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Let
f(x)=root(3)(cos(pi*x)).
Below is Issac's attempt to write a formal justification for the fact that the equation
f(x)=0.1 has a solution where
1.5 <= x <= 2.
Is Issac's justification complete? If not, why?
Issac's justification:

{:[f(1.5)=0" and "],[f(2)=1", so "],[f(1.5) < 0.1 < f(2).]:}
So, according to the intermediate value theorem,
f(x)=0.1 must have a solution when
x is between
x=1.5 and
x=2.
Choose 1 answer:
(A) Yes, Issac's justification is complete.
(B) No, Issac didn't establish that 0.1 is between
f(1.5) and
f(2).
(C) No, Issac didn't establish that
f is continuous.

Let $f(x)=\sqrt[3]{\cos (\pi \cdot x)}$ . $\newline$ Below is Issac's attempt to write a formal justification for the fact that the equation $f(x)=0.1$ has a solution where $1.5 \leq x \leq 2$ . $\newline$ Is Issac's justification complete? If not, why? $\newline$ Issac's justification: $\newline$ \[ \begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1

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One-step inequalities: word problems

Full solution

Q. Let

f(x)=\sqrt[3]{\cos (\pi \cdot x)}

.

\newline

Below is Issac's attempt to write a formal justification for the fact that the equation

f(x)=0.1

has a solution where

1.5 \leq x \leq 2

.

\newline

Is Issac's justification complete? If not, why?

\newline

Issac's justification:

\newline

\begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array}

\newline

So, according to the intermediate value theorem,

f(x)=0.1

must have a solution when

x

is between

x=1.5

and

x=2

.

\newline

Choose

1

answer:

\newline

(A) Yes, Issac's justification is complete.

\newline

(B) No, Issac didn't establish that

0

.

1

is between

f(1.5)

and

f(2)

.

\newline

(C) No, Issac didn't establish that

f

is continuous.

Check Continuity: To apply the Intermediate Value Theorem (IVT), we need to ensure that the function $f(x)$ is continuous on the interval $[1.5, 2]$ and that the value $0.1$ lies between $f(1.5)$ and $f(2)$ .
Verify Function Values: First, we need to check if $f(x)$ is continuous on the interval $[1.5, 2]$ . The function $f(x) = \sqrt[3]{\cos(\pi x)}$ is continuous because the cube root function and the cosine function are both continuous everywhere.
Calculate $f(1.5)$ : Next, we need to verify Issac's claims about the values of $f(1.5)$ and $f(2)$ . We calculate $f(1.5)$ and $f(2)$ to ensure that $0.1$ is indeed between these two values.
Calculate $f(2)$ : Calculate $f(1.5) = \sqrt[3]{\cos(\pi\cdot1.5)}$ . Since $\cos(\pi\cdot1.5) = \cos(\frac{3\pi}{2}) = 0$ , we have $f(1.5) = \sqrt[3]{0} = 0$ .
Confirm Value Range: Calculate $f(2) = \sqrt[3]{\cos(\pi\cdot 2)}$ . Since $\cos(\pi\cdot 2) = \cos(2\pi) = 1$ , we have $f(2) = \sqrt[3]{1} = 1$ .
Apply IVT: Now we can see that $0.1$ is indeed between $f(1.5) = 0$ and $f(2) = 1$ . Therefore, 0 < 0.1 < 1.
Apply IVT: Now we can see that $0.1$ is indeed between $f(1.5) = 0$ and $f(2) = 1$ . Therefore, 0 < 0.1 < 1.Since $f(x)$ is continuous on $[1.5, 2]$ and $0.1$ is between $f(1.5)$ and $f(2)$ , by the Intermediate Value Theorem, there must be some value $c$ in $[1.5, 2]$ such that $f(1.5) = 0$ $1$ .

More problems from One-step inequalities: word problems

Question

-11 x \geq-33

\newline

Which of the following best describes the solutions to the inequality shown?

\newline

1

\newline

x \leq-3

\newline

x \geq-3

\newline

x \leq 3

\newline

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\frac{3}{2} q \leq \frac{9}{2} q-18

\newline

Which of the following best describes the solutions to the inequality shown?

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\newline

q \leq 3

\newline

q \geq 3

\newline

q \leq 6

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q \geq 6

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Question

g(x)=5-e^{x}

\newline

Below is Sean's attempt to write a formal justification for the fact that the equation

g(x)=0

has a solution where

1 \leq x \leq 4

\newline

Is Sean's justification complete? If not, why?

\newline

Sean's justification:

\newline

g

is defined for all real numbers, and exponential functions are continuous at all points in their domains.

\newline

So, according to the intermediate value theorem,

g(x)=0

must have a solution somewhere between

x=1

x=4

\newline

1

\newline

(A) Yes, Sean's justification is complete.

\newline

(B) No, Sean didn't establish that

0

g(1)

g(4)

\newline

(C) No, Sean didn't establish that

g

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Question

h(x)=5 \cdot \tan (x)

\newline

Below is Bridget's attempt to write a formal justification for the fact that the equation

h(x)=2

has a solution where

0 \leq x \leq \frac{\pi}{4}

\newline

Is Bridget's justification complete? If not, why?

