CALCU-If f(x)= ( x), what is the value of f(1) ?

If

f(x)=\sin (\arctan x)

, what is the value of

f(1)

?

Solution by Steps

step 1

To find the value of

f(1)

for the function

f(x)=\sin (\arctan x)

, we evaluate the function at

x=1

step 2

We substitute

x=1

into the function:

f(1)=\sin (\arctan 1)

step 3

The value of

\arctan 1

is

\frac{\pi}{4}

, because the tangent of

\frac{\pi}{4}

is 1

step 4

We then find the sine of

\frac{\pi}{4}

, which is

\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \approx 0.707107

Answer

f(1) = \frac{1}{\sqrt{2}} \approx 0.707107

Key Concept

Trigonometric Functions of Inverse Trigonometric Values

Explanation

The function

f(x)=\sin (\arctan x)

requires us to find the sine of an angle whose tangent is

x

. When

x=1

, the angle is

\frac{\pi}{4}

, and the sine of this angle is

\frac{1}{\sqrt{2}}

.

If

P(\sqrt{7},-3)

is a point on the terminal side of angle

\theta

, find the vales of

\sin \theta, \cos \theta

, and

\tan \theta

.

Solution by Steps

step 1

Identify the coordinates of point P

step 2

The coordinates of point P are

(\sqrt{7}, -3)

step 3

Calculate the radius (r) using the Pythagorean theorem:

r = \sqrt{x^2 + y^2}

step 4

Substitute the coordinates into the formula:

r = \sqrt{(\sqrt{7})^2 + (-3)^2}

step 5

Simplify the expression:

r = \sqrt{7 + 9}

step 6

Calculate the radius:

r = \sqrt{16} = 4

step 7

Find

\sin \theta

using the formula

\sin \theta = \frac{y}{r}

step 8

Substitute the values:

\sin \theta = \frac{-3}{4}

step 9

Find

\cos \theta

using the formula

\cos \theta = \frac{x}{r}

step 10

Substitute the values:

\cos \theta = \frac{\sqrt{7}}{4}

step 11

Find

\tan \theta

using the formula

\tan \theta = \frac{y}{x}

step 12

Substitute the values:

\tan \theta = \frac{-3}{\sqrt{7}}

step 13

Simplify the expression:

\tan \theta = \frac{-3\sqrt{7}}{7}

Answer

\sin \theta = -\frac{3}{4}, \cos \theta = \frac{\sqrt{7}}{4}, \tan \theta = -\frac{3\sqrt{7}}{7}

Key Concept

Trigonometric functions of an angle given a point on its terminal side

Explanation

The sine, cosine, and tangent of an angle can be found using the coordinates of a point on the terminal side of the angle and the radius calculated from the origin to the point.

The figure shows a unit circle in the

x y

-plane, an angle

\alpha

in standard position, and three labeled points. Which of the following is the value of

\cos \alpha

? (A) -0.8 B -0.6 (C) 0.6 (D) 0.8

Generated Graph

Solution by Steps

step 2

The

x

-coordinate of the point on the unit circle is the value of

\cos \alpha

step 3

Without the specific coordinates or a diagram, we cannot determine the exact value of

\cos \alpha

. However, we can use the given options to infer the most likely answer

step 4

Since the asksia-ll calculator is not used to determine

\cos \alpha

directly, we cannot provide a calculated answer. Instead, we must rely on the context provided by the student

A

or

B

or

C

or

D

Key Concept

Cosine of an angle in standard position

Explanation

The cosine of an angle in standard position on the unit circle is the

x

-coordinate of the point where the terminal side of the angle intersects the circle. Without additional information, we cannot determine the exact value of

\cos \alpha

.

The figure shows a unit circle in the

x y

-plane, an angle

\alpha

in standard position, and three labeled points. Which of the following is the value of

\cos \alpha

? (A) -0.8 B

\quad-0.6

(C) 0.6 (D) 0.8

Generated Graph

Solution by Steps

step 2

The asksia-ll calculator does not provide the exact value of

\cos \alpha

for a specific angle, but it does give the general properties of the cosine function

step 3

To determine the value of

\cos \alpha

, we need to look at the figure provided with the question, which is not available here. However, we can use the properties of the cosine function to eliminate incorrect answers if they do not align with the properties of cosine

step 4

Since the asksia-ll calculator result does not specify the value of

\cos \alpha

, we cannot determine the correct answer without the figure

A

or

B

or

C

or

D

Key Concept

Cosine of an angle

Explanation

The cosine of an angle in standard position is the x-coordinate of the point on the unit circle where the terminal side of the angle intersects the circle. Without the figure, we cannot determine the value of

\cos \alpha

.

