatives of Polynomials
Instructor:
Paul Siegel
Thanks to the rules above, differentiating polynomial functions
is no harder than differentiating the function f (x) = x n where
n is a non-negative integer. To do this, we’ll need formulas for
(x0 + h)n :
• (x0 + h)0 = 1
• (x0 + h)1 = x0 + h
• (x0 + h)2 = x02 + 2x0 h + h2
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
• n = 3:
(x0 + h)3 = x03 + 3x02 h + 3x0 h2 + h3
= x03 + 3x02 h + h2 (3x0 + h)
• n = 4:
(x0 + h)4 = x04 + 4x03 h + 6x02 h2 + 4x0 h3 + h4
= x04 + 4x03 h + h2 (6x02 + 4x0 h + h2 )
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
In general:
(x0 + h)n = x0n + nx0n−1 h + h2 p(x0 , h)
where p(x0 , h) is a polynomial in a and h
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Derivative of f (x) = x n at any point x0 :
f (x0 + h) − f (x0 )
h→0
h
(x0 + h)n − x0n
= lim
h→0
h
n
x + nx0n−1 h + h2 p(x0 , h) − x0n
= lim 0
h→0
h
= lim nx0n−1 + hp(x0 , h)
f 0 (x0 ) = lim
=
h→0
nx0n−1
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Thus the derivative of f (x) = x n at any point x0 is nx0n−1 . The
calculation in the previous slide only works when n is a positive
integer, but in fact the result is true for any other power:
Theorem
Let f (x) = x p where p is any real number. Then
f 0 (x0 ) = px0p−1
for any point x0 in the domain of f except possibly x0 = 0
(where f is not differentiable if p < 1 and p 6= 0).
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
First few examples:
• The derivative of f (x) = 1 at x0 is 0
• The derivative of f (x) = x at x0 is 1
• The derivative of f (x) = x 9 at x0 is 9x08
• The derivative of f (x) =
√
−1/2
x at x0 is 12 x0
√1
2 x0
−2
x03
=
• The derivative of f (x) = x12 at x0 is −2x0−3 =
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Find the equation of the line which is tangent to
f (x) = 3x 5 − x 2 + 4 at x = 1.
• Show that the function f (x) = x 3 − 6x 2 + 21x + 3 is
increasing at every point.
• Find all points on the graph of f (x) = x 2 whose tangent
line passes through the point (0, −1).
Calculus I,
Columbia
University,
Fall 2014
The Derivative as a Function
Instructor:
Paul Siegel
So far we have only considered the derivative of a function at a
single point x0 . But many functions are differentiable at lots of
points in their domain; some functions (e.g. polynomials) are
even differentiable everywhere.
Given a function f , we can define a new function f 0 by defining:
f 0 (x) = slope of tangent line to f at x
(if the tangent line exists)
f 0 is simply called the derivative of f . The domain of f 0 is the
set of all points where f is differentiable; note that this can be
smaller than the domain of f .
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Example: if f (x) = x n then f 0 (x) = nx n−1 .
Example: find the derivative of f (x) = |x|.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Key idea: relate properties of f to properties of f 0 .
Theorem
Let f be a function which is differentiable at every point in an
interval [a, b].
• If f 0 (x) > 0 for all x ∈ [a, b] then f is increasing on [a, b],
meaning f (x) < f (y ) whenever x < y .
• If f 0 (x) < 0 for all x ∈ [a, b] then f is decreasing on [a, b],
meaning f (x) > f (y ) whenever x < y .
(Similarly, if f 0 (x) ≥ 0 then f is non-decreasing and if
f 0 (x) ≤ 0 then f is non-increasing.)
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Applications:
• If f and g are differentiable functions on an interval [a, b]
such that f (a) ≥ g (a) and f 0 (x) ≥ g 0 (x) for all x ∈ [a, b]
then f (x) ≥ g (x) because f − g is an increasing function
which satisfies (f − g )(a) ≥ 0.