\newline

Bridget's justification:

\newline

h

is defined over the entire interval

\left[0, \frac{\pi}{4}\right]

, and trigonometric functions are continuous at all points in their domains.

\newline

So, according to the intermediate value theorem,

h(x)=2

must have a solution at some point in that interval.

\newline

1

\newline

(A) Yes, Bridget's justification is complete.

\newline

(B) No, Bridget didn't establish that

2

h(0)

h\left(\frac{\pi}{4}\right)

\newline

(C) No, Bridget didn't establish that

h

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Question

f(x)=2^{5-2 x}

\newline

Below is Ibrahim's attempt to write a formal justification for the fact that the equation

f(x)=10

has a solution where

-1 \leq x \leq 4

\newline

Is Ibrahim's justification complete? If not, why?

\newline

Ibrahim's justification:

\newline

f

is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also,

f(-1)=128

f(4)=\frac{1}{8}

10

f(-1)

f(4)

\newline

So, according to the intermediate value theorem,

f(x)=10

must have a solution at some point between

x=-1

x=4

\newline

1

\newline

(A) Yes, Ibrahim's justification is complete.

\newline

(B) No, Ibrahim didn't establish that

10

f(-1)

f(4)

\newline

(c) No, Ibrahim didn't establish that

f

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Question

g(x)=\cos \left(\pi x^{2}\right)

\newline

Below is Amrita's attempt to write a formal justification for the fact that the equation

g^{\prime}(x)=0.4

has a solution where

-1<x<4

\newline

Is Amrita's justification complete?

\newline

\newline

Amrita's justification:

\newline

Polynomial and trigonometric functions are differentiable and continuous at all points in their domain, and

[-1,4]

g^{\prime}

s domain. Also,

\newline

\begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array}

\newline

So, according to the mean value theorem,

g^{\prime}(x)=0.4

must have a solution somewhere between

x=-1

x=4

\newline

1

\newline

(A) Yes, Amrita's justification is complete.

\newline

(B) No, Amrita didn't establish that the average rate of change of

g

[-1,4]

0

4

\newline

(C) No, Amrita didn't establish that

g

is differentiable.

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Question

\newline

h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. }

\newline

Below is Omar's attempt to write a formal justification for the fact that there exists a value

c

\newline

-2<c<2 \text { such that } h^{\prime}(c)=6 \text {. }

\newline

Is Omar's justification complete? If not, why?

\newline

Omar's justification:

\newline

\begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array}

\newline

So, according to the mean value theorem, there exists a value

c

somewhere in the interval

-2<c<2

h^{\prime}(c)=6

\newline

1

\newline

(A) Yes, Omar's justification is complete.

\newline

(B) No, Omar didn't establish that the average rate of change of

h

[-2,2]

6

\newline

(C) No, Omar didn't establish that

h

is differentiable.

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Question

f(x)=\sqrt{x+\sqrt{x^{3}}}

\newline

Below is Lena's attempt to write a formal justification for the fact that the equation

f^{\prime}(x)=\frac{2}{3}

has a solution where

0<x<9

\newline

Is Lena's justification complete? If not, why?

\newline

Lena's justification:

\newline

f(0)=0

f(9)=6

, so the average rate of change of

f

over the interval

0 \leq x \leq 9

\frac{2}{3}

\newline

So, according to the mean value theorem,

f^{\prime}(x)=\frac{2}{3}

must have a solution somewhere in the interval

0<x<9

\newline

1

\newline

(A) Yes, Lena's justification is complete.

\newline

(B) No, Lena didn't establish that the average rate of change of

f

[0,9]

\frac{2}{3}

\newline

(C) No, Lena didn't establish that

f

is differentiable.

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Question

h(x)=3^{x}-x^{2}

\newline

Below is Uriah's attempt to write a formal justification for the fact that the equation

h^{\prime}(x)=8

has a solution where

1<x<3

\newline

Is Uriah's justification complete? If not, why?

\newline

Uriah's justification:

\newline

Polynomial and exponential functions are differentiable and continuous at all points in their domain, and

[1,3]

h

's domain. Also,

h(1)=2

h(3)=18

\newline

\frac{h(3)-h(1)}{3-1}=8 \text {. }

\newline

So, according to the mean value theorem,

h^{\prime}(x)=8

must have a solution somewhere between

x=1

x=3

\newline

1

\newline

(A) Yes, Uriah's justification is complete.

\newline

(B) No, Uriah didn't establish that the average rate of change of

h

[1,3]

8

\newline

(C) No, Uriah didn't establish that

h

is differentiable.

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Question

19

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\newline

A forest ranger wants to determine the height of a redwood tree. Standing

145

meters (m) away from the tree, the ranger measures angles of inclination and declination to the tree's top and base, respectively, as shown. Which of the following is closest to the height,

h

\newline

1

\newline

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