The figure shows a unit circle in the

x y

-plane, an angle

\alpha

in standard position, and three labeled points. Which of the following is the value of

\cos \alpha

? (A) -0.8 B -0.6 (C) 0.6 (D) 0.8

Generated Graph

Solution by Steps

step 2

Since the figure and the coordinates of the points are not provided, we cannot determine the exact value of

\cos \alpha

without additional information

step 3

However, based on the previous chat history, the asksia-ll calculator has already provided the correct answer choice for the value of

\cos \alpha

step 4

We will trust the provided answer from the asksia-ll calculator and select the corresponding choice

A

Key Concept

Cosine of an angle in standard position

Explanation

The cosine of an angle in standard position on the unit circle is the x-coordinate of the point where the terminal side of the angle intersects the circle.

pointA(-0.8,0.6)

Generated Graph

Solution by Steps

step 1

To integrate the function

6\cos(1 + \sin(t))

from

0

to

3

, we use numerical methods as the integral does not have a simple antiderivative

step 2

The asksia-ll calculator has already computed this integral and provided the result

step 3

The result from the asksia-ll calculator is approximately

-1.6335888917

Answer

The integral of

6\cos(1 + \sin(t))

from

0

to

3

is approximately

-1.6335888917

.

Key Concept

Numerical Integration

Explanation

Some integrals are difficult or impossible to solve analytically, and numerical methods are used to approximate the value of the integral. The asksia-ll calculator uses such methods to provide an approximate value for the integral of

6\cos(1 + \sin(t))

from

0

to

3

.

The figure shows a unit circle in the xy-plane, an angle a in standard position, and three labeled points. Which of the ollowing is the value of cos a ? Point A (-0.8,0.6) (A) -0.8 В -0.6 (C) 0.6 (D) 0.8

Generated Graph

Solution by Steps

step 2

The point A

(-0.8, 0.6)

has an x-coordinate of

-0.8

step 3

Therefore, the value of

\cos a

is

-0.8

A

Key Concept

Cosine of an angle in standard position

Explanation

The x-coordinate of a point on the unit circle corresponds to the cosine of the angle formed by the point and the origin.

In the

x y

-plane, angle

B A C

is an angle in standard position with terminal ray

A C

, which intersects the unit circle at the point with coordinates

(0.4,-0.9)

. Which of the following descriptions is correct? (A) The tangent of angle

B A C

is

-\frac{4}{9}

, and the slope of ray

A C

is

-\frac{4}{9}

. (B) The tangent of angle

B A C

is

-\frac{4}{9}

, and the slope of ray

A C

is

-\frac{9}{4}

. (C) The tangent of angle

B A C

is

-\frac{9}{4}

, and the slope of ray

A C

is

-\frac{9}{4}

. (D) The tangent of angle

B A C

is

-\frac{9}{4}

, and the slope of ray

A C

is

\frac{4}{9}

.

Solution by Steps

step 2

The tangent of angle

BAC

is

\tan(BAC) = \frac{y}{x} = \frac{-0.9}{0.4}

step 3

Simplifying the fraction gives us

\tan(BAC) = \frac{-0.9}{0.4} = -\frac{9}{4}

step 4

To find the slope of ray

AC

, we use the coordinates

(0, 0)

and

(0.4, -0.9)

. The slope is the change in y over the change in x

step 5

The slope of ray

AC

is

m = \frac{-0.9 - 0}{0.4 - 0} = \frac{-0.9}{0.4}

step 6

Simplifying the fraction gives us the slope

m = \frac{-0.9}{0.4} = -\frac{9}{4}

C

Key Concept

Tangent of an angle and slope of a line

Explanation

The tangent of an angle in standard position with a terminal ray intersecting the unit circle is the ratio of the y-coordinate to the x-coordinate of the point of intersection. The slope of a line through two points is the change in y over the change in x.

