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Which of the following is a solution to the inequality below? $\newline$ j < -8 $\newline$ Choices: $\newline$ (A) $j = -5$ $\newline$ (B) $j = -8$ $\newline$ (C) $j = -12$ $\newline$ (D) $j = -4$

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Solutions to inequalities

Full solution

Q. Which of the following is a solution to the inequality below?

\newline

j < -8

\newline

Choices:

\newline

(A)

j = -5

\newline

(B)

j = -8

\newline

(C)

j = -12

\newline

(D)

j = -4

Understand the inequality: Understand the inequality. $\newline$ The inequality j < -8 means that we are looking for values of $j$ that are less than negative eight.
Evaluate choices: Evaluate each choice to see if it satisfies the inequality. $\newline$ Start with choice (A) $j = -5$ . $\newline$ Check if -5 < -8. $\newline$ This is not true because $-5$ is greater than $-8$ .
Check choice (A): Move to choice (B) $j = -8$ . $\newline$ Check if -8 < -8. $\newline$ This is not true because $-8$ is equal to $-8$ , not less than.
Check choice (B): Evaluate choice (C) $j = -12$ . Check if -12 < -8. This is true because $-12$ is less than $-8$ .
Check choice (C): There is no need to check choice (D) $j = -4$ because we have already found a correct answer in choice (C). However, for completeness, we will check it. $\newline$ Check if -4 < -8. $\newline$ This is not true because $-4$ is greater than $-8$ .

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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

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

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Sean's justification:

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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,

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must have a solution when

x

x=1.5

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\newline

1

\newline

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

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(B) No, Issac didn't establish that

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(C) No, Issac didn't establish that

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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

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\newline

So, according to the intermediate value theorem,

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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

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

\newline

(C) No, Bridget didn't establish that

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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

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has a solution where

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\newline

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

\newline

Ibrahim's justification:

\newline

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\newline

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1

\newline

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

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(B) No, Ibrahim didn't establish that

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(c) No, Ibrahim didn't establish that

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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

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0

4

\newline

(C) No, Amrita didn't establish that

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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

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\newline

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c

somewhere in the interval

-2<c<2

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\newline

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

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

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\newline

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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

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\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

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h(1)=2

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\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

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8

\newline

(C) No, Uriah didn't establish that

h

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Related topics

Algebra - Order of Operations Algebra - Distributive Property `X` and `Y` Axes Geometry - Scalene Triangle Common Multiple Geometry - Quadrant