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Does the equation 9x+cos (3x)=8 have a solution in the domain (0,1)?
Related Topics
Wize University Calculus 1 Textbook > Limits
The Intermediate Value Theorem
3 Activities
Does the equation
9
x
+
cos
(
3
x
)
=
8
9x+\cos\ (3x)=8
9
x
+
cos
(
3
x
)
=
8
have a solution in the domain (0,1)?
Yes
\text{Yes}
Yes
No
\text{No}
No
I don't know
Check Submission
More The Intermediate Value Theorem Questions:
The Intermediate Value Theorem
Show that there is at least one solution to the equation
ln
x
=
e
sin
(
π
x
2
)
\ln x=e^{\sin\left(\frac{\pi x}{2}\right)}
ln
x
=
e
s
i
n
(
2
π
x
)
.
The Intermediate Value Theorem
Use the Intermediate Value Theorem to show that the function
f
(
x
)
=
x
3
−
4
x
+
2
f\left(x\right)=x^3-4x+2
f
(
x
)
=
x
3
−
4
x
+
2
crosses the
x
−
x-
x
−
axis in the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
.
In intermediate Value Theorem
Does
2
x
3
−
2
+
cos
(
x
)
=
0
2x^3-2+\cos(x)=0
2
x
3
−
2
+
cos
(
x
)
=
0
have a root in (0,1) ?
In intermediate Value Theorem
Does
2
x
3
−
2
+
cos
(
x
)
=
0
2x^3-2+\cos(x)=0
2
x
3
−
2
+
cos
(
x
)
=
0
have a root in (0,1) ?
The Intermediate Value Theorem: At Least One Solution
Show that the function
f
(
x
)
=
x
3
−
7
x
2
+
6
x
−
13
=
0
f(x)=x^3-7x^2+6x-13=0
f
(
x
)
=
x
3
−
7
x
2
+
6
x
−
13
=
0
must possess at least one real-valued solution.
In intermediate Value Theorem
Does
2
x
3
−
2
+
cos
(
x
)
=
0
2x^3-2+\cos(x)=0
2
x
3
−
2
+
cos
(
x
)
=
0
have a root in (0,1) ?
Intermediate Value Theorem
If
f
(
x
)
f\left(x\right)
f
(
x
)
is a continuous function on
[
−
1
,
1
]
\left[-1,1\right]
[
−
1
,
1
]
, such that
f
(
−
1
)
=
−
π
f\left(-1\right)=-\pi
f
(
−
1
)
=
−
π
and
f
(
1
)
=
π
f\left(1\right)=\pi
f
(
1
)
=
π
, which statement is always true?
The Intermediate Value Theorem
Show that there is at least one solution to the equation
ln
x
=
e
sin
(
π
x
2
)
\ln x=e^{\sin\left(\frac{\pi x}{2}\right)}
ln
x
=
e
s
i
n
(
2
π
x
)
.
Indetermiante Value Theorem
Does
x
3
−
1
+
sin
(
x
)
=
0
x^3-1+\sin\left(x\right)=0
x
3
−
1
+
sin
(
x
)
=
0
have a root in
[
0
,
1
]
\left[0,1\right]
[
0
,
1
]
?
The Intermediate Value Theorem
Show that the equation
sin
x
=
x
2
−
x
−
1
\sin x=x^2-x-1
sin
x
=
x
2
−
x
−
1
has at least one root in the interval
(
0
,
3
)
(0,3)
(
0
,
3
)
,
i.e.
, there exists a number
c
∈
(
0
,
3
)
c \in (0,3)
c
∈
(
0
,
3
)
such that
sin
c
=
c
2
−
c
−
1
\sin c=c^2-c-1
sin
c
=
c
2
−
c
−
1
.
The Intermediate Value Theorem
Use the Intermediate Value Theorem to show that the function
f
(
x
)
=
x
3
−
4
x
+
2
f\left(x\right)=x^3-4x+2
f
(
x
)
=
x
3
−
4
x
+
2
crosses the
x
−
x-
x
−
axis in the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
.
The Intermediate Value Theorem
Determine if the function
f
(
x
)
=
x
3
−
6
x
+
4
f(x)=x^3-6x+4
f
(
x
)
=
x
3
−
6
x
+
4
has a root in the domain
[
0
,
1
]
[0,1]
[
0
,
1
]
.
Does
3
x
2
+
cos
x
=
3
3x^2+\cos{x}=3
3
x
2
+
cos
x
=
3
have a root in
(
0
,
1
)
(0,1)
(
0
,
1
)
?
In intermediate Value Theorem
Does
2
x
3
−
2
+
cos
(
x
)
=
0
2x^3-2+\cos(x)=0
2
x
3
−
2
+
cos
(
x
)
=
0
have a root in (0,1) ?
The Intermediate Value Theorem: At Least One Solution
Show that the function
f
(
x
)
=
x
3
−
7
x
2
+
6
x
−
13
=
0
f(x)=x^3-7x^2+6x-13=0
f
(
x
)
=
x
3
−
7
x
2
+
6
x
−
13
=
0
must possess at least one real-valued solution.