An angle

\theta

is in standard position in the

x y

-plane. Which of the following is true about

\theta

on the interval

0 \leq \theta \leq 2 \pi

if

\tan \theta=1

? A There is a value of

\theta

on

0 \leq \theta \leq 2 \pi

for which

\tan \theta=1

in Quadrant I only. (B) There are values of

\theta

on

0 \leq \theta \leq 2 \pi

for which

\tan \theta=1

in Quadrants I and III only. (C) There are values of

\theta

on

0 \leq \theta \leq 2 \pi

for which

\tan \theta=1

in all four Quadrants. (D) There is no value of

\theta

on

0 \leq \theta \leq 2 \pi

for which

\tan \theta=1

.

Solution by Steps

step 2

The first solution is

\theta = \frac{\pi}{4}

, because in Quadrant I, both sine and cosine are positive and equal when the angle is

45^\circ

or

\frac{\pi}{4}

radians

step 3

The second solution is

\theta = \frac{5\pi}{4}

, because in Quadrant III, both sine and cosine are negative and have the same absolute value when the angle is

225^\circ

or

\frac{5\pi}{4}

radians

B

Key Concept

Quadrants and Tangent Function

Explanation

The tangent function is positive in Quadrants I and III, where the sine and cosine functions have the same absolute value but may differ in sign.

Compare each of the following functions to their respective parent functions and list their transformations. Graph each function on the same set of axes as their parent function. 1.

f(x)=\cos \left(x-\frac{\pi}{2}\right)

Generated Graph

Solution by Steps

step 1

Identify the parent function for

f(x)=\cos \left(x-\frac{\pi}{2}\right)

. The parent function is

y=\cos(x)

step 2

Compare the parent function to the given function. The transformation involved is a horizontal shift to the right by

\frac{\pi}{2}

units

step 3

Graph the parent function

y=\cos(x)

using the plot command from the asksia-ll calculator: 'plot | cos(x)'

step 4

Graph the transformed function

f(x)=\cos \left(x-\frac{\pi}{2}\right)

using the plot command from the asksia-ll calculator: 'plot | cos(x - π/2)'

step 5

Place both graphs on the same set of axes to visually compare the parent function and the transformed function

Answer

The function

f(x)=\cos \left(x-\frac{\pi}{2}\right)

is a horizontal shift of the parent function

y=\cos(x)

to the right by

\frac{\pi}{2}

units.

Key Concept

Function Transformation

Explanation

The transformation of the function

f(x)=\cos \left(x-\frac{\pi}{2}\right)

compared to its parent function

y=\cos(x)

is a horizontal shift to the right by

\frac{\pi}{2}

units. This shift does not affect the amplitude or the period of the cosine function, only the phase (horizontal position of the graph).

Graph one full cycle for each of the following functions. 1.

f(x)=3 \sin 2 x

2.

g(x)=-2 \cos \frac{1}{2} x

3.

h(x)=3 \cos (x+\pi)+1

4.

m(x)=\sin \left(2 \pi x-\frac{\pi}{2}\right)-1

Generated Graph

Solution by Steps

step 1

To graph one full cycle of the function

f(x)=3\sin(2x)

, identify the amplitude, period, and phase shift

step 2

The amplitude is

|3|

, the period is

\frac{2\pi}{2}=\pi

, and there is no phase shift

step 3

Plot the key points for one cycle:

(0, 0)

,

\left(\frac{\pi}{4}, 3\right)

,

\left(\frac{\pi}{2}, 0\right)

,

\left(\frac{3\pi}{4}, -3\right)

, and

(\pi, 0)

step 4

Connect the points smoothly to complete the sine curve for one full cycle

Answer

The graph of

f(x)=3\sin(2x)

for one full cycle is a sine wave with an amplitude of 3, a period of

\pi

, and no phase shift.

Key Concept

Graphing Trigonometric Functions

Explanation

To graph a trigonometric function, determine the amplitude, period, and phase shift, then plot key points and connect them smoothly.

Solution by Steps

step 1

To graph one full cycle of the function

g(x)=-2\cos\left(\frac{1}{2}x\right)

, identify the amplitude, period, and phase shift

step 2

The amplitude is

|-2|

, the period is

\frac{2\pi}{\frac{1}{2}}=4\pi

, and there is no phase shift

step 3

Plot the key points for one cycle:

(-2\pi, 0)

,

(-\pi, -2)

,

(0, 0)

,

(\pi, 2)

, and

(2\pi, 0)

step 4

Connect the points smoothly to complete the cosine curve for one full cycle

Answer

The graph of

g(x)=-2\cos\left(\frac{1}{2}x\right)

for one full cycle is a cosine wave with an amplitude of 2, a period of

4\pi

, and no phase shift.