• A continuous function f on an interval [a, b] is invertible if
and only if it is increasing or decreasing. So if f is
differentiable on [a, b] and f 0 (x) > 0 for all x or f 0 (x) < 0
for all x then f is invertible.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Let f (x) = x 3 − 3x 2 and g (x) = 23 x 2 − 27
2 . Show that
f (x) ≥ g (x) for x ≥ 3.
• Find the largest interval containing x = 12 on which
f (x) = x 4 − 2x 2 is invertible.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
If f is differentiable at x, it may or may not be the case that f 0
is also differentiable at x, but if it is we can take the derivative
of f 0 at x and call it f 00 (x), the second derivative of f .
In general, we denote by f (n) the function obtained by
differentiating f repeatedly n times. Note that the existence of
the nth derivative for any n does not guarantee the existence of
the (n + 1)st derivative.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Comment: the notation f 0 (x), f 00 (x), and so on for the
derivatives of a function f is due to Newton (more or less).
Leibniz, who developed calculus close to the same time but
independently of Newton used different notation.
2
d y
0
Setting y = f (x), he wrote dy
dx instead of f (x), dx 2 instead of
00
f (x), and so on. This notation emphasizes the fact that the
derivative calculates slopes of tangent lines and helps clarify
certain properties of derivatives.
In this class we will pass freely back and forth between
Newton’s and Leibniz’s notation. In other classes you may see
other notational conventions for derivatives.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Physical interpretation: assume f (t) represents the total
distance travelled by a moving object at time t. Then:
• f 0 (t) represents the velocity of the object at time t (the
rate at which total distance changes)
• f 00 (t) represents the acceleration of the object at time t
(the rate at which velocity changes)
Calculus I,
Columbia
University,
Fall 2014
The plan
Instructor:
Paul Siegel
We already know how to differentiate polynomial functions.
The next step is to understand how to differentiate more
functions. First:
• Differentiate logarithmic and exponential functions
• Differentiate trigonometric functions
Then:
• Differentiate products and quotients
• Differentiate compositions and inverses
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Derivatives of logarithm and
exponential functions
Logarithmic and exponential functions come up very naturally
in calculus because their derivatives have a particularly simple
form. Let’s start with f (x) = loga (x).
loga (x + h) − loga x
h→0
h
1
x +h
= lim loga
h→0 h
x
h 1/h
= lim loga 1 +
h→0
x
f 0 (x) = lim
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Substituting k = xh , we get:
f 0 (x) = lim loga (1 + k)1/(kx)
k→0
1
1/k
= loga lim (1 + k)
k→0
x
Theorem
lim (1 + k)1/k = e = 2.71828 . . .
k→0
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Note that loga e =
loge e
loge a
=
1
ln a ,
so we may write:
d
1 1
loga x =
dx
loge a x
Hence,
d
1 1
loga x =
dx
ln a x
In particular:
d
1
ln x =
dx
x
Calculus I,
Columbia
University,
Fall 2014
Now we’ll differentiate f (x) = ax . We have:
Instructor:
Paul Siegel
ah − 1
ax+h − ax
= ax lim
h→0
h→0
h
h
f 0 (x) = lim
Homework:
ah − 1
= ln a
h→0
h
lim
Consequently,
d x
a = ln a · ax
dx
In particular,
d x
e = ex
dx
Thus the function f (x) = e x is its own derivative. In fact, e x is
the only function with this property, and because of this it is
ubiquitous in mathematical models of real world systems.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Find the second derivative of the function
f (x) = 3x − x 3 − x.
• Does the curve y = 2e x + 3x + 5x 2 have a tangent line
with slope 2?
• For any real number c > 0, show that e cx ≥ cx + 1
whenever x ≥ 0.
