CALCU-Find the point of inflection of the graph of

Find the point of inflection of the graph of the function f(x) = 3sin(x over 6) on the interval (0, 12pi). Options: a. (0,0) b. (3pi,0) c. (7pi,0) d. (6pi,0) e. (6pi,3)

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Solution by Steps

step 2

The first derivative is

f'(x) = \frac{d}{dx} \left( 3 \sin\left(\frac{x}{6}\right) \right) = \frac{1}{2} \cos\left(\frac{x}{6}\right)

step 3

The second derivative is

f''(x) = \frac{d}{dx} \left( \frac{1}{2} \cos\left(\frac{x}{6}\right) \right) = -\frac{1}{12} \sin\left(\frac{x}{6}\right)

step 4

To find the inflection points, we set

f''(x) = 0

:

-\frac{1}{12} \sin\left(\frac{x}{6}\right) = 0

step 5

Solving for

x

, we get

\sin\left(\frac{x}{6}\right) = 0

. This occurs when

\frac{x}{6} = n\pi

for

n \in \mathbb{Z}

step 6

Therefore,

x = 6n\pi

step 7

On the interval

(0, 12\pi)

, the values of

x

that satisfy this are

x = 6\pi

and

x = 12\pi

step 8

Evaluating

f(x)

at these points, we get

f(6\pi) = 3 \sin\left(\frac{6\pi}{6}\right) = 3 \sin(\pi) = 0

and

f(12\pi) = 3 \sin\left(\frac{12\pi}{6}\right) = 3 \sin(2\pi) = 0

step 9

Therefore, the inflection points are

(6\pi, 0)

and

(12\pi, 0)

D

Key Concept

Inflection Points

Explanation

Inflection points occur where the second derivative changes sign, indicating a change in concavity.

Find the points of inflection and discuss the concavity of the function f(x) = sin x minus cos x on the interval (0, 2pi).

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Solution by Steps

step 1

To find the points of inflection and discuss the concavity of the function

f(x) = \sin(x) - \cos(x)

on the interval

(0, 2\pi)

, we first need to find the second derivative

f''(x)

step 2

The first derivative is

f'(x) = \frac{d}{dx} (\sin(x) - \cos(x)) = \cos(x) + \sin(x)

step 3

The second derivative is

f''(x) = \frac{d}{dx} (\cos(x) + \sin(x)) = -\sin(x) + \cos(x)

step 4

To find the points of inflection, we set the second derivative equal to zero:

-\sin(x) + \cos(x) = 0

step 5

Solving

-\sin(x) + \cos(x) = 0

gives

\cos(x) = \sin(x)

step 6

Dividing both sides by

\cos(x)

, we get

\tan(x) = 1

step 7

The solutions to

\tan(x) = 1

in the interval

(0, 2\pi)

are

x = \frac{\pi}{4}

and

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

step 8

To determine the concavity, we analyze the sign of

f''(x) = -\sin(x) + \cos(x)

around the points

x = \frac{\pi}{4}

and

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

step 9

For 0 < x < \frac{\pi}{4}, f''(x) > 0 (concave up)

step 10

For \frac{\pi}{4} < x < \frac{5\pi}{4}, f''(x) < 0 (concave down)

step 11

For \frac{5\pi}{4} < x < 2\pi, f''(x) > 0 (concave up)

Answer

The points of inflection are

x = \frac{\pi}{4}

and

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

. The function is concave up on

(0, \frac{\pi}{4})

and

(\frac{5\pi}{4}, 2\pi)

, and concave down on

(\frac{\pi}{4}, \frac{5\pi}{4})

.

Key Concept

Points of inflection and concavity

Explanation

Points of inflection occur where the second derivative changes sign. The concavity of the function is determined by the sign of the second derivative.

Find the points of inflection and discuss the concavity of the function f(x) = sin x minus cos x on the interval (0, 2pi). options: a. no inflection points. concave up on (0,2pi). b. concave down on (0,pi/2); concave up on (5pi/2 , 2pi); inflection point at (0,pi/2). c. no inflection points. concave down on (0,2pi). d. concave up on (0,pi/2) ; concave down on (5pi/2 , 2pi) ; inflection point at (0, pi/2). e. none of the above

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Solution by Steps

step 2

Next, we find the second derivative:

f''(x) = -\sin(x) + \cos(x)

step 3

Set the second derivative equal to zero to find potential inflection points:

-\sin(x) + \cos(x) = 0

. This simplifies to

\sin(x) = \cos(x)

, which occurs at

x = \frac{\pi}{4} + n\pi

for

n \in \mathbb{Z}

step 4

Within the interval

(0, 2\pi)

, the solutions are

x = \frac{\pi}{4}

and

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

step 5

To determine concavity, we analyze the sign of

f''(x)

on the intervals

(0, \frac{\pi}{4})

,

(\frac{\pi}{4}, \frac{5\pi}{4})

, and

(\frac{5\pi}{4}, 2\pi)

step 6

On

(0, \frac{\pi}{4})

, f''(x) > 0 (concave up). On

(\frac{\pi}{4}, \frac{5\pi}{4})

, f''(x) < 0 (concave down). On

(\frac{5\pi}{4}, 2\pi)

, f''(x) > 0 (concave up)

step 7

Therefore, the function is concave up on

(0, \frac{\pi}{4})

and

(\frac{5\pi}{4}, 2\pi)

, and concave down on

(\frac{\pi}{4}, \frac{5\pi}{4})

. The inflection points are at

x = \frac{\pi}{4}

and

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

d

Key Concept

Inflection Points and Concavity

Explanation

Inflection points occur where the second derivative changes sign. Concavity is determined by the sign of the second derivative: positive for concave up and negative for concave down.