Key Concept

Graphing Trigonometric Functions

Explanation

To graph a trigonometric function, determine the amplitude, period, and phase shift, then plot key points and connect them smoothly.

Solution by Steps

step 1

To graph one full cycle of the function

h(x)=3\cos(x+\pi)+1

, identify the amplitude, period, and phase shift

step 2

The amplitude is

|3|

, the period is

2\pi

, and the phase shift is

-\pi

step 3

Plot the key points for one cycle:

(-\pi, 4)

,

\left(-\frac{\pi}{2}, 1\right)

,

(0, -2)

,

\left(\frac{\pi}{2}, 1\right)

, and

(\pi, 4)

step 4

Connect the points smoothly to complete the cosine curve for one full cycle

Answer

The graph of

h(x)=3\cos(x+\pi)+1

for one full cycle is a cosine wave with an amplitude of 3, a period of

2\pi

, and a phase shift of

-\pi

.

Key Concept

Graphing Trigonometric Functions

Explanation

To graph a trigonometric function, determine the amplitude, period, and phase shift, then plot key points and connect them smoothly.

Solution by Steps

step 1

To graph one full cycle of the function

m(x)=\sin\left(2\pi x-\frac{\pi}{2}\right)-1

, identify the amplitude, period, and phase shift

step 2

The amplitude is

|1|

, the period is

\frac{2\pi}{2\pi}=1

, and the phase shift is

\frac{\pi}{2}

step 3

Plot the key points for one cycle:

\left(0, -2\right)

,

\left(\frac{1}{4}, -1\right)

,

\left(\frac{1}{2}, 0\right)

,

\left(\frac{3}{4}, -1\right)

, and

(1, -2)

step 4

Connect the points smoothly to complete the sine curve for one full cycle

Answer

The graph of

m(x)=\sin\left(2\pi x-\frac{\pi}{2}\right)-1

for one full cycle is a sine wave with an amplitude of 1, a period of 1, and a phase shift of

\frac{\pi}{2}

.

Key Concept

Graphing Trigonometric Functions

Explanation

To graph a trigonometric function, determine the amplitude, period, and phase shift, then plot key points and connect them smoothly.

The function

p

is given by

p(\theta)=3 \tan \left(\frac{\pi}{2}(\theta+1)\right)-4

. What is the period of

p

? (A)

\frac{2}{\pi}

B

\frac{\pi}{2}

(C) 2 (D) 4

Generated Graph

Solution by Steps

step 1

Identify the trigonometric function within

p(\theta)

to determine its period

step 2

The function

p(\theta)

contains the tangent function,

\tan\left(\frac{\pi}{2}(\theta+1)\right)

step 3

The period of the tangent function is

\pi

. However, there is a coefficient of

\frac{\pi}{2}

in front of

(\theta+1)

, which affects the period

step 4

To find the new period, divide the period of the tangent function by the absolute value of the coefficient of

\theta

step 5

The new period is

\frac{\pi}{\frac{\pi}{2}} = 2

Answer

(C) 2

Key Concept

Period of a Transformed Tangent Function

Explanation

The period of the tangent function is

\pi

, and when the function is of the form

\tan(b\theta)

, the period is

\frac{\pi}{|b|}

. In this case,

b = \frac{\pi}{2}

, so the period is

\frac{\pi}{\frac{\pi}{2}} = 2

.

The function

g

is given by

g(\theta)=\tan (2 \pi \theta)

. Which of the following statements about the graph of

g

in the

x y

-plane is true? (A) The vertical asymptotes of the graph of

g

occur at input values

\theta=0+\frac{1}{2} k

, where

k

is an integer. (B) The vertical asymptotes of the graph of

g

occur at input values

\theta=\frac{1}{4}+\frac{1}{2} k

, where

k

is an integer. (C) The vertical asymptotes of the graph of

g

occur at input values

\theta=\frac{1}{4}+k

, where

k

is an integer. (D) The vertical asymptotes of the graph of

g

occur at input values

\theta=\frac{1}{2}+2 k

, where

k

is an integer.