Calculus I,
Columbia
University,
Fall 2014
The product rule
Instructor:
Paul Siegel
As we saw before, the process of differentiation distributes over
addition and scalar multiplication:
df
dg
d
(af + bg ) = a
+b
dx
dx
dx
However, differentiation does NOT in general distribute over
products:
d
df dg
(fg ) 6=
dx
dx dx
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Still, it is possible to express the derivative of fg in terms of the
derivatives of f and g . Look at the difference quotient for fg :
f (x + h)g (x + h) − f (x)g (x)
h
g (x + h) − g (x)
f (x + h) − f (x)
+ g (x)
h
h
Take a limit as h → 0 and we get:
= f (x + h)
(fg )0 (x) = f (x)g 0 (x) + g (x)f 0 (x)
This is called the product rule or sometimes the Leibniz rule.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Differentiate (3x 2 − 2) ln x.
• Differentiate (x − 1)e x
• Given that f (0) = 1, f 0 (0) = 2, and f 00 (0) = 3, calculate
g 00 (0) where g (x) = x 2 f (x).
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
From the product rule one can derive a formula for the
derivative of the quotient of two functions (homework):
0
g (x)f 0 (x) − f (x)g 0 (x)
f
(x) =
g
g (x)2
Examples:
2
• Find the derivative of xx 3 −2
+1
x +x
• Find the derivative of lne x+1
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Derivatives of Trigonometric
Functions
We now know how to differentiate polynomials, exponential
functions, logarithms, and products / quotients of these
functions.
Next: trigonometric functions. We’ll start with sine and cosine,
and get the rest using the quotient rule.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Let’s differentiate f (x) = sin x. We will use the angle sum
formula for sine:
sin(x + h) = sin x cos h + cos x sin h
This gives:
sin(x + h) − sin x
h→0
h
sin x cos h + cos x sin h − sin x
= lim
h→0
h
sin h
cos h − 1
+ cos x lim
= sin x lim
h→0 h
h→0
h
f 0 (x) = lim
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
So in order to differentiate sin x we need to calculate two limits:
cos h − 1
h→0
h
lim
and
lim
h→0
sin h
h
We’ll start by computing the second limit using some geometry
and the squeeze law.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
• Area of triangle 0BC : 21 sin θ cos θ
sin θ
• Area of triangle 0AD: 21 cos
θ
• Area of sector 0AB: 21 θ
Calculus I,
Columbia
University,
Fall 2014
Comparing areas, we get:
Instructor:
Paul Siegel
1
1 sin θ
1
sin θ cos θ ≤ θ ≤
2
2
2 cos θ
Hence:
cos θ ≤
As θ → 0, cos θ → 1 and
θ
1
≤
sin θ
cos θ
1
cos θ
→ 1. By the squeeze theorem:
θ
=1
θ→0 sin θ
lim
Consequently,
sin θ
=1
θ→0 θ
lim
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Homework:
cos θ − 1
=0
θ→0
θ
lim
Therefore:
sin x lim
h→0
Conclusion:
cos h − 1
sin h
+ cos x lim
= cos x
h→0 h
h
d
sin x = cos x
dx
Calculus I,
Columbia
University,
Fall 2014
To differentiate cosine, use the identity:
Instructor:
Paul Siegel
cos(x + h) = cos x cos h − sin x sin h
Get:
d
cos(x + h) − cos x
cos x = lim
h→0
dx
h
cos x cos h − sin x sin h − cos x
= lim
h→0
h
cos h − 1
sin h
= cos x lim
− sin x lim
h→0
h→0 h
h
We calculated these limits earlier, so we get:
d
cos x = − sin x
dx
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Find the equation of the tangent line to f (x) = x cos x at
x = 0.
• Differentiate f (x) = cot x.
• Show that sin(x) ≤ x for 0 ≤ x ≤ π2 .
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Table of derivatives of trigonometric functions (using the
quotient rule):
d
• dx
sin x = cos x
d
• dx
cos x = − sin x
d
• dx
tan x = sec2 x
d
• dx
cot x = − csc2 x
d
• dx
sec x = sec x tan x
d
• dx
csc x = − csc x cot x
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Composition of functions and the
chain rule
Recall: if f : A → B and g : B → C are two functions, the
composition of f and g is the function g ◦ f : A → C given by
(g ◦ f )(a) = g (f (a))
The chain rule is a tool for relating the derivative of g ◦ f to
the derivatives of g and f .
