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Let
h(x)=5*tan(x).
Below is Bridget's attempt to write a formal justification for the fact that the equation
h(x)=2 has a solution where
0 <= x <= (pi)/(4).
Is Bridget's justification complete? If not, why?
Bridget's justification:

h is defined over the entire interval
[0,(pi)/(4)], and trigonometric functions are continuous at all points in their domains.
So, according to the intermediate value theorem,
h(x)=2 must have a solution at some point in that interval.
Choose 1 answer:
(A) Yes, Bridget's justification is complete.
(B) No, Bridget didn't establish that 2 is between
h(0) and
h((pi)/(4)).
(C) No, Bridget didn't establish that
h is continuous.

Let $h(x)=5 \cdot \tan (x)$ . $\newline$ 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. $\newline$ Choose $1$ answer: $\newline$ (A) Yes, Bridget's justification is complete. $\newline$ (B) No, Bridget didn't establish that $2$ is between $h(0)$ and $h\left(\frac{\pi}{4}\right)$ . $\newline$ (C) No, Bridget didn't establish that $h$ is continuous.

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One-step inequalities: word problems

Full solution

Q. Let

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

Choose

1

answer:

\newline

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

\newline

(B) No, Bridget didn't establish that

2

is between

h(0)

and

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

.

\newline

(C) No, Bridget didn't establish that

h

is continuous.

Define Function: Bridget states that the function $h(x) = 5\tan(x)$ is defined and continuous over the interval $[0, (\pi)/(4)]$ . To use the intermediate value theorem to guarantee a solution to $h(x) = 2$ , we need to check that the value $2$ is between $h(0)$ and $h((\pi)/(4))$ . Let's calculate $h(0)$ and $h((\pi)/(4))$ .
Calculate $h(0)$ : Calculate $h(0)$ by substituting $x = 0$ into the function $h(x)$ . $\newline$ $h(0) = 5\tan(0) = 5\cdot 0 = 0$ .
Calculate $h\left(\frac{\pi}{4}\right)$ : Calculate $h\left(\frac{\pi}{4}\right)$ by substituting $x = \frac{\pi}{4}$ into the function $h(x)$ . $\newline$ $h\left(\frac{\pi}{4}\right) = 5\tan\left(\frac{\pi}{4}\right) = 5\cdot 1 = 5$ .
Apply Intermediate Value Theorem: Now we have $h(0) = 0$ and $h\left(\frac{\pi}{4}\right) = 5$ . Since $2$ is between $0$ and $5$ , and the function $h(x)$ is continuous on $\left[0, \frac{\pi}{4}\right]$ , by the intermediate value theorem, there must be some value $c$ in $\left[0, \frac{\pi}{4}\right]$ such that $h(c) = 2$ .
Incomplete Justification: Bridget's justification is not complete because she did not explicitly establish that the value $2$ is between $h(0)$ and $h(\frac{\pi}{4})$ . This step is crucial for the intermediate value theorem to be applicable.

More problems from One-step inequalities: word problems

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-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

x \geq 3

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Question

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

\newline

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

\newline

1

\newline

q \leq 3

\newline

q \geq 3

\newline

q \leq 6

\newline

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

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

x=1.5

x=2

\newline

1

\newline

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

\newline

(B) No, Issac didn't establish that

0

1

f(1.5)

f(2)

\newline

(C) No, Issac didn't establish that

f

Posted 9 months ago

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

Posted 9 months ago

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.

Posted 9 months ago

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.

Posted 9 months ago

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

You pick a card at random, put it back, and then pick another card at random.

\newline

4

\newline

5

\newline

6

\newline

7

\newline

What is the probability of picking a number greater than

5

and then picking a

4

\newline

Write your answer as a percentage.

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