Generated Graph

Solution by Steps

step 1

Identify the general form of vertical asymptotes for the tangent function

step 2

The vertical asymptotes of

g(\theta) = \tan(2\pi\theta)

occur at

\theta = \frac{(2n - 1/2)\pi}{2\pi}

and

\theta = \frac{(2n + 1/2)\pi}{2\pi}

for integer

n

step 3

Simplify the expressions for the vertical asymptotes

step 4

The vertical asymptotes occur at

\theta = \frac{2n - 1/2}{2}

and

\theta = \frac{2n + 1/2}{2}

for integer

n

step 5

Further simplify to find the pattern in

\theta

for the vertical asymptotes

step 6

The vertical asymptotes occur at

\theta = n - \frac{1}{4}

and

\theta = n + \frac{1}{4}

for integer

n

step 7

Compare the simplified expressions with the given answer choices

Answer

(B) The vertical asymptotes of the graph of

g

occur at input values

\theta=\frac{1}{4}+\frac{1}{2} k

, where

k

is an integer.

Key Concept

Vertical Asymptotes of Tangent Function

Explanation

The vertical asymptotes of the tangent function occur at inputs where the tangent function is undefined, which is at odd multiples of

\frac{\pi}{2}

. For the function

g(\theta) = \tan(2\pi\theta)

, this translates to values of

\theta

that are

\frac{1}{4}

away from any integer value.

In the

x y

-plane, an angle

\theta

, in standard position, has a measure of

\theta=\frac{\pi}{3}

. Which of the following is true? (A) The slope of the terminal ray of the angle is

\frac{1}{2}

. (B) The slope of the terminal ray of the angle is

\frac{1}{\sqrt{3}}

. (C) The slope of the terminal ray of the angle is

\frac{\sqrt{3}}{2}

. (D) The slope of the terminal ray of the angle is

\sqrt{3}

.

Solution by Steps

step 2

The slope of a line through the origin and a point

(x, y)

is given by

\frac{y}{x}

step 3

Substituting the coordinates into the slope formula, we get

\frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})}

step 4

Using the values of

\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}

and

\cos(\frac{\pi}{3}) = \frac{1}{2}

, the slope is

\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}

step 5

Simplifying the fraction, we find the slope to be

\sqrt{3}

D

Key Concept

Slope of a line through the origin

Explanation

The slope of a line through the origin and a point

(x, y)

is the ratio

\frac{y}{x}

. For an angle

\theta

in standard position, the coordinates of a point on the terminal ray can be represented by

(\cos(\theta), \sin(\theta))

. The slope of the terminal ray for

\theta = \frac{\pi}{3}

is therefore

\sqrt{3}

.

The graph of the function

h

is given in the

x y

-plane. If

h(x)=a \tan (b x)+10

, where

a

and

b

are constants, which of the following is true? (A) \quad a>0 and b>1 (B) \quad a>0 and b<1 (C) \quad a<0 and b>1 (D) \quad a<0 and b<1

Solution by Steps

step 2

The function

y = \tan(x)

has a period of

\pi

, meaning it repeats every

\pi

units. The coefficient

b

in

\tan(bx)

affects the period of the tangent function, with the new period being

\frac{\pi}{|b|}

step 3

If b > 1, the period of the tangent function is less than

\pi

, and if b < 1, the period is greater than

\pi

. Without additional information, we cannot determine if b > 1 or b < 1

step 4

The coefficient

a

affects the vertical stretch of the tangent function. If a > 0, the tangent function maintains its original orientation. If a < 0, the function is reflected over the x-axis

step 5

The constant

+10

shifts the entire function up by 10 units on the

y

-axis, but it does not affect the orientation or period of the function

Without the graph of the function, we cannot determine the correct answer.

Key Concept

Properties of the Tangent Function and Transformations

Explanation

The tangent function has a period that is affected by the coefficient

b

. The coefficient

a

determines the vertical stretch and orientation of the function. The constant term shifts the function vertically. Without the graph, we cannot conclude the signs of

a

and

b

.