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Chain rule: (g ◦ f )0 (x) = g 0 (f (x)) · f 0 (x)
In Leibniz notation:
dy
dy du
=
·
dx
du dx
where y = g (f (x)) and u = f (x).
(The chain rule is easy to remember in Leibniz notation
because it looks as if one simply ”cancels” the du’s. However
du does not have any independent meaning, so this doesn’t
actually make sense.)
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples: find the derivatives of each of the following
functions
• y = sin(x 2 )
• y = sin2 x
• y = sinh(e t ) cosh(e −t )
• y =
t−2
2t+1
9
2
• y = e 3 sec(t )
Calculus I,
Columbia
University,
Fall 2014
Implicit differentiation
Instructor:
Paul Siegel
All of our techniques for calculating derivatives so far rely on
having an explicit formula for the function f (x) to be
differentiated.
Sometimes functions are defined implicitly by an equation, such
as:
x2 + y2 = 1
(Technical remark: it is not at the outset obvious that there is
a function y = f (x) which satisfies an equation such as the one
above. In most of the examples we will consider the existence
of such a function is guaranteed by a very hard theorem called
the implicit function theorem; we will simply ignore this issue.)
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
To calculate the derivative of an implicitly defined function,
simply differentiate both sides of its defining equation with
respect to x:
d
d 2
(x + y 2 ) =
(1)
dx
dx
d 2
2x +
(y ) = 0
dx
Since we think of y as a function of x, we must use the chain
rule to differentiate y 2 :
d 2
dy
(y ) = 2y
dx
dx
Hence,
dy
=0
dx
dy
x
=−
dx
y
2x + 2y
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Find the derivative of the function implicitly defined by the
equation e x/y = x − y
• Find an equation of the tangent line to the curve
x 2 + xy + y 2 = 3 at the point (1, 1)
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
As a special case, suppose a function f is invertible and let g
denote its inverse. Writing y = g (x), we see that f is implicitly
defined by the equation f (y ) = x. Differentiating, we get:
f 0 (y )
But since y = g (x) we have
g 0 (x) =
dy
=1
dx
dy
dx
= g 0 (x), so:
1
f 0 (y )
=
1
f 0 (g (x))
This is a useful formula for the derivative of the inverse of a
function. This formula is part of the inverse function theorem.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Use the inverse function theorem to calculate the
derivative of f (x) = ln x.
• Use the inverse function theorem to calculate the
derivative of f (x) = tan−1 (x).
• Find an equation of the tangent line to the graph of
f (x) = cosh−1 (x) at (1, 0).
Calculus I,
Columbia
University,
Fall 2014
Recap: Basic examples
Instructor:
Paul Siegel
d p
• dx
x = px p−1 for any real number p
d
• dx
loga x = ln1a x1
d x
• dx
a = ln a · ax
d
d
• dx
sin x = cos x and dx
cos x = − sin x
Calculus I,
Columbia
University,
Fall 2014
Recap: Techniques
Instructor:
Paul Siegel
(x)
• The definition: f 0 (x) = limh→0 f (x+h)−f
h
• Basic laws: (f + g )0 (x) = f 0 (x) + g 0 (x) and
(cf )0 (x) = cf 0 (x)
• Product rule: (fg )0 (x) = f (x)g 0 (x) + g (x)f 0 (x)
• Quotient rule:
0
f
g
g (x)f 0 (x)−f (x)g 0 (x)
g (x)2
0
g (f (x)) · f 0 (x)
(x) =
• Chain rule: (g ◦ f )(x) =
1
• Inverse function theorem: (f −1 )0 (x) = f 0 (f −1
(x))
Calculus I,
Columbia
University,
Fall 2014
Applications of Differentiation
Instructor:
Paul Siegel
With the basic properties of derivatives in hand, we now turn
to some applications. Specifically, we will use derivatives to:
• Reason with rates of change
• Analyze the geometry of curves
• Solve optimization problems
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Rates of change
Suppose a variable x changes over time, so that x is a function
of a time variable t: x = x(t). Then x 0 (t) represents the rate
at which x is changing at the time t.