Determine the open intervals on which the graph of f(x) = 5x^2 minus 7x minus 8 is concave downward or concave upward.

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Solution by Steps

step 1

To determine the concavity of the function

f(x) = 5x^2 - 7x - 8

, we first need to find the second derivative

step 2

The first derivative of

f(x)

is

f'(x) = \frac{d}{dx}(5x^2 - 7x - 8) = 10x - 7

step 3

The second derivative of

f(x)

is

f''(x) = \frac{d}{dx}(10x - 7) = 10

step 4

Since

f''(x) = 10

is a constant and always positive, the graph of

f(x)

is concave upward on the entire real line

Answer

The graph of

f(x) = 5x^2 - 7x - 8

is concave upward on the entire real line.

Key Concept

Concavity of a function

Explanation

The concavity of a function is determined by the sign of its second derivative. If the second derivative is positive, the function is concave upward; if it is negative, the function is concave downward.

Find all points of inflection of the graph of the function f(x) = 2sin(4x) + sin(8x) on the interval (0,1.571). Round your answer to three decimal places wherever applicable.

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Solution by Steps

step 1

To find the inflection points of the function

f(x) = 2\sin(4x) + \sin(8x)

on the interval

(0, 1.571)

, we first need to find the second derivative of the function

step 2

The first derivative of

f(x)

is

f'(x) = 8\cos(4x) + 8\cos(8x)

step 3

The second derivative of

f(x)

is

f''(x) = -32\sin(4x) - 64\sin(8x)

step 4

To find the inflection points, we set the second derivative equal to zero:

-32\sin(4x) - 64\sin(8x) = 0

step 5

Solving

-32\sin(4x) - 64\sin(8x) = 0

for

x

in the interval

(0, 1.571)

, we get the following points:

x \approx 0.455869

,

x \approx 0.785398

, and

x \approx 1.11493

step 6

These points are the inflection points of the function

f(x) = 2\sin(4x) + \sin(8x)

on the interval

(0, 1.571)

Answer

The inflection points are approximately at

x \approx 0.456

,

x \approx 0.785

, and

x \approx 1.115

.

Key Concept

Inflection Points

Explanation

Inflection points occur where the second derivative changes sign, indicating a change in concavity.

Find all points of inflection, if any exist, of the graph of the function f(x) = x sqrt(x + 21). Round your answers to two decimal places.

Generated Graph

Solution by Steps

step 1

To find the points of inflection of the function

f(x) = x \sqrt{x + 21}

, we first need to find the second derivative of the function

step 2

The first derivative

f'(x)

is found using the product rule:

f'(x) = \frac{d}{dx} \left( x \sqrt{x + 21} \right) = \sqrt{x + 21} + \frac{x}{2\sqrt{x + 21}}

step 3

Simplify the first derivative:

f'(x) = \sqrt{x + 21} + \frac{x}{2\sqrt{x + 21}} = \frac{2(x + 21) + x}{2\sqrt{x + 21}} = \frac{3x + 42}{2\sqrt{x + 21}}

step 4

Now, find the second derivative

f''(x)

using the quotient rule:

f''(x) = \frac{d}{dx} \left( \frac{3x + 42}{2\sqrt{x + 21}} \right)

f''(x) = \frac{(3)(2\sqrt{x + 21}) - (3x + 42)\left(\frac{1}{\sqrt{x + 21}}\right)}{(2\sqrt{x + 21})^2}

f''(x) = \frac{6\sqrt{x + 21} - \frac{3x + 42}{\sqrt{x + 21}}}{4(x + 21)}

f''(x) = \frac{6(x + 21) - (3x + 42)}{4(x + 21)\sqrt{x + 21}}

f''(x) = \frac{6x + 126 - 3x - 42}{4(x + 21)\sqrt{x + 21}}

f''(x) = \frac{3x + 84}{4(x + 21)\sqrt{x + 21}}

step 5

Set the second derivative equal to zero to find the points of inflection:

\frac{3x + 84}{4(x + 21)\sqrt{x + 21}} = 0

3x + 84 = 0

x = -28

step 6

Verify if

x = -28

is within the domain of the original function

f(x) = x \sqrt{x + 21}

. Since

x + 21 \geq 0

,

x \geq -21

. Therefore,

x = -28

