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Continuity
Related Topics
Wize University Calculus 1 Textbook > Limits
Continuity
3 Activities
Find
A
A
A
so that the following function is continuous at
x
=
1
x=1
x
=
1
:
f
(
x
)
=
{
2
x
2
−
x
+
A
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1
{f(x)=\left\{\begin{array}{ll} \displaystyle 2x^2-x+A\quad\quad\,\,\text{if }\,0\leq x\leq 1\\ \\ \displaystyle\frac{x^2+2x-3}{x^2-1}\,\,\,\quad\quad\text{if }\,x> 1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
2
x
2
−
x
+
A
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1
Answer
I don't know
Check Submission
More Continuity Questions:
Continuity
For what values of
a
a
a
is the following function continuous on
(
−
∞
,
∞
)
(-\infty,\infty)
(
−
∞
,
∞
)
.
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
<
2
x
2
−
(
2
+
a
)
x
+
2
a
x
−
1
x
≥
2
f(x)= \begin{cases} ax^3-x^2+\sin(\pi x) & x<2 \\ \frac{x^2-(2+a)x+2a}{x-1} & x\geq 2 \end{cases}
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
−
1
x
2
−
(
2
+
a
)
x
+
2
a
x
<
2
x
≥
2
Continuity
For what values of
a
a
a
is the following function continuous on
(
−
∞
,
∞
)
(-\infty,\infty)
(
−
∞
,
∞
)
.
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
<
2
x
2
−
(
2
+
a
)
x
+
2
a
x
−
1
x
≥
2
f(x)= \begin{cases} ax^3-x^2+\sin(\pi x) & x<2 \\ \frac{x^2-(2+a)x+2a}{x-1} & x\geq 2 \end{cases}
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
−
1
x
2
−
(
2
+
a
)
x
+
2
a
x
<
2
x
≥
2
Continuity
Find
A
A
A
and
B
B
B
so that the following function is continuous on
R
\mathbb{R}
R
:
f
(
x
)
=
{
A
(
1
−
cos
x
sin
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1
{f(x)=\left\{\begin{array}{ll} \displaystyle A\left(\frac{1-\cos{x}}{\sin^{2}{x}}\right)\quad\text{if}\,x<0 \\ \\ \displaystyle 2x^2-x+B\quad\quad\,\,\text{if}\,0\leq x\leq 1\\ \\ \displaystyle\frac{x^2+2x-3}{x^2-1}\,\,\,\quad\quad\text{if}\,x> 1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
A
(
sin
2
x
1
−
cos
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1
Continuity
Consider the piecewise function
h
(
x
)
=
{
C
x
−
17
x
<
10
C
+
3
x
x
=
10
2
C
−
4
x
>
10
h(x) = \begin{cases} \frac{C}{x-17} & x<10 \\ C+3x & x=10 \\ 2C-4 & x>10 \end{cases}
h
(
x
)
=
⎩
⎨
⎧
x
−
17
C
C
+
3
x
2
C
−
4
x
<
10
x
=
10
x
>
10
where
C
is the same real number in each formula.
Continuous Functions
For what values of
a
a
a
and
b
b
b
is the following function continuous? Answers are given in the form
(
a
,
b
)
\left(a,b\right)
(
a
,
b
)
.
f
(
x
)
=
{
x
2
−
2
x
+
a
,
x
≤
0
a
x
+
b
,
0
<
x
<
4
x
+
21
,
4
≤
x
f(x) = \left \{ \begin{array}{ll} x^2 - 2x + a, & x \leq 0\\ ax + b, & 0 < x < 4\\ \sqrt{x + 21}, & 4 \leq x \end{array} \right .
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
x
+
a
,
a
x
+
b
,
x
+
21
,
x
≤
0
0
<
x
<
4
4
≤
x
Continuity
For what values of
a
a
a
is the following function continuous on
(
−
∞
,
∞
)
(-\infty,\infty)
(
−
∞
,
∞
)
.
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
<
2
x
2
−
(
2
+
a
)
x
+
2
a
x
−
1
x
≥
2
f(x)= \begin{cases} ax^3-x^2+\sin(\pi x) & x<2 \\ \frac{x^2-(2+a)x+2a}{x-1} & x\geq 2 \end{cases}
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
−
1
x
2
−
(
2
+
a
)
x
+
2
a
x
<
2
x
≥
2
Continuity
For what values of
a
a
a
is the following function continuous on
(
−
∞
,
∞
)
(-\infty,\infty)
(
−
∞
,
∞
)
.
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
<
2
x
2
−
(
2
+
a
)
x
+
2
a
x
−
1
x
≥
2
f(x)= \begin{cases} ax^3-x^2+\sin(\pi x) & x<2 \\ \frac{x^2-(2+a)x+2a}{x-1} & x\geq 2 \end{cases}
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
−
1
x
2
−
(
2
+
a
)
x
+
2
a
x
<
2
x
≥
2
Special Limit: Continuity
Evaluate
lim
x
→
1
+
arctan
(
e
(
π
x
−
1
)
)
\lim_{x\to1^+}\arctan\left(e^{\left(\frac{\pi}{x-1}\right)}\right)
lim
x
→
1
+
arctan
(
e
(
x
−
1
π
)
)
.
Continuous Functions
For what values of
a
a
a
and
b
b
b
is the following function continuous? Answers are given in the form
(
a
,
b
)
(a,b)
(
a
,
b
)
.
f
(
x
)
=
{
x
2
−
2
x
+
a
,
x
≤
0
a
x
+
b
,
0
<
x
<
4
x
+
21
,
4
≤
x
f(x) = \left \{ \begin{array}{ll} x^2 - 2x + a, & x \leq 0\\ ax + b, & 0 < x < 4\\ \sqrt{x + 21}, & 4 \leq x \end{array} \right .