Now suppose y is another variable which depends on x, so that
y = f (x) for some function f . Using the chain rule, we can
express the rate of change of y in terms of the rate of change
of x:
y 0 (t) = f 0 (x)x 0 (t)
or in Leibniz notation:
dy
dy dx
=
dt
dx dt
Of course, if y is implicitly defined in terms of x using an
equation, we could use implicit differentiation to achieve a
similar result.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples:
• Air is pumped into a spherical balloon at a rate of
100cm3 /2. How quickly is the radius changing when the
diameter is 50cm?
• A 10 foot latter rests against a wall and is pulled away so
that the bottom of the latter slides at a rate of 1 foot per
second. How fast is the top of the ladder moving when the
bottom is 6 feet away from the wall? What about when it
is 10 feet away?
• The length of a rectangle increases at 8cm/s while the
width increases at 3cm/s. When the length is 20cm and
the width is 10cm, how fast is the area changing?
• You’re riding a 100 foot tall ferris wheel which completes
two full revolutions per minute. How fast are you
ascending when you are 75 feet above the ground?
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
General strategy for ”related rates” problems:
1
Assign names to the relevant variables, and try to
understand (e.g. with a picture) how they are related.
2
Formulate a model (i.e. an equation or set of equations)
which relates the variables to each other.
3
Specify which quantities and rates of change are given and
which are required
4
Use the model together with implicit differentiation to
solve for the unknown quantities and rates of change.
Calculus I,
Columbia
University,
Fall 2014
Optimization
Instructor:
Paul Siegel
Question: given a continuous function f (x) what are the
largest and smallest values that it takes?
This question, as it is posed above, does not necessarily have
an answer: for instance, the function f (x) = x 3 takes arbitrarily
large positive and negative values.
Calculus I,
Columbia
University,
Fall 2014
However, if we are only interested in values of x which lie in a
closed interval [a, b], then for all such x we have that
Instructor:
Paul Siegel
x 3 ≥ a3
and
x 3 ≤ b3
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Of course, the maximum or minimum value of a function on a
closed interval does not necessarily occur at one of the
endpoints of the interval:
Calculus I,
Columbia
University,
Fall 2014
Recall:
Instructor:
Paul Siegel
Theorem (Extreme Value Theorem)
Let f : [a, b] → R be a continuous function defined on a closed
interval. Then there are numbers xmin and xmax in [a, b] such
that
f (xmin ) ≤ f (x) ≤ f (xmax )
for every x in [a, b].
(Note: there may be many possible choices for xmin and xmax ;
the theorem just asserts that f takes its minimum and
maximum value somewhere.)
The theorem doesn’t say anything about how to actually find
xmin or xmax . For an arbitrary continuous function this is
basically hopeless, but if f is differentiable then we have some
techniques available.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Definition
A function f has a local maximum at a point c if f (x) ≤ f (c)
for all x sufficiently close to c. Similarly, f has a local
minimum at c if f (x) ≥ f (c) for all x sufficiently close to c.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Theorem
Let f be a differentiable function. If f has a local maximum or
local minimum at a point c, then f 0 (c) = 0
Warning 1: f can also have local maxima or minima at points
where it is not differentiable (e.g. f (x) = |x|)
Warning 2: The converse to the theorem is false: the derivative
of x 3 at 0 is 0, but x 3 does not have a local maximum there.
Definition
If f is a differentiable function and c is a point such that
f 0 (c) = 0, then c is called a critical point of f .
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples: find the global maximum and minimum values of
each of the following functions on the given interval:
• f (x) = x 3 − 6x 2 + 5, [−3, 5]
x
• f (x) = x 2 −x+1
, [0, 3]
• f (x) = 2 cos t + sin(2t), [0, π/2]
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
More entertaining examples:
• You need to make an open box whose base is a square
which holds 25 cubic feet of water. What is the least
amount of material needed to build the box?
• Find an equation of the line through the point (3, 5) which
cuts the smallest possible area from the first quadrant.
