AP Calculus BC
Thursday, 03 September 2015
• OBJECTIVE TSW (1) determine continuity at a point
and continuity on an open interval; (2) determine onesided limits and continuity on a closed interval; (3) use
properties of continuity; and (4) understand and use the
Intermediate Value Theorem.
• T-Shirt sales:
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2XL – 4XL
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– Due by Tuesday, 08 September 2015.
Sec. 2.6: Continuity
Sec. 2.6: Continuity
Informally, a function f is continuous at x = c if
there is no “interruption” in the graph of f at
x = c.
That is, the graph of f has no holes, jumps, or
gaps at c.
Sec. 2.6: Continuity
Sec. 2.6: Continuity
Sec. 2.6: Continuity
Types of discontinuities:
Hole
Gap
Hole
NOTE: The third type of discontinuity is an asymptote.
Sec. 2.6: Continuity
Ex:
Discuss the continuity of the following:
1
a) f  x  
x
f has a domain of (–∞, 0) ᴜ (0, ∞), so f is continuous
at every x-value in its domain.
f has a nonremovable discontinuity (asymptote) at
x = 0.
Sec. 2.6: Continuity
Ex:
Discuss the continuity of the following:
x2  1
b) g  x  
x 1
g has a domain of (–∞, 1) ᴜ (1, ∞), so g is
continuous at every x-value in its domain.
g has a removable discontinuity (hole) at
x = 1.
Sec. 2.6: Continuity
Ex:
Discuss the continuity of the following:
 x  1, x  0
c) h  x    2
 x  1, x  0
h has a domain of (–∞, ∞), so h is continuous at
every x-value.
h is everywhere continuous.
Sec. 2.6: Continuity
Ex:
Discuss the continuity of the following:
d) y  sin x
y has a domain of (–∞, ∞), so y is continuous at
every x-value.
y is everywhere continuous
Sec. 2.6: Continuity
One-Sided Limits
A one-sided limit considers only one side of a
given value.
Ex:
The limit from the right means that x
approaches c from values greater than c.
lim  f  x   L
x c
A similar meaning is given to the limit from
the left.
lim f  x   L
x  c
Sec. 2.6: Continuity
Ex:
Find the following:
lim
x   2
4  x2
lim
x   2
4  x2  0
Sec. 2.6: Continuity
Ex:
Using the definition of continuity,
determine if f is continuous at x = 2.
 x  4, x  2
f x  
3 x  1, x  2
i) f  2   6, so f  2  is defined
ii)
lim  f  x   6
x 2

lim  f  x   7  lim f  x  DNE
x2
 f is not continuous at x  2.
x 2
Sec. 2.6: Continuity
Ex:
Using the definition of continuity,
determine if f is continuous at x = 2.
 x  5, x  2
f x  
3 x  1, x  2
i) f  2   7, so f  2  is defined
ii)
lim  f  x   7
x 2

lim  f  x   7  lim f  x   7
x2
iii) f  2  lim f  x 
x 2
 f is continuous at x  2.
x 2
lim f  x  exists
x 2
Sec. 2.6: Continuity
The two one-sided limits must equal each other.
Sec. 2.6: Continuity
Sec. 2.6: Continuity
Functions that are continuous at every point in
their domain:
a)
b)
c)
d)
Polynomial functions
Rational functions
Radical functions
Trigonometric functions
Sec. 2.6: Continuity
Sec. 2.6: Continuity
Sec. 2.6: Continuity
Combining these properties with continuous
functions allows you to state that many
functions are continuous.
Ex:
State why the following are continuous:
a)
f(x) = x2 + 1 – sin x
f is the sum of a polynomial and a
trig function
Sec. 2.6: Continuity
Ex:
State why the following are continuous:
f (x)  x  4
b)
f is a composition of a radical
function and a polynomial
Sec. 2.6: Continuity
Ex:
State why the following are continuous:
c)
f ( x )  ( x 3  3 x )(tan x )
f is the product of a polynomial
function and a trig function
Sec. 2.6: Continuity
Sec. 2.6: Continuity
f is continuous on [a, b].
 three c’s  f(c) = k.
f is not continuous on [a, b].
 no c’s  f(c) = k.
Sec. 2.6: Continuity
Ex:
Use the Intermediate Value Theorem to
show that f(x) = x3 +2x – 1 has at least one
zero in [0, 1].
i)
ii)
f(x) is continuous (because it’s a polynomial)
f(0) = –1
f(1) = 2
f(0) < 0 < f(1)
By the IVT,  at least one zero in [0, 1]  f(c) = 0.
Sec. 2.6: Continuity
Ex:
(a) Verify that the Intermediate Value
Theorem applies in the indicated interval for
f(x). (b) Find the value of c guaranteed by
the theorem.
f(x) = x2 – 6x + 8, [0, 3], f(c) = 0
(a)
i) f is continuous on [0, 3] (polynomial)
ii) f(0) = 8 and f(3) = –1
f(0) > f(c) = 0 > f(3)
By the IVT,  at least one zero in [0, 3]
 f(c) = 0.
Sec. 2.6: Continuity
Ex:
(a) Verify that the Intermediate Value
Theorem applies in the indicated interval for
f(x). (b) Find the value of c guaranteed by
the theorem.
f(x) = x2 – 6x + 8, [0, 3], f(c) = 0
(b)
Find c.
x2 – 6x + 8 = 0
(x – 4)(x – 2) = 0
x = 4, x = 2
Since x = 2 is in the interval and f(2) = 0,
c=2