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
x
+
a
,
a
x
+
b
,
x
+
21
,
x
≤
0
0
<
x
<
4
4
≤
x
Continuity
Use the graph below to find
Continuity
Use the graph below to find
Continuity
Determine the values of
a
a
a
and
b
b
b
such that
f
(
x
)
f(x)
f
(
x
)
is continuous, where
f
(
x
)
=
{
a
x
+
b
,
if
x
≤
1
2
b
+
1
,
if
1
<
x
≤
2
4
a
x
−
13
b
,
if
x
>
2
f(x)= \begin{cases} ax+b & , \text{if } x \leq 1\\ 2b +1& , \text{if } 1 <x\le2 \\ 4ax-13b&, \text{if } x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
x
+
b
2
b
+
1
4
a
x
−
13
b
,
if
x
≤
1
,
if
1
<
x
≤
2
,
if
x
>
2
Continuity of Piecewise Functions
Find the values of 𝑎 and 𝑏 that make 𝑓 continuous everywhere
f
(
x
)
=
{
x
2
−
2
if
x
<
2
a
x
if
2
≤
x
<
3
2
x
−
b
if
x
≥
3
f\left(x\right)= \begin{cases} x^2-2 & \text{if }\space x<2\\ ax & \text{if }\space 2\le x<3\\ 2x-b & \text{if }\space x\ge3 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
a
x
2
x
−
b
if
x
<
2
if
2
≤
x
<
3
if
x
≥
3
Continuous Functions
For what values of
a
a
a
and
b
b
b
is the following function continuous?
f
(
x
)
=
{
x
2
−
2
x
+
a
,
x
≤
0
a
x
+
b
,
0
<
x
<
4
x
+
21
,
4
≤
x
f(x) = \left \{ \begin{array}{ll} x^2 - 2x + a, & x \leq 0\\ ax + b, & 0 < x < 4\\ \sqrt{x + 21}, & 4 \leq x \end{array} \right .
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
x
+
a
,
a
x
+
b
,
x
+
21
,
x
≤
0
0
<
x
<
4
4
≤
x
Special Limit: Continuity
Evaluate
lim
x
→
1
+
arctan
(
e
(
π
x
−
1
)
)
\lim_{x\to1^+}\arctan\left(e^{\left(\frac{\pi}{x-1}\right)}\right)
lim
x
→
1
+
arctan
(
e
(
x
−
1
π
)
)
.
Continuity of Piecewise Functions
Practice: Continuity of Piecewise Functions
Find the values of 𝑎 and 𝑏 that make 𝑓 continuous everywhere
f
(
x
)
=
{
x
2
−
2
if
x
<
2
a
x
if
2
≤
x
<
3
2
x
−
b
if
x
≥
3
f\left(x\right)= \begin{cases} x^2-2 & \text{if }\space x<2\\ ax & \text{if }\space 2\le x<3\\ 2x-b & \text{if }\space x\ge3 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
a
x
2
x
−
b
if
x
<
2
if
2
≤
x
<
3
if
x
≥
3
Continuous Functions
For what values of
a
a
a
and
b
b
b
is the following function continuous?
f
(
x
)
=
{
x
2
−
2
x
+
a
,
x
≤
0
a
x
+
b
,
0
<
x
<
4
x
+
21
,
4
≤
x
f(x) = \left \{ \begin{array}{ll} x^2 - 2x + a, & x \leq 0\\ ax + b, & 0 < x < 4\\ \sqrt{x + 21}, & 4 \leq x \end{array} \right .
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
x
+
a
,
a
x
+
b
,
x
+
21
,
x
≤
0
0
<
x
<
4
4
≤
x
Special Limit: Continuity
Evaluate
lim
x
→
1
+
arctan
(
e
(
π
x
−
1
)
)
\lim_{x\to1^+}\arctan\left(e^{\left(\frac{\pi}{x-1}\right)}\right)
lim
x
→
1
+
arctan
(
e
(
x
−
1
π
)
)
.
Continuity
Determine the values of
a
a
a
and
b
b
b
such that
f
(
x
)
f(x)
f
(
x
)
is continuous, where
f
(
x
)
=
{
a
x
+
b
,
if
x
≤
1
2
b
+
1
,
if
1
<
x
≤
2
4
a
x
−
13
b
,
if
x
>
2
f(x)= \begin{cases} ax+b & , \text{if } x \leq 1\\ 2b +1& , \text{if } 1 <x\le2 \\ 4ax-13b&, \text{if } x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
x
+
b
2
b
+
1
4
a
x
−
13
b
,
if
x
≤
1
,
if
1
<
x
≤
2
,
if
x
>
2
Continuity
Determine the values of
a
a
a
and
b
b
b
such that
f
(
x
)
f(x)
f
(
x
)
is continuous, where
f
(
x
)
=
{
a
x
+
b
,
if
x
≤
1
2
b
+
1
,
if
1
<
x
≤
2
4
a
x
−
13
b
,
if
x
>
2
f(x)= \begin{cases} ax+b & , \text{if } x \leq 1\\ 2b +1& , \text{if } 1 <x\le2 \\ 4ax-13b&, \text{if } x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
x
+
b
2
b
+
1
4
a
x
−
13
b
,
if
x
≤
1
,
if
1
<
x
≤
2
,
if
x
>
2
Continuity of Piecewise Functions
Practice: Continuity of Piecewise Functions
On which intervals is the function
f
(
x
)
=
{
e
+
ln
(
x
+
2
)
if
x
<
−
1
arcsin
x
+
e
if
−
1
≤
x
<
0
e
x
2
+
1
if
x
≥
0
f\left(x\right)= \begin{cases} e+\ln(x+2) & \text{if }\space x<-1\\ \arcsin x+e &\text{if }\space -1\le x<0\\ e^{x^2+1}&\text{if }\space x\ge0 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
e
+
ln
(
x
+
2
)
arcsin
x
+
e
e
x
2
+
1
if
x
<
−
1
if
−
1
≤
x
<
0
if
x
≥
0
Practice: Evaluating limits
Practice: Evaluating limits
The following is the graph of the function
f
(
x
)
f\left(x\right)
f
(
x
)
. Determine when the function
g
(
x
)
=
1
ln
(
f
(
x
)
)
g\left(x\right)=\frac{1}{\ln\left(f\left(x\right)\right)}
g
(
x
)
=
l
n
(
f
(
x
)
)
1
is continuous.
f
(
x
)
f\left(x\right)
f
(
x
)
is discontinuous at
x=0
Special Limit: Continuity
Evaluate
lim
x
→
1
+
arctan
(
e
(
π
x
−
1
)
)
\lim_{x\to1^+}\arctan\left(e^{\left(\frac{\pi}{x-1}\right)}\right)
lim
x
→
1
+
arctan
(
e
(
x
−
1
π
)
)
.
Continuous Functions
Practice: Continuous Functions
Suppose that
f
(
x
)
f\left(x\right)
f
(
x
)
and
g
(
x
)
g\left(x\right)
g
(
x
)
are continuous functions on
(
−
3
,
2
)
\left(-3,\ 2\right)
(
−
3
,
2
)
. If
g
(
1
)
=
−
3
g\left(1\right)=-3
g
(
1
)
=
−
3
and
lim
x
→
1
(
ln
(
4
+
g
(
x
)
)
−
g
(
x
)
e
f
(
x
)
)
=
1
\displaystyle\lim_{x\to1}\left(\ln\left(4+g\left(x\right)\right)-\frac{g\left(x\right)}{e^{^{f\left(x\right)}}}\right)=1
x
→
1
lim
(
ln
(
4
+
g
(
x
)
)
−
e
f
(
x
)
g
(
x
)
)
=
1
, determine the value of
f
(
1
)
f\left(1\right)
f
(
1
)
.
Continuity of Piecewise Functions
Practice: Continuity of Piecewise Functions
Find the values of 𝑎 and 𝑏 that make 𝑓 continuous everywhere
f
(
x
)
=
{
x
2
−
2
if
x
<
2
a
x
if
2
≤
x
<
3
2
x
−
b
if
x
≥
3
f\left(x\right)= \begin{cases} x^2-2 & \text{if }\space x<2\\ ax & \text{if }\space 2\le x<3\\ 2x-b & \text{if }\space x\ge3 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
a
x
2
x
−
b
if
x
<
2
if
2
≤
x
<
3
if
x
≥
3
Logarithmic Function & Continuity
The following is the graph of
y
=
f
(
x
)
y=f\left(x\right)
y
=
f
(
x
)
. On what interval is
ln
(
f
(
x
)
−
1
)
\ln\left(f\left(x\right)-1\right)
ln
(
f
(
x
)
−
1
)
continuous?
Continuity of Piecewise Functions
Practice: Continuity of Piecewise Functions
On which intervals is the function
f
(
x
)
=
{
e
+
ln
(
x
+
2
)
if
x
<
−
1
arcsin
x
+
e
if
−
1
≤
x
<
0
e
x
2
+
1
if
x
≥
0
f\left(x\right)= \begin{cases} e+\ln(x+2) & \text{if }\space x<-1\\ \arcsin x+e &\text{if }\space -1\le x<0\\ e^{x^2+1}&\text{if }\space x\ge0 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
e
+
ln
(
x
+
2
)
arcsin
x
+
e
e
x
2
+
1
if
x
<
−
1
if
−
1
≤
x
<
0
if
x
≥
0
Continuity
Find all the points at which the following function is continuous:
f
(
x
)
=
{
x
3
−
8
x
2
−
4
+
1
if
x
≠
−
2
4
if
x
=
2
5
if
x
=
−
2
f(x)=\left\{\begin{array}{ll} \displaystyle \frac{x^3-8}{x^2-4}+1\quad\text{if}\,x\neq -2 \\ \\ 4\quad\qquad\quad\,\,\,\,\,\,\,\text{if}\,x=2\\ \\ 5\quad\qquad\quad\,\,\,\, \,\,\,\text{if}\,x=-2 \end{array}\right.
f
(
x
)
=
⎩
⎨
⎧
x
2
−
4
x
3
−
8
+
1
if
x
=
−
2
4
if
x
=
2
5
if
x
=
−
2
Practice: Remove Discontinuity
Q.
\textbf{Q.}
Q.
Consider the function
f
(
x
)
=
cos
x
−
1
x
\displaystyle f(x)=\frac{\cos{x}-1}{x}
f
(
x
)
=
x
cos
x
−
1
. Define a function
g
(
x
)
g(x)
g
(
x
)
which is equal to
f
f
f
everywhere where
f
f
f
is defined, and that is continuous on
R
\mathbb{R}
R
.
Continuity
Assume
a
,
b
∈
R
a,\ b \in \mathbb{R}
a
,
b
∈
R
and
f
(
x
)
f(x)
f
(
x
)
is the function given by
f
(
x
)
=
{
e
x
−
1
,
if
x
>
1
a
,
if
x
=
1
−
x
2
+
b
,
if
x
<
1
f(x)=\begin{cases} e^{x-1},& \text{if}\;x>1\\ a,&\text{if}\;x=1\\ -x^2+b,&\text{if}\;x<1 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
e
x
−
1
,
a
,
−
x
2
+
b
,
if
x
>
1
if
x
=
1
if
x
<
1
Continuity
Find
A
A
A
and
B
B
B
so that the following function is continuous on
R
\mathbb{R}
R
:
f
(
x
)
=
{
A
(
1
−
cos
x
sin
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1
{f(x)=\left\{\begin{array}{ll} \displaystyle A\left(\frac{1-\cos{x}}{\sin^{2}{x}}\right)\quad\text{if}\,x<0 \\ \\ \displaystyle 2x^2-x+B\quad\quad\,\,\text{if}\,0\leq x\leq 1\\ \\ \displaystyle\frac{x^2+2x-3}{x^2-1}\,\,\,\quad\quad\text{if}\,x> 1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
A
(
sin
2
x
1
−
cos
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1
Q.
\textbf{Q.}
Q.
Consider the function
f
(
x
)
=
{
x
2
−
1
x
+
1
if
x
<
−
1
x
2
−
3
if
x
≥
−
1
{f(x)=\left\{\begin{array}{ll} \displaystyle\dfrac{x^2-1}{x+1}\quad\,\,\,\,\text{if}\,x<-1 \\\text{}\\ x^2-3\quad\quad\text{if}\,x\geq-1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
x
+
1
x
2
−
1
if
x
<
−
1
x
2
−
3
if
x
≥
−
1
Is it continuous on the interval
(
−
4
,
4
)
(-4,4)
(
−
4
,
4
)
?
Consider the function
f
(
x
)
=
{
x
2
−
1
x
+
1
if
x
<
−
1
x
2
−
3
if
x
≥
−
1
{f(x)=\left\{\begin{array}{ll} \displaystyle\dfrac{x^2-1}{x+1}\quad\,\,\,\,\text{if}\,x<-1 \\\text{}\\ x^2-3\quad\quad\text{if}\,x\geq-1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
x
+
1
x
2
−
1
if
x
<
−
1
x
2
−
3
if
x
≥
−
1
Is it continuous on the interval
(
−
4
,
4
)
(-4,4)
(
−
4
,
4
)
?
Consider the function
f
(
x
)
=
cos
2
x
−
1
x
2
f(x)=\dfrac{\cos^2{x}-1}{x^2}
f
(
x
)
=
x
2
cos
2
x
−
1
. Define a function
g
(
x
)
g(x)
g
(
x
)
which is equal to
f
f
f
everywhere where
f
f
f
is defined and that is continuous on all
R
\mathbb{R}
R
.
Find values of
a
a
a
and
b
b
b
such that the following function
f
f
f
is continuous at
x
=
0
x=0
x
=
0
and
x
=
1
x=1
x
=
1
:
f
(
x
)
=
{
a
(
1
−
cos
2
x
sin
(
2
x
)
)
,
if
x
<
0
2
x
2
−
x
+
b
,
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
,
if
x
>
1
f(x)=\begin{cases} a\left(\dfrac{\sqrt{1-\cos^2x}}{\sin(2x)}\right)&,\text{ if }x<0\\\\ 2x^2-x+b&,\text{ if }0\le x\le1\\\\ \dfrac{x^2+2x-3}{x^2-1}&,\text{ if }x>1 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
(
sin
(
2
x
)
1
−
cos
2
x
)
2
x
2
−
x
+
b
x
2
−
1
x
2
+
2
x
−
3
,
if
x
<
0
,
if
0
≤
x
≤
1
,
if
x
>
1
Practice: Continuity of piecewise functions
What do the constants
a
and
b
have to be such that the piecewise-define function is continuous at
x
=
−
2
?
x = -2\ ?
x
=
−
2
?
f
(
x
)
=
{
x
2
+
6
x
+
8
x
+
2
a
sin
(
x
+
2
)
b
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
f(x)=\begin{cases} \frac{x^2+6x+8}{x+2}\\ a\\ \frac{\sin(x+2)}{b(x+2)} \end{cases}\begin{array}{l} if\ x\ <\ -2\\ if\ x=\ -2\\ if\ x\ >\ -2 \end{array}
f
(
x
)
=
⎩
⎨
⎧
x
+
2
x
2
+
6
x
+
8
a
b
(
x
+
2
)
s
i
n
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
Practice: Continuity of piecewise functions
What do the constants
a
and
b
have to be such that the piecewise-define function is continuous at
x
=
−
2
?
x = -2\ ?
x
=
−
2
?
f
(
x
)
=
{
x
2
+
6
x
+
8
x
+
2
a
sin
(
x
+
2
)
b
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
f(x)=\begin{cases} \frac{x^2+6x+8}{x+2}\\ a\\ \frac{\sin(x+2)}{b(x+2)} \end{cases}\begin{array}{l} if\ x\ <\ -2\\ if\ x=\ -2\\ if\ x\ >\ -2 \end{array}
f
(
x
)
=
⎩
⎨
⎧
x
+
2
x
2
+
6
x
+
8
a
b
(
x
+
2
)
s
i
n
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
Practice: Continuity of piecewise functions
What do the constants
a
and
b
have to be such that the piecewise-define function is continuous at
x
=
−
2
?
x = -2\ ?
x
=
−
2
?
f
(
x
)
=
{
x
2
+
6
x
+
8
x
+
2
a
sin
(
x
+
2
)
b
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
f(x)=\begin{cases} \frac{x^2+6x+8}{x+2}\\ a\\ \frac{\sin(x+2)}{b(x+2)} \end{cases}\begin{array}{l} if\ x\ <\ -2\\ if\ x=\ -2\\ if\ x\ >\ -2 \end{array}
f
(
x
)
=
⎩
⎨
⎧
x
+
2
x
2
+
6
x
+
8
a
b
(
x
+
2
)
s
i
n
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
Piecewise Functions
Practice: Piecewise Functions
What value of 'a' would make the following piecewise function
continuous
?
f
(
x
)
=
{
x
2
+
10
;
x
<
1
−
x
+
7
a
;
x
≥
1
f(x)=\begin{cases} x^2+10;&x<1\\ -x+7a;&x\geq1 \end{cases}
f
(
x
)
=
{
x
2
+
10
;
−
x
+
7
a
;
x
<
1
x
≥
1
Piecewise Functions
Practice: Piecewise Functions
What value of 'a' would make the following piecewise function
continuous
?
f
(
x
)
=
{
−
2
x
2
+
7
;
x
<
−
4
3
x
+
a
;
x
≥
−
4
f(x)=\begin{cases} -2x^2+7;&x<-4\\ 3x+a;&x\geq-4 \end{cases}
f
(
x
)
=
{
−
2
x
2
+
7
;
3
x
+
a
;
x
<
−
4
x
≥
−
4
Continuity and Piecewise Functions
What values of
a
a
a
and
b
b
b
make the following function
continuous
?
f
(
x
)
=
{
a
x
2
−
4
;
x
<
−
2
a
x
+
b
;
−
2
≤
x
≤
3
a
x
2
+
b
x
+
10
;
x
>
3
f(x)= \begin{cases} ax^2-4;&&x<-2\\\\ ax+b;&&-2\leq{}x\leq3\\\\ ax^2+bx+10;&&x>3 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
x
2
−
4
;
a
x
+
b
;
a
x
2
+
b
x
+
10
;
x
<
−
2
−
2
≤
x
≤
3
x
>
3
Continuity
Determine whether the following function is continuous at
x
=
0
x = 0
x
=
0
using the definition of continuity. You must full justify your solution.
g
(
x
)
=
{
x
2
+
4
−
2
x
,
x
<
0
0
,
x
≥
0
\displaystyle g(x) = \begin{cases} \frac{\sqrt{x^2 + 4} - 2}{x} & \text{ , } x < 0 \\ 0 & \text{ , } x \geq 0 \end{cases}
g
(
x
)
=
{
x
x
2
+
4
−
2
0
,
x
<
0
,
x
≥
0
Continuity
Let
f
(
x
)
f(x)
f
(
x
)
be a continuous function on the open interval
(
a
,
b
)
(a, b)
(
a
,
b
)
. Which of the following four statements are always true.
Select all that apply.
Continuity
Where is
f
(
x
)
=
cos
(
π
x
2
)
1
−
x
2
f(x) = \frac{\cos\left(\frac{\pi x}{2} \right)}{\sqrt{1 - x^2}}
f
(
x
)
=
1
−
x
2
c
o
s
(
2
π
x
)
continuous?
Infinite Limits and Continuity
Given the function
f
(
x
)
=
{
arctan
x
+
π
/
2
if
x
≤
0
2
x
if
0
<
x
≤
e
1
ln
x
if
x
>
e
f(x)=\begin{cases} \arctan x+\pi/2&\text{if }x \le 0\\ \frac{2}{x}&\text{if }0<x\le e\\ \frac{1}{\ln x}&\text{if }x>e \end{cases}
f
(
x
)
=
⎩
⎨
⎧
arctan
x
+
π
/2
x
2
l
n
x
1
if
x
≤
0
if
0
<
x
≤
e
if
x
>
e
, which of the following statements is true about
f
(
x
)
f(x)
f
(
x
)
?
f
(
x
)
=
{
a
2
x
2
+
1
i
f
x
<
1
x
+
4
i
f
x
=
1
−
5
(
a
+
x
)
i
f
x
>
1
f\left(x\right)= \begin{cases} a^2x^2+1&if\space x<1\\ x+4 & if \space x=1\\ -5(a+x)&if~x>1 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
2
x
2
+
1
x
+
4
−
5
(
a
+
x
)
i
f
x
<
1
i
f
x
=
1
i
f
x
>
1
a) For what value(s) of
a
a
a
does
lim
x
→
1
f
(
x
)
\displaystyle \lim_{x\to1}f(x)
x
→
1
lim
f
(
x
)
exist?
b) For what value(s) of
a
a
a
is
f
(
x
)
f(x)
f
(
x
)
continuous at
x
=
1
x=1
x
=
1
?
Continuity
Use the graph below to find
Find
A
A
A
and
B
B
B
such that the following function is continuous:
f
(
x
)
=
{
A
(
1
−
c
o
s
2
x
sin
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1.
f(x)=\left\{ \begin{array}{ll} \displaystyle A\left(\dfrac{\sqrt{1-cos^2x}}{\sin{2x}}\right)\quad\text{if}\,x<0 \\ 2x^2-x+B\quad\,\,\,\,\,\,\,\text{if}\,0\leq x\leq 1\\ \displaystyle\dfrac{x^2+2x-3}{x^2-1}\quad\,\,\,\,\, \,\,\,\text{if}\,x>1. \end{array} \right.
f
(
x
)
=
⎩
⎨
⎧
A
(
sin
2
x
1
−
co
s
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1.
Find
A
A
A
and
B
B
B
such that the following function is continuous:
f
(
x
)
=
{
A
(
1
−
c
o
s
2
x
sin
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1.
f(x)=\left\{ \begin{array}{ll} \displaystyle A\left(\dfrac{\sqrt{1-cos^2x}}{\sin{2x}}\right)\quad\text{if}\,x<0 \\ 2x^2-x+B\quad\,\,\,\,\,\,\,\text{if}\,0\leq x\leq 1\\ \displaystyle\dfrac{x^2+2x-3}{x^2-1}\quad\,\,\,\,\, \,\,\,\text{if}\,x>1. \end{array} \right.
f
(
x
)
=
⎩
⎨
⎧
A
(
sin
2
x
1
−
co
s
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1.
Practice: Continuity of Piecewise Functions
Consider the function
f
(
x
)
=
{
ln
(
x
2
)
+
x
if
x
<
−
1
cos
π
x
2
if
−
1
≤
x
<
1
2
e
x
if
x
≥
1
2
f\left(x\right)= \begin{cases} \ln(x^2)+x & \text{if }\space x<-1\\ \cos \frac{{\pi}x}{2}&\text{if }\space -1\le x<\frac{1}{2}\\ e^{x}&\text{if }\space x\ge \frac{1}{2} \end{cases}
f
(
x
)
=
⎩
⎨
⎧
ln
(
x
2
)
+
x
cos
2
π
x
e
x
if
x
<
−
1
if
−
1
≤
x
<
2
1
if
x
≥
2
1
Limits: Continuity
Find all the points at which the following function is continuous:
f
(
x
)
=
{
x
3
−
8
x
2
−
4
+
1
if
x
≠
−
2
,
x
≠
2
4
if
x
=
2
5
if
x
=
−
2
f(x) = \begin{cases} \frac{x^3-8}{x^2-4}+1 & \text{ if } x\neq -2, \, x \neq 2 \\ 4 & \text{ if } x = 2 \\ 5 & \text{ if } x = -2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
−
4
x
3
−
8
+
1
4
5
if
x
=
−
2
,
x
=
2
if
x
=
2
if
x
=
−
2
Limits: Continuity
Find
A
and
B
so that the following function is continuous at 0 and 1:
f
(
x
)
=
{
A
(
1
−
cos
2
x
sin
2
x
)
x
<
0
2
x
2
−
x
+
B
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
x
>
1
f(x) = \begin{cases} A\left(\frac{\sqrt{1-\cos^2 x}}{\sin 2x}\right) & x < 0 \\ 2x^2-x+B & 0 \leq x \leq 1 \\ \frac{x^2+2x-3}{x^2-1} & x > 1 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
A
(
s
i
n
2
x
1
−
c
o
s
2
x
)
2
x
2
−
x
+
B
x
2
−
1
x
2
+
2
x
−
3
x
<
0
0
≤
x
≤
1
x
>
1
Enter your answer as
A,B
. For instance, if you think
A=0, B=1
, enter 0,1
If
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
2
x=-2
x
=
−
2
and if
lim
x
→
−
2
(
2
f
(
x
)
−
x
)
2
∣
f
(
x
)
∣
2
+
1
=
4
\lim\limits_{x\rightarrow -2}\frac{(2f(x)-x)^2}{|f(x)|^2+1}=4
x
→
−
2
lim
∣
f
(
x
)
∣
2
+
1
(
2
f
(
x
)
−
x
)
2
=
4
, then
Continuity
Find all the points at which the following function is continuous:
f
(
x
)
=
{
x
3
−
8
x
2
−
4
+
1
if
x
≠
−
2
4
if
x
=
2
5
if
x
=
−
2
f(x)=\left\{\begin{array}{ll} \displaystyle \frac{x^3-8}{x^2-4}+1\quad\text{if}\,x\neq -2 \\ \\ 4\quad\qquad\quad\,\,\,\,\,\,\,\text{if}\,x=2\\ \\ 5\quad\qquad\quad\,\,\,\, \,\,\,\text{if}\,x=-2 \end{array}\right.
f
(
x
)
=
⎩
⎨
⎧
x
2
−
4
x
3
−
8
+
1
if
x
=
−
2
4
if
x
=
2
5
if
x
=
−
2
Continuity
Find
A
A
A
and
B
B
B
so that the following function is continuous on
R
\mathbb{R}
R
:
f
(
x
)
=
{
A
(
1
−
cos
x
sin
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1
{f(x)=\left\{\begin{array}{ll} \displaystyle A\left(\frac{1-\cos{x}}{\sin^{2}{x}}\right)\quad\text{if}\,x<0 \\ \\ \displaystyle 2x^2-x+B\quad\quad\,\,\text{if}\,0\leq x\leq 1\\ \\ \displaystyle\frac{x^2+2x-3}{x^2-1}\,\,\,\quad\quad\text{if}\,x> 1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
A
(
sin
2
x
1
−
cos
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1
Q.
\textbf{Q.}
Q.
Consider the function
f
(
x
)
=
{
x
2
−
1
x
+
1
if
x
<
−
1
x
2
−
3
if
x
≥
−
1
{f(x)=\left\{\begin{array}{ll} \displaystyle\dfrac{x^2-1}{x+1}\quad\,\,\,\,\text{if}\,x<-1 \\\text{}\\ x^2-3\quad\quad\text{if}\,x\geq-1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
x
+
1
x
2
−
1
if
x
<
−
1
x
2
−
3
if
x
≥
−
1
Is it continuous on the interval
(
−
4
,
4
)
(-4,4)
(
−
4
,
4
)
?
Practice: Remove Discontinuity
Q.
\textbf{Q.}
Q.
Consider the function
f
(
x
)
=
cos
x
−
1
x
\displaystyle f(x)=\frac{\cos{x}-1}{x}
f
(
x
)
=
x
cos
x
−
1
. Define a function
g
(
x
)
g(x)
g
(
x
)
which is equal to
f
f
f
everywhere where
f
f
f
is defined, and that is continuous on
R
\mathbb{R}
R
.
Continuity
Assume
a
,
b
∈
R
a,\ b \in \mathbb{R}
a
,
b
∈
R
and
f
(
x
)
f(x)
f
(
x
)
is the function given by
f
(
x
)
=
{
e
x
−
1
,
if
x
>
1
a
,
if
x
=
1
−
x
2
+
b
,
if
x
<
1
f(x)=\begin{cases} e^{x-1},& \text{if}\;x>1\\ a,&\text{if}\;x=1\\ -x^2+b,&\text{if}\;x<1 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
e
x
−
1
,
a
,
−
x
2
+
b
,
if
x
>
1
if
x
=
1
if
x
<
1
Continuity
Find
A
A
A
so that the following function is continuous at
x
=
1
x=1
x
=
1
:
f
(
x
)
=
{
2
x
2
−
x
+
A
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1
{f(x)=\left\{\begin{array}{ll} \displaystyle 2x^2-x+A\quad\quad\,\,\text{if }\,0\leq x\leq 1\\ \\ \displaystyle\frac{x^2+2x-3}{x^2-1}\,\,\,\quad\quad\text{if }\,x> 1 \end{array}\right.}
f
(
x
)
=
⎩
⎨
⎧
2
x
2
−
x
+
A
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1
Find all number(s)
C
that make(s)
f
(
x
) continuous.
f
(
x
)
=
{
x
−
C
1
+
C
x
≤
0
2
x
2
+
C
x
>
0
f(x)=\begin{cases}\displaystyle\frac{x-C}{1+C}&x\leq0\\2x^2+C&x>0\end{cases}
f
(
x
)
=
⎩
⎨
⎧
1
+
C
x
−
C
2
x
2
+
C
x
≤
0
x
>
0
Discuss where the function is continuous:
f
(
x
)
=
sin
(
x
+
cos
x
)
f(x)=\sin(x+\cos x)
f
(
x
)
=
sin
(
x
+
cos
x
)
.
If
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
2
x=2
x
=
2
and if
lim
x
→
2
3
f
(
x
)
+
2
x
f
(
x
)
−
2
=
4
\lim\limits_{x\rightarrow 2}\frac{3f(x)+2}{xf(x)-2}=4
x
→
2
lim
x
f
(
x
)
−
2
3
f
(
x
)
+
2
=
4
, then
Continuity
Find
a
a
a
such that
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
1
x = 1
x
=
1
f
(
x
)
=
{
4
x
2
−
3
ln
x
x
≥
1
3
x
3
−
4
a
x
x
<
1
f(x)= \begin{cases} 4x^2 - 3\ln{x} & x \ge 1 \\ 3x^3 - 4ax & x < 1 \end{cases}
f
(
x
)
=
{
4
x
2
−
3
ln
x
3
x
3
−
4
a
x
x
≥
1
x
<
1
Continuity
For what values of
a
a
a
is the following function continuous on
(
−
∞
,
∞
)
(-\infty,\infty)
(
−
∞
,
∞
)
.
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
<
2
x
2
−
(
2
+
a
)
x
+
2
a
x
−
1
x
≥
2
f(x)= \begin{cases} ax^3-x^2+\sin(\pi x) & x<2 \\ \frac{x^2-(2+a)x+2a}{x-1} & x\geq 2 \end{cases}
f
(
x
)
=
{
a
x
3
−
x
2
+
sin
(
π
x
)
x
−
1
x
2
−
(
2
+
a
)
x
+
2
a
x
<
2
x
≥
2
Continuity
Consider the piecewise function
h
(
x
)
=
{
C
x
−
17
x
<
10
C
+
3
x
x
=
10
2
C
−
4
x
>
10
h(x) = \begin{cases} \frac{C}{x-17} & x<10 \\ C+3x & x=10 \\ 2C-4 & x>10 \end{cases}
h
(
x
)
=
⎩
⎨
⎧
x
−
17
C
C
+
3
x
2
C
−
4
x
<
10
x
=
10
x
>
10
where
C
is the same real number in each formula.