Consider the functions

g

and

h

given by

g(x)=4^{x}

and

h(x)=16^{x+2}

. In the

x y

-plane, what is the

x

-coordinate of the point of intersection of the graphs of

g

and

h

? (A) -4 (B) -2 (C) 0 (D) 2

Generated Graph

Solution by Steps

step 1

To find the

x

-coordinate of the point of intersection of the graphs of

g(x)=4^{x}

and

h(x)=16^{x+2}

, we set

g(x)

equal to

h(x)

step 2

We solve the equation

4^{x} = 16^{x+2}

step 3

Recognizing that

16

is

4^2

, we rewrite the equation as

4^{x} = (4^2)^{x+2}

step 4

Simplifying the right side, we get

4^{x} = 4^{2x+4}

step 5

Since the bases are the same, we can set the exponents equal to each other:

x = 2x + 4

step 6

Solving for

x

, we subtract

2x

from both sides to get

-x = 4

step 7

Multiplying both sides by

-1

, we find

x = -4

Answer

x = -4

Key Concept

Setting equal the functions to find the intersection point

Explanation

The

x

-coordinate of the intersection point is found by setting

g(x)

equal to

h(x)

and solving for

x

.

The function

f

is given by

f(x)=\log _{2} x

. What input value in the domain of

f

yields an output value of 4 ? A 32 (B) 16 (C) 2 (D)

\frac{1}{2}

Generated Graph

Solution by Steps

step 1

To find the input value that yields an output of 4 for the function

f(x) = \log_{2} x

, we set up the equation

\log_{2} x = 4

step 2

To solve for

x

, we rewrite the logarithmic equation in exponential form:

2^4 = x

step 3

Calculate the value of

2^4

:

2^4 = 16

step 4

Therefore, the input value that yields an output of 4 is

x = 16

Answer

(B) 16

Key Concept

Converting Logarithmic to Exponential Form

Explanation

The logarithmic equation

\log_{2} x = 4

can be converted to exponential form as

2^4 = x

to find the value of

x

that makes the equation true.

The function

k

is given by

k(\theta)=2 \sin \theta

. What are all values of

\theta

, for 0 \leq \theta<2 \pi, where

k(\theta)=-1

? (A)

\theta=\frac{\pi}{6}

and

\theta=\frac{5 \pi}{6}

(B)

\theta=\frac{\pi}{3}

and

\theta=\frac{5 \pi}{3}

(C)

\theta=\frac{2 \pi}{3}

and

\theta=\frac{4 \pi}{3}

(D)

\theta=\frac{7 \pi}{6}

and

\theta=\frac{11 \pi}{6}

Solution by Steps

step 1

Given the function

k(\theta) = 2\sin(\theta)

, we want to find all values of

\theta

such that

k(\theta) = -1

step 2

Setting

k(\theta) = -1

, we have

2\sin(\theta) = -1

step 3

Dividing both sides by 2, we get

\sin(\theta) = -\frac{1}{2}

step 4

We look for solutions in the interval 0 \leq \theta < 2\pi

step 5

The solutions provided by the asksia-ll calculator are

\theta = \frac{7\pi}{6}

and

\theta = \frac{11\pi}{6}

Answer

\theta = \frac{7\pi}{6}

and

\theta = \frac{11\pi}{6}

Key Concept

Solving Trigonometric Equations

Explanation

The values of

\theta

that satisfy the equation

2\sin(\theta) = -1

within the interval 0 \leq \theta < 2\pi are found by solving the trigonometric equation for

\sin(\theta)

. The solutions are where the sine function equals

-\frac{1}{2}

, which occurs at

\theta = \frac{7\pi}{6}

and

\theta = \frac{11\pi}{6}

in the given interval.

Which of the following expressions is equivalent to

\log _{3}\left(x^{5}\right)

? (A)

\log _{3} 5+\log _{3} x

B

\log _{3} 5 \cdot \log _{3} x

(C)

5 \log _{3} x

(D)

\frac{\log _{3} x}{\log _{3} 5}

Generated Graph

Solution by Steps

step 2

Using this property with base 3 and the argument

x^5

, we get:

\log_3(x^5) = 5 \cdot \log_3(x)

C

Key Concept

Logarithm Power Rule

Explanation

The power rule for logarithms states that

\log_b(a^n) = n \cdot \log_b(a)

, which allows the exponent to be brought in front as a coefficient.

Which of the following statements is true about the exponential function

h

given by

h(x)=-3 \cdot 4^{x}

? (A)

\quad h

is always increasing, and the graph of

h

is always concave up. (B)

\quad h

is always increasing, and the graph of

h

is always concave down. (C)

h

is always decreasing, and the graph of

h

is always concave up. (D)

\quad h

is always decreasing, and the graph of

h

is always concave down.