Practice: IVT & Equation Solutions
Practice: IVT & Equation Solutions
Show that there is at least one solution to the equation
ln
x
=
e
sin
(
π
x
2
)
\ln x=e^{\sin\left(\frac{\pi x}{2}\right)}
ln
x
=
e
s
i
n
(
2
π
x
)
.
The Intermediate Value Theorem
Let
f
(
x
)
f(x)
f
(
x
)
be a function so that
f
(
x
)
,
f
′
(
x
)
,
f
′
′
(
x
)
f(x), f'(x), f''(x)
f
(
x
)
,
f
′
(
x
)
,
f
′′
(
x
)
exist and are continuous for all
x
x
x
, and
∣
f
(
x
)
−
sin
(
x
)
∣
≤
1
/
4
|f(x) - \sin(x)| \leq 1/4
∣
f
(
x
)
−
sin
(
x
)
∣
≤
1/4
for all
x
x
x
.
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.
Alex is thinking about
tan
x
\tan x
tan
x
function. He knows that
tan
1
>
0
\tan1>0
tan
1
>
0
and
tan
2
<
0
\tan2<0
tan
2
<
0
He concludes that
tan
x
\tan x
tan
x
has a root in (1, 2). Justify if he is right or wrong.
The intermediate value theorem
f
(
x
)
=
1
5
x
5
+
4
3
x
3
+
3
x
−
2
f(x)=\frac{1}{5}x^5+\frac{4}{3}x^3+3x-2
f
(
x
)
=
5
1
x
5
+
3
4
x
3
+
3
x
−
2
has
f
(
x
)
:
[
0
,
1
]
→
[
1
,
3
]
f(x):[0,1]\rightarrow[1,3]
f
(
x
)
:
[
0
,
1
]
→
[
1
,
3
]
is continuous. Does
f
(
x
)
f(x)
f
(
x
)
intersect with
y
=
2
x
+
1
y=2x+1
y
=
2
x
+
1
?
final114
Megan was thinking about the function
f
(
x
)
=
3
(
x
−
3
)
3
f(x)=\dfrac{3}{(x-3)^3}
f
(
x
)
=
(
x
−
3
)
3
3
and she thought
f
(
0
)
<
0
f(0)<0
f
(
0
)
<
0
and also
f
(
4
)
>
0
f(4)>0
f
(
4
)
>
0
.So, by the Intermediate Value Theorem, this function has a root in the interval
(
0
,
4
)
(0,4)
(
0
,
4
)
. Is she right?
Indetermiante Value Theorem
Does
x
3
−
1
+
sin
(
x
)
=
0
x^3-1+\sin\left(x\right)=0
x
3
−
1
+
sin
(
x
)
=
0
have a root in
[
0
,
1
]
\left[0,1\right]
[
0
,
1
]
?
The Intermediate Value Theorem
Determine if the function
f
(
x
)
=
x
3
−
6
x
+
4
f(x)=x^3-6x+4
f
(
x
)
=
x
3
−
6
x
+
4
has a root in the domain
[
0
,
1
]
[0,1]
[
0
,
1
]
.
Does
3
x
2
+
cos
x
=
3
3x^2+\cos{x}=3
3
x
2
+
cos
x
=
3
have a root in
(
0
,
1
)
(0,1)
(
0
,
1
)
?
In intermediate Value Theorem
Does
2
x
3
−
2
+
cos
(
x
)
=
0
2x^3-2+\cos(x)=0
2
x
3
−
2
+
cos
(
x
)
=
0
have a root in (0,1) ?
The Intermediate Value Theorem: At Least One Solution
Show that the function
f
(
x
)
=
x
3
−
7
x
2
+
6
x
−
13
=
0
f(x)=x^3-7x^2+6x-13=0
f
(
x
)
=
x
3
−
7
x
2
+
6
x
−
13
=
0
must possess at least one real-valued solution.
The Intermediate Value Theorem
Use the Intermediate Value Theorem to show that the function
f
(
x
)
=
x
3
−
4
x
+
2
f\left(x\right)=x^3-4x+2
f
(
x
)
=
x
3
−
4
x
+
2
crosses the
x
−
x-
x
−
axis in the interval
[
0
,
2
]
[0,2]
[
0
,
2
]
.
The Intermediate Value Theorem
Show that the equation
sin
x
=
x
2
−
x
−
1
\sin x=x^2-x-1
sin
x
=
x
2
−
x
−
1
has at least one root in the interval
(
0
,
3
)
(0,3)
(
0
,
3
)
,
i.e.
, there exists a number
c
∈
(
0
,
3
)
c \in (0,3)
c
∈
(
0
,
3
)
such that
sin
c
=
c
2
−
c
−
1
\sin c=c^2-c-1
sin
c
=
c
2
−
c
−
1
.
🦊
TRICKY!
Megan Mah was thinking about the
y
=
tan
x
y=\tan x
y
=
tan
x
function. She thought tan (1) > 0 and also tan(2) < 0, so, by the Intermediate Value Theorem, this function has a root in the interval (1,2). Is she right?