• In the following diagram, where should P be chosen to
maximize θ? (Assume a = 2, b = 3, c = 5.)
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
General strategy for ”optimization” problems:
1
Identify the quantity to be optimized (maximized or
minimized) and the variable(s) on which it depends.
2
Find a model which expresses the quantity to be optimized
in terms of the other variables. Simplify the model so that
it expresses the quantity as a function of just one
independent variable.
3
Find the domain of the function that you constructed in
the previous step. Analyze the behavior of the function
near the endpoints of the domain (taking limits if
necessary).
4
User derivatives to find all local extrema.
Calculus I,
Columbia
University,
Fall 2014
Calculus and Geometry
Instructor:
Paul Siegel
What can we infer about the shape of the graph of a function
given information about its derivatives?
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
We have already observed that the sign of the derivative
determines whether the function is increasing or decreasing:
Theorem
Let f (x) be a differentiable function on an open interval (a, b).
• If f 0 (x) > 0 for all x in (a, b) then f is increasing:
f (x) < f (y ) if x < y .
• If f 0 (x) < 0 for all x in (a, b) then f is decreasing:
f (x) > f (y ) if x < y .
We have also seen that local maxima / minima can be detected
using derivatives:
Theorem
Let f (x) be a differentiable function on an open interval (a, b).
If f has a local maximum or minimum at a point c in (a, b),
then f 0 (c) = 0.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
The second derivative of a function is related to the concavity
of its graph.
Definition
Let f be a continuous function.
• f is concave up on (a, b) if the line between (x, f (x)) and
(y , f (y )) lies above the graph of f for every x, y in (a, b).
• f is concave down on (a, b) if the line between (x, f (x))
and (y , f (y )) lies below the graph of f for every x, y in
(a, b).
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Theorem
Let f be a twice differentiable function on an open interval
(a, b).
• If f 00 (x) > 0 for every x in (a, b) then f is concave up on
(a, b).
• If f 00 (x) < 0 for every x in (a, b) then f is concave down
on (a, b).
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Critical points of a function f are useful because they might
indicate where f changes from increasing to decreasing (or
vice-versa). It is also useful to look for points where f changes
from accelerating to decelerating (or vice versa).
Definition
Let f be a continuous function on an open interval (a, b). A
point c in (a, b) is an inflection point if the concavity of f
changes at c. In other words, there is some ε > 0 such that f
is concave up on (c − ε, c] and concave down on [c, c + ε) (or
vice-versa).
Theorem
Let f be a twice differentiable function on an open interval
(a, b). If c in(a, b) is an inflection point then f 00 (c) = 0.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
If f 0 (c) = 0 and f 00 (c) = 0, anything can happen:
• f (x) = x 3 :
• f (x) = x 4 :
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Using what we now know about the first two derivatives, we
can make a reasonably detailed sketch of a (twice
differentiable) function f (x):
1
2
3
Find all x intercepts, critical points, and inflection points
of f .
Calculate values of f 0 between the critical points to
determine where f is increasing and decreasing.
Calculate values of f 00 between the inflection points to
determine where f is concave up and concave down.
4
Determine the end behavior of f .
5
Find and classify vertical asymptotes (if any).
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples: sketch graphs of each of the following functions.
• f (x) = x(x − 4)2
4
• f (x) = xx 2 +4
+1
2
x
• f (x) = 1−x
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
An application of concavity: if c is a critical point for f and f
is concave down in a neighborhood of c then f (x) must be
smaller than f (c) for x close to c. Thus, f is locally maximized
at c.
Similarly, if f is concave up in a neighborhood of c then f is
locally minimized at c.
Theorem (Second derivative test)
Suppose that f is twice differentiable near a point c and that
f 0 (c) = 0.
• If f 00 (c) < 0 then f has a local maximum at c.
• If f 00 (c) > 0 then f has a local minimum at c.
Calculus I,
Columbia
University,
Fall 2014
Instructor:
Paul Siegel
Examples: classify all critical points for each of the following
functions: