Monday, January 9, 2017
Tangent and Secant Lines
• Suppose you are given some curve, in this
case a circle, and a point P.
L
P
Q1
L1
• The line L passing through the point P is
called a tangent line.
• Given another point Q1 on the curve, let L1
be the line passing through the points P
and Q1.
• L1 is called a secant line and is at times
written PQ1.
Monday, January 9, 2017
Tangent and Secant Lines
L2
L
• Pick a point Q2 closer to P.
Q2
P
Q1
• Notice that the slope of the secant line L2 is
closer to the slope of L compared to the
slope of L1.
L1
• In fact, as we pick points closer and closer
to P, the slope of the associated secant
lines will get closer and closer to the slope
of L.
Q: So, how can we differentiate between secant lines and
tangent lines? Is it enough to say that a tangent line goes
through one point, while a secant line goes through two
points?
A: No!
Tangent and Secant Lines
• L is the tangent line to P,
even though it intersects
the curve more than
once.
P
L
• However, we can remedy
this by saying that there is
a neighborhood of the
point P for which L
doesn’t intersect the
curve at any other point
within this neighborhood.
Q: Is this enough? Do we have enough
information now to distinguish between
tangent and secant lines?
A: Not quite.
Tangent and Secant Lines
L1
P
• L1 is a secant line for the
curve through the point P.
• We can find a
neighborhood of P such
that L1 doesn’t intersect
any other point of the
curve within the
neighborhood.
What then is the real distinction between
tangent lines and secant lines?
Tangent and Secant Lines
P
• If we zoom in on the point P, the
curve at P starts to look more and
more like the tangent line at P.
L
• We can think of the tangent line as
a linear approximation of the curve
at P.
• The notion of linear approximation
is a really powerful tool in math. We
will study it a bit later in the course.
Warning: The tangent line to a curve at a given point
may NOT exist. This is equivalent to saying that there
is no good linear approximation of the curve at the
given point.
Ex: Cusps
P
L
No matter how close we
zoom in, L will never give
a linear approximation of
the curve at the point P.
Tangent and Secant Lines
L
P
• If we zoom in on the point P, the
curve at P starts to look more and
more like the tangent line at P.
• We can think of the tangent line as
a linear approximation of the curve
at P.
• The notion of linear approximation
is a really powerful tool in math. We
will study it a bit later in the course.
P
L1
• A secant line through the point P,
on the other hand, doesn’t give a
linear approximation of the curve
at P.
• Nonetheless, secant lines have
their use!
Quick Review of Lines
L
• Recall that the equation of a
line is of the form y = mx + b,
where m is the slope and b is
the y-intercept.
• Let L be a line, (x1,y1) and
(x2,y2) two points on L such
that x2 > x1.
y2
(x2,y2)
y2 – y1
y1
(x1,y1)
• The slope m = (y2 - y1)/(x2 – x1).
• To find the y-intercept, plug in
either (x1,y1) or (x2,y2) into the
equation y = mx + b and solve
for b.
x2 – x1
x1
x2
Quick Review of Lines
Ex: Find the equation of the line
passing through the points (5,7)
and (1,8).
• Since 5 > 1, the slope is given
by:
m = (7-8)/(5-1) = -1/4
• Thus, the equation of the line is
going to be of the form:
y = (-1/4)x + b
• Since the point (1,8) is on the
line, it must be that
8 = (-1/4)(1) + b
which means that
b = 33/4
• Hence, the equation of the line
is given by y = (-1/4)x + 33/4
Quick Review of Lines: Class Problems.
Question 1: What is the slope of the line passing through (1,1) and
(2,4)?
Question 2: What is the slope of the line passing through (1,1) and
(a,a2), for a ≠ 1?
Calculating Tangent and Secant Lines
f
f(b)
• Suppose you are given the graph of a
function f and a secant line through
points (a,f(a)) and (b,f(b)).
• We can calculate the equation of the
secant line using the technique we
reviewed in the previous slide.
f(a)
• The slope is given by:
m=(f(b) – f(a))/(b – a)
a
b
Calculating Tangent and Secant Lines
f
f(a + h)
• Suppose you are given the graph of a
function f and a secant line through
points (a,f(a)) and (b,f(b)).
• We can calculate the equation of the
secant line using the technique we
reviewed in the previous slide.
f(a)
• The slope is given by:
m=(f(b) – f(a))/(b – a)
a
a+h
• Since b = a + h, for some quantity h, we
can replace the formula for the slope by
m = (f(a + h) – f(a))/((a + h) – a)
= (f(a + h) – f(a))/h
Slope of the secant line =
Calculating Tangent and Secant Lines
Slope of the secant line =
f(a)
a
• If we allow the quantity h to get smaller, the
point (a+h,f(a+h)) is getting closer to
(a,f(a)), which in turn means that the slope
of the secant line through (a+h,f(a+h)) is
getting closer to the slope of the tangent
line at (a,f(a)).
• Thus, the slope of the tangent line is given
by taking the limit as h goes to 0.
Slope of the tangent line at (a,f(a)) =
Calculating Tangent and Secant Lines
Ex: Calculate the equation of the tangent
line of f(x)=x3 at the point (2,8).
• The slope of the tangent line is given by:
limh→0
! !!! !!(!)
= limh→0
= limh→0
!
• Given that the slope is 12, we can
plug in the point (2,8) to find the y(2 + ℎ)! −2!
intercept
ℎ
8 = 12(2) + b, and so
b = -16
!!!"#! !"! ! !! !!
= limh→0
!
!"#! !"! ! !!
!
= limh→0 (12 + 6h + h2)
= 12
• Thus, the equation of the line is given
by y = 12x -16.
Calculating Tangent and Secant Lines:
Class Problems
Consider the function f(x) = x2
Question 3: What is the slope of the line passing through (1,1) and (1 + h,(1 + ℎ)! )?
Question 4: What is the slope and equation of the tangent line at (1,1)?
Question 5: What is the slope and equation of the tangent line at (a,a2)?
Limits of Functions
L
f
• Given a function f, and a
number a, the limit of f as x
approaches a is the value, L, f(x)
approaches as x goes to a.
• In this case, we write
limx→a f(x) = L
a
Limits of Functions
f(a)
L
f
a
Note that L doesn’t have to
be the same as f(a).
Limits of Functions
• We can take limits from the
left hand side and the right
hand side.
f
L2
• Here, the limit from the left
hand side is L1. We denote
this by
limx→a- f(x) = L1
L1
• The limit from the right
hand side is L2. We denote
this by
a
limx→a+ f(x) = L2
Note: If, as in this picture, the left hand limit and the right hand
limit as x approaches a are different, then the limit of f as x
approaches a does not exist.
Limits of Functions
• It’s possible that the limit of f
as x approaches a is not a
number.
• In this picture
limx→a f(x) = ∞
f
a
• The line x = a is an example
of a vertical asymptote.
• In general, we say that f has a
vertical asymptote at x = a if
the limit as x approaches a
from the left, right, or both is
∞ or -∞.
Limits of Functions
In this picture, we have
limx→a- f(x) = -∞
limx→a+ f(x) = ∞
f
And so
limx→a f(x) does not exist.
a
Limits of Functions
Some Limit Laws
• limx→a [f(x) + g(x)] = limx→a f(x) + limx→a g(x)
• limx→a [f(x) - g(x)] = limx→a f(x) - limx→a g(x)
• limx→a [f(x) · g(x)] = limx→a f(x) · limx→a g(x)
• limx→a
!(!)
!(!)
=
limx→a f(x)
, as long as limx→a g(x) ≠ 0
limx→a g(x)
• limx→a [𝑓 𝑥 ]! = [limx→a f(x)]! for n ∈ ℤ
Limits of Functions
Limit Theorems
Direct Substitution property: If f is a polynomial or a
rational function defined at a, then limx→a f(x) = f(a)
Theorem: If f(x) ≤ g(x) in some neighborhood of a
and limx→a f(x) and limx→a g(x) exist, then
limx→a f(x) ≤ limx→a g(x).
g
f
a
Limits of Functions
Limit Theorems
Squeeze Theorem: If f(x) ≤ h(x) ≤ g(x)
in some neighborhood of a and
limx→a f(x) = limx→a g(x) = L,
then limx→a h(x) = L
a
Limits of Functions: Class Problems
Question 6:
Next Class
• Next time, we will give a precise definition of limits.
• Make sure to visit Canvas and do the assignments for
next class.
Math 3 Group Work Solutions, 1/9
1. What is the slope of the line passing through (1, 1) and (2, 4)? [Answer: 3]
Solution:
The slope of the line is
y2 y 1
x2 x1
=
4 1
2 1
=
3
1
= 3.
2. What is the slope of the line passing through (1, 1) and (a, a2 ), for a 6= 1?
[Answer: a + 1]
Solution:
The slope of the line is
y2 y 1
x2 x1
=
a2 1
a 1
=
(a+1)(a 1)
.
a 1
Since a 6= 1, this is equal to a + 1.
3. What is the slope of the line passing through (1, 1) and (1 + h, (1 + h)2 ), where
h > 0? [Answer: h + 2]
Solution:
The slope of the line is
y 2 y1
x2 x1
=
(1h )2 1
1+h 1
=
h2 +2h
.
h
Since h 6= 01, we can rewrite this as h + 2.
4. What is the slope and equation of the tangent line at (1, 1)? [Answer: 2; f (x) =
2x 1]
Solution:
We can look at the slope of the line through (1, 1) and (1 + h, (1 + h)2 ) as h gets closer
2
and closer to 0, In problem 3, we calculated this to be h +2h
, and this goes to 2 as h gets
h
closer to 0. Thus, the slope is 2.
Now, we know the slope of our desired line is 2, we know a point on it, namely, (1, 1), so
we can write the equation in point-slope form y 1 = 2(x 1), or y = 2x 1. Thus the
equation of the tangent line is f (x) = 2x 1.
5. What is the slope and equation of the tangent line at (a, a2 )? [Answer: 2a; f (x) =
2ax a2 ]
Solution:
Using the same method as for the point (1, 1), we look at the slope of the line through
(a, a2 ) and (a + h, (a + h)2 ). In other words, we look at the slope of the secant lines through
the points (a, a2 ) and (a + h, (a + h)2 ), a point very close to (a, a2 ). The slope of this line is
2
(a+h)2 a2
= 2ah+h
. This gets closer and closer to 2a as h gets closer to 0. Thus, the slope is
a+h a
h
2a.
1
Again, using point slope form, we know that the point on the tangent line is (a, a2 ) and
the slope is 2a, so we get y a2 = 2a(x a), or y = 2ax a2 . Thus, the equation of the
tangent line at (a, a) is f (x) = 2ax a2 .
6. Determine whether the following limits exist and if so, calculate the value:
i. limx!2 2x3 + 1 [answer: 17]
ii. The limit as x ! 1: [answer: -1]
iii. limx!⇡ sin(1/x) [answer:does not exist]
iv. limx!0 1/x [answer: does not exist]
v. limx!1
x3 6x2 +3x+10
x 2
[answer: -8]
Solution:
i. This is a polynomial that is defined at x = 2, so here, we can just plug in x = 2 to get
2(23 ) + 1 = 17.
ii. Again here, we can just look at the value of f (1), which is
points with x-value near 1 have f (x) near 1.
1. This is because the
iii. The limit does not exist: we can always find values of x close to 0 with sin(1/x) = 1
and values of x close to 0 with sin(1/x) = 0, namely, for any n, we can take x = 1/⇡n, and
sin(1/x) = sin(⇡n) = 0. But, for m odd, we can take x = 2/⇡m, and sin(1/x) = 1 or 1.
The values of m and n can be as big as we want, so x can be as close to 0 as we want and
we will get very di↵erent values of f (x).
2
iv. The value of 1/x gets arbitrarily large as x gets closer and closer to 0, so it does not
approach any limit point. This limit does not exist.
v. This one is easiest to do by plugging in values of x close to 1 to see what value f (x)
takes. However, a systematic way to do this is to factor the polynomial in the numerator as
5)
(x 2)(x + 1)(x 5). Then, we calculate limx!1 (x 2)(x+1)(x
= lim(x!1 (x + 1)(x 5) =
x 2
(1 + 1)(1 5) = 2( 4) = 8. We can do this because we are only concerned about points
near x = 1.
3