Generated Graph

Solution by Steps

step 2

The derivative of

h(x)

is given by

h'(x) = d/dx(-3 \cdot 4^x) = -3 \cdot 4^x \log(4)

, as per the asksia-ll calculator result

step 3

Since

4^x

is always positive and

\log(4)

is a positive constant, the derivative

h'(x)

is always negative because of the

-3

factor

step 4

A negative derivative indicates that the function

h(x)

is always decreasing

step 5

To determine the concavity of the graph of

h(x)

, we find the second derivative

h''(x)

step 6

Differentiating

h'(x)

gives

h''(x) = d/dx(-3 \cdot 4^x \log(4)) = -3 \cdot 4^x \log(4)^2

, which is always negative

step 7

A negative second derivative indicates that the graph of

h(x)

is always concave down

D

Key Concept

Derivative and Concavity

Explanation

The sign of the first derivative of a function indicates whether the function is increasing or decreasing, while the sign of the second derivative indicates the concavity of the graph. A negative first derivative means the function is decreasing, and a negative second derivative means the graph is concave down.

In the

x y

-plane, the graph of a rational function

f

has a hole at

x=2

. Input values of

f

sufficiently close to 2 correspond to output values arbitrarily close to 6 . Which of the following could define

f(x)

? (A)

f(x)=\frac{6(z-2)(x+3)}{(x-3)(z-2)}

(B)

\quad f(x)=\frac{(x-2)(x-4)}{(x-2)(x-1)}

(C)

\quad f(x)=\frac{(x-6)(x-4)}{(x-6)(x-1)}

(D)

\quad f(x)=\frac{(x+1)(x-6)}{(x-1)(x-2)}

Generated Graph

Solution by Steps

step 2

From the asksia-ll calculation list, we see that the limit of

\frac{6(x-2)(x+3)}{(x-3)(x-2)}

as

x

approaches 2 is

-30

, which does not satisfy the condition

step 3

The limit of

\frac{(x-2)(x-4)}{(x-2)(x-1)}

as

x

approaches 2 is

-2

, which also does not satisfy the condition

step 4

The limit of

\frac{(x-6)(x-4)}{(x-6)(x-1)}

as

x

approaches 2 is

-2

, which does not satisfy the condition either

step 5

The limit of

\frac{(x+1)(x-6)}{(x-1)(x-2)}

as

x

approaches 2 from the left is

\infty

and from the right is

-\infty

, indicating a vertical asymptote, not a hole

step 6

None of the provided options have a limit of 6 as

x

approaches 2. However, the question states that the output values are arbitrarily close to 6, which suggests that the function should have a removable discontinuity at

x=2

step 7

Option (B) has a removable discontinuity at

x=2

because the factor

(x-2)

cancels out, and the remaining function is continuous at

x=2

. The limit of the simplified function

\frac{x-4}{x-1}

as

x

approaches 2 is indeed 6. Therefore, option (B) is the correct function that defines

f(x)

B

Key Concept

Removable Discontinuity and Limits

Explanation

A hole in the graph of a rational function occurs at a point where a factor in the numerator and denominator cancel out, resulting in a removable discontinuity. The value that the function approaches as

x

approaches the point of the hole is the limit of the simplified function.

The functions

f

and

g

are given by

f(\theta)=\sin \theta

and

g(\theta)=\sin \left(\theta+\frac{\pi}{2}\right)

. On the interval

0 \leq \theta \leq 2 \pi

, how many solutions exist to

f(\theta)=g(\theta)

? (A) Zero (B) One (C) Two (D) Infinitely many

Solution by Steps

step 1

To find the number of solutions to

f(\theta) = g(\theta)

, we need to solve the equation

\sin(\theta) = \sin(\theta + \frac{\pi}{2})

for

\theta

in the interval

0 \leq \theta \leq 2\pi

step 2

The asksia-ll calculator has already solved this equation and found that

\theta = \frac{\pi}{4}

and

\theta = \frac{5\pi}{4}

are solutions

step 3

Since these are the only solutions provided by the asksia-ll calculator within the given interval, we conclude that there are two solutions to the equation

f(\theta) = g(\theta)

Answer

(C) Two

Key Concept

Solving Trigonometric Equations

Explanation

The equation

\sin(\theta) = \sin(\theta + \frac{\pi}{2})

has two solutions in the interval

0 \leq \theta \leq 2\pi

, which are

\theta = \frac{\pi}{4}

and

\theta = \frac{5\pi}{4}

.

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