Continuity
Discuss where the following function is continuous:
h
(
x
)
=
log
(
x
+
1
−
2
)
h(x)=\log(\sqrt{x+1}-2)
h
(
x
)
=
lo
g
(
x
+
1
−
2
)
Consider
f
(
x
)
=
sin
2
x
x
\displaystyle f(x)=\frac{\sin{2x}}{x}
f
(
x
)
=
x
sin
2
x
. Does
lim
x
→
0
sin
2
x
x
\displaystyle \lim_{x\rightarrow 0}\frac{\sin{2x}}{x}
x
→
0
lim
x
sin
2
x
exist? Is
f
f
f
continuous at
x
=
0
x=0
x
=
0
?
When is the following function continuous:
g
(
x
)
=
x
2
−
1
x
2
−
1
\displaystyle g(x)=\frac{x^2-1}{x^2-1}
g
(
x
)
=
x
2
−
1
x
2
−
1
?
Where is the following function continuous?
f
(
x
)
=
ln
(
x
2
−
1
)
f(x)=\ln(x^2-1)
f
(
x
)
=
ln
(
x
2
−
1
)
Find
a
a
a
,
b
b
b
such that
f
(
x
)
f(x)
f
(
x
)
is continuous everywhere.
f
(
x
)
=
{
x
2
+
x
+
a
x
>
1
x
−
b
0
≤
x
≤
1
x
3
−
4
x
+
5
x
<
0
f(x)=\begin{cases}x^2+x+a&x>1\\x-b&0\leq x\leq1\\x^3-4x+5&x<0\end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
+
x
+
a
x
−
b
x
3
−
4
x
+
5
x
>
1
0
≤
x
≤
1
x
<
0
🦊
TRICKY!
Find
A
> 0 and
B
so that the following function is continuous at
x
= 0 and
x
= 1:
f
(
x
)
=
{
x
2
tan
2
A
x
if
x
<
0
3
x
2
+
2
x
+
B
if
0
≤
x
≤
1
3
x
2
−
7
x
−
20
x
2
−
3
x
−
4
if
x
>
1
f(x)=\begin{cases} \frac{x^2}{\tan^2Ax}&\text{if}&x<0\\ 3x^2+2x+B&\text{if}&0\leq x\leq 1\\ \frac{3x^2-7x-20}{x^2-3x-4}&\text{if}&x>1\end{cases}
f
(
x
)
=
⎩
⎨
⎧
t
a
n
2
A
x
x
2
3
x
2
+
2
x
+
B
x
2
−
3
x
−
4
3
x
2
−
7
x
−
20
if
if
if
x
<
0
0
≤
x
≤
1
x
>
1
Continuity
Consider the function
f
(
x
)
=
cos
2
x
−
1
x
2
f(x)=\dfrac{\cos^2{x}-1}{x^2}
f
(
x
)
=
x
2
cos
2
x
−
1
. Define a function
g
(
x
)
g(x)
g
(
x
)
which is equal to
f
f
f
everywhere where
f
f
f
is defined and that is continuous on all
R
\mathbb{R}
R
.
Practice: Continuity of piecewise functions
What do the constants
a
and
b
have to be such that the piecewise-define function is continuous at
x
=
−
2
?
x = -2\ ?
x
=
−
2
?
f
(
x
)
=
{
x
2
+
6
x
+
8
x
+
2
a
sin
(
x
+
2
)
b
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
f(x)=\begin{cases} \frac{x^2+6x+8}{x+2}\\ a\\ \frac{\sin(x+2)}{b(x+2)} \end{cases}\begin{array}{l} if\ x\ <\ -2\\ if\ x=\ -2\\ if\ x\ >\ -2 \end{array}
f
(
x
)
=
⎩
⎨
⎧
x
+
2
x
2
+
6
x
+
8
a
b
(
x
+
2
)
s
i
n
(
x
+
2
)
i
f
x
<
−
2
i
f
x
=
−
2
i
f
x
>
−
2
Limits: Continuity of piecewise functions
Identify the type of discontinuity (or state the function is continuous) at
x
=
5
,
10
,
15
x=5, 10, 15
x
=
5
,
10
,
15
. Is there another point of discontinuity in the domain other than at the stated
x
-values?
f
(
x
)
=
{
x
+
3
x
−
6
x
−
6
+
7
ln
(
15
−
x
)
sin
x
i
f
−
∞
<
x
≤
5
i
f
5
<
x
≤
10
i
f
10
<
x
<
15
i
f
15
≤
x
<
∞
f(x)=\begin{cases} x+3\\ \frac{x-6}{x-6}+7\\ \ln(15-x)\\ \sin x \end{cases}\begin{array}{l} if -\infty < x \leq 5\\ if\ 5 < x \leq 10\\ if 10 < x< 15\\ if 15 \leq x < \infty \end{array}
f
(
x
)
=
⎩
⎨
⎧
x
+
3
x
−
6
x
−
6
+
7
ln
(
15
−
x
)
sin
x
i
f
−
∞
<
x
≤
5
i
f
5
<
x
≤
10
i
f
10
<
x
<
15
i
f
15
≤
x
<
∞
Find
A
A
A
and
B
B
B
such that the following function is continuous:
f
(
x
)
=
{
A
(
1
−
c
o
s
2
x
sin
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
+
2
x
−
3
x
2
−
1
if
x
>
1.
f(x)=\left\{ \begin{array}{ll} \displaystyle A\left(\dfrac{\sqrt{1-cos^2x}}{\sin{2x}}\right)\quad\text{if}\,x<0 \\ 2x^2-x+B\quad\,\,\,\,\,\,\,\text{if}\,0\leq x\leq 1\\ \displaystyle\dfrac{x^2+2x-3}{x^2-1}\quad\,\,\,\,\, \,\,\,\text{if}\,x>1. \end{array} \right.
f
(
x
)
=
⎩
⎨
⎧
A
(
sin
2
x
1
−
co
s
2
x
)
if
x
<
0
2
x
2
−
x
+
B
if
0
≤
x
≤
1
x
2
−
1
x
2
+
2
x
−
3
if
x
>
1.