Does
x
3
−
1
+
sin
(
x
)
=
0
x^3-1+\sin\left(x\right)=0
x
3
−
1
+
sin
(
x
)
=
0
have at least one root in
[
0
,
1
]
\left[0,1\right]
[
0
,
1
]
?
Show that the equation
sin
x
=
x
2
−
x
−
1
\sin x=x^2-x-1
sin
x
=
x
2
−
x
−
1
has at least one root in the interval
(
0
,
3
)
(0,3)
(
0
,
3
)
,
i.e.
, there exist a number
c
∈
(
0
,
3
)
c \in (0,3)
c
∈
(
0
,
3
)
such that
sin
c
=
c
2
−
c
−
1
\sin c=c^2-c-1
sin
c
=
c
2
−
c
−
1
. State the name of the theorem you use.
🦊
TRICKY!
Use the Intermediate Value Theorem to show that the function
f
(
x
)
=
x
3
−
4
x
+
2
f\left(x\right)=x^3-4x+2
f
(
x
)
=
x
3
−
4
x
+
2
crosses the
x
-axis at least twice in the interval [0,2].
Use Intermediate Value Theorem to conclude that
x
2
+
x
−
3
=
2
\sqrt{x^2+x-3}=2
x
2
+
x
−
3
=
2
has at least one solution in the interval
x
∈
(
2
,
3
)
x\in(2,3)
x
∈
(
2
,
3
)
.
Does the equation
3
x
2
+
cos
x
=
3
3x^2+\cos{x}=3
3
x
2
+
cos
x
=
3
have a solution in the domain
(
0
,
1
)
(0,1)
(
0
,
1
)
?
Does the equation
2
x
3
−
2
+
cos
(
x
)
=
0
2x^3-2+\cos(x)=0
2
x
3
−
2
+
cos
(
x
)
=
0
have at least one solution in the domain (0,1)?
Determine if the polynomial
x
3
−
6
x
x^3-6x
x
3
−
6
x
will attain the value of -4 in the domain [0,1].
Intermediate Value Theorem
Suppose
f
is a continuous function and
f
(0)=0 and
f
(1)=2. Is there a constant
c
∈
(
0
,
1
)
c\in (0,1)
c
∈
(
0
,
1
)
such that
f
(
c
)
=
2
−
c
5
f(c)=\sqrt[5]{2-c}
f
(
c
)
=
5
2
−
c
? HINT: Apply the Intermediate Value Theorem on the function
g
(
x
)
=
[
f
(
x
)
]
5
+
x
g(x)=[f(x)]^5+x
g
(
x
)
=
[
f
(
x
)
]
5
+
x
.
The Intermediate Value Theorem
Show that there is at least one solution to the equation
ln
x
=
e
sin
(
π
x
2
)
\ln x=e^{\sin\left(\frac{\pi x}{2}\right)}
ln
x
=
e
s
i
n
(
2
π
x
)
.
Intermediate Value Theorem
If
f
(
x
)
f\left(x\right)
f
(
x
)
is a continuous function on
[
−
1
,
1
]
\left[-1,1\right]
[
−
1
,
1
]
, such that
f
(
−
1
)
=
−
π
f\left(-1\right)=-\pi
f
(
−
1
)
=
−
π
and
f
(
1
)
=
π
f\left(1\right)=\pi
f
(
1
)
=
π
, which statement is always true?
Intermediate Value Theorem
Let
f
(
x
)
f(x)
f
(
x
)
be an even, continuous function and suppose that
f
(
−
2019
)
=
1
2
f(-2019) = \frac{1}{2}
f
(
−
2019
)
=
2
1
and that
Intermediate Value Theorem
Find an interval of length at most 1 that contains a solution to the equation
e
x
−
2
=
0
e^x - 2 = 0
e
x
−
2
=
0
Intermediate Value Theorem
Show that the function
p
(
x
)
=
x
4
+
x
2
−
2
p(x)=x^4+x^2-2
p
(
x
)
=
x
4
+
x
2
−
2
has a solution on the interval
[
0
,
2
]
[0, 2]
[
0
,
2
]
.
The Intermediate Value Theorem
Does
3
x
2
+
cos
x
=
3
3x^2+\cos x=3
3
x
2
+
cos
x
=
3
has a root in (0, 1)?
f
(
x
)
:
[
0
,
1
]
→
[
1
,
3
]
f(x):[0,1]\rightarrow[1,3]
f
(
x
)
:
[
0
,
1
]
→
[
1
,
3
]
is continuous. Does
f
(
x
)
f(x)
f
(
x
)
intersect with
y
=
2
x
+
1
y=2x+1
y
=
2
x
+
1
?
Alex is thinking about
tan
x
\tan x
tan
x
function. He knows that
tan
1
>
0
\tan1>0
tan
1
>
0
and
tan
2
<
0
\tan2<0
tan
2
<
0
He concludes that
tan
x
\tan x
tan
x
has a root in (1, 2). Justify if he is right or wrong.