Continuity
Find the value of
a
a
a
such that the function
f
(
x
)
=
{
x
2
+
1
i
f
x
≤
−
3
a
x
−
1
i
f
x
>
−
3
f\left(x\right)= \begin{cases} x^2+1&if\space x\le -3\\ ax-1 & if \space x>-3 \end{cases}
f
(
x
)
=
{
x
2
+
1
a
x
−
1
i
f
x
≤
−
3
i
f
x
>
−
3
is continuous everywhere.
Special Limit: Continuity
Evaluate
lim
x
→
1
+
arctan
(
e
(
π
x
−
1
)
)
\lim_{x\to1^+}\arctan\left(e^{\left(\frac{\pi}{x-1}\right)}\right)
lim
x
→
1
+
arctan
(
e
(
x
−
1
π
)
)
.
Continuity of Piecewise Functions
Practice: Continuity of Piecewise Functions
On which intervals is the function
f
(
x
)
=
{
e
+
ln
(
x
+
2
)
if
x
<
−
1
arcsin
x
+
e
if
−
1
≤
x
<
0
e
x
2
+
1
if
x
≥
0
f\left(x\right)= \begin{cases} e+\ln(x+2) & \text{if }\space x<-1\\ \arcsin x+e &\text{if }\space -1\le x<0\\ e^{x^2+1}&\text{if }\space x\ge0 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
e
+
ln
(
x
+
2
)
arcsin
x
+
e
e
x
2
+
1
if
x
<
−
1
if
−
1
≤
x
<
0
if
x
≥
0
Continuity of Piecewise Functions
Practice: Continuity of Piecewise Functions
Find the values of 𝑎 and 𝑏 that make 𝑓 continuous everywhere
f
(
x
)
=
{
x
2
−
2
if
x
<
2
a
x
if
2
≤
x
<
3
2
x
−
b
if
x
≥
3
f\left(x\right)= \begin{cases} x^2-2 & \text{if }\space x<2\\ ax & \text{if }\space 2\le x<3\\ 2x-b & \text{if }\space x\ge3 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
a
x
2
x
−
b
if
x
<
2
if
2
≤
x
<
3
if
x
≥
3
Practice: Evaluating limits
The following is the graph of the function
f
(
x
)
f\left(x\right)
f
(
x
)
, which is defined only defined on
[
−
5
,
9
]
[-5,9]
[
−
5
,
9
]
. Determine when the function
g
(
x
)
=
1
ln
(
f
(
x
)
)
g\left(x\right)=\dfrac{1}{\ln\left(f\left(x\right)\right)}
g
(
x
)
=
ln
(
f
(
x
)
)
1
is continuous.
Practice: Evaluating limits
Practice: Evaluating limits
The following is the graph of the function
f
(
x
)
f\left(x\right)
f
(
x
)
. Determine when the function
g
(
x
)
=
1
ln
(
f
(
x
)
)
g\left(x\right)=\frac{1}{\ln\left(f\left(x\right)\right)}
g
(
x
)
=
l
n
(
f
(
x
)
)
1
is continuous.
f
(
x
)
f\left(x\right)
f
(
x
)
is discontinuous at
x=0
Continuous Functions
Practice: Continuous Functions
Suppose that
f
(
x
)
f\left(x\right)
f
(
x
)
and
g
(
x
)
g\left(x\right)
g
(
x
)
are continuous functions on
(
−
3
,
2
)
\left(-3,\ 2\right)
(
−
3
,
2
)
. If
g
(
1
)
=
−
3
g\left(1\right)=-3
g
(
1
)
=
−
3
and
lim
x
→
1
(
ln
(
4
+
g
(
x
)
)
−
g
(
x
)
e
f
(
x
)
)
=
1
\displaystyle\lim_{x\to1}\left(\ln\left(4+g\left(x\right)\right)-\frac{g\left(x\right)}{e^{^{f\left(x\right)}}}\right)=1
x
→
1
lim
(
ln
(
4
+
g
(
x
)
)
−
e
f
(
x
)
g
(
x
)
)
=
1
, determine the value of
f
(
1
)
f\left(1\right)
f
(
1
)
.
Logarithmic Function & Continuity
The following is the graph of
y
=
f
(
x
)
y=f\left(x\right)
y
=
f
(
x
)
. On what interval is
ln
(
f
(
x
)
−
1
)
\ln\left(f\left(x\right)-1\right)
ln
(
f
(
x
)
−
1
)
continuous?
Each question is worth 3 marks.
Continuity
What value of
a
a
a
makes the following function continuous?
f
(
x
)
=
{
2
x
+
4
,
x
≤
2
a
x
2
+
x
−
2
,
2
<
x
f(x) = \left \{ \begin{array}{ll} 2^x + 4, & x \leq 2\\ ax^2 + x - 2, & 2 < x \end{array} \right .
f
(
x
)
=
{
2
x
+
4
,
a
x
2
+
x
−
2
,
x
≤
2
2
<
x
Continuous Functions
For what values of
a
a
a
and
b
b
b
is the following function continuous?
f
(
x
)
=
{
x
2
−
2
x
+
a
,
x
≤
0
a
x
+
b
,
0
<
x
<
4
x
+
21
,
4
≤
x
f(x) = \left \{ \begin{array}{ll} x^2 - 2x + a, & x \leq 0\\ ax + b, & 0 < x < 4\\ \sqrt{x + 21}, & 4 \leq x \end{array} \right .
f
(
x
)
=
⎩
⎨
⎧
x
2
−
2
x
+
a
,
a
x
+
b
,
x
+
21
,
x
≤
0
0
<
x
<
4
4
≤
x
Continuity
Determine the values of
a
a
a
and
b
b
b
such that
f
(
x
)
f(x)
f
(
x
)
is continuous, where
f
(
x
)
=
{
a
x
+
b
,
if
x
≤
1
2
b
+
1
,
if
1
<
x
≤
2
4
a
x
−
13
b
,
if
x
>
2
f(x)= \begin{cases} ax+b & , \text{if } x \leq 1\\ 2b +1& , \text{if } 1 <x\le2 \\ 4ax-13b&, \text{if } x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
x
+
b
2
b
+
1
4
a
x
−
13
b
,
if
x
≤
1
,
if
1
<
x
≤
2
,
if
x
>
2
When is the following function continuous?
e
3
x
+
1
sin
(
π
x
)
\displaystyle \frac{e^{3x}+1}{\sin(\pi x)}
sin
(
π
x
)
e
3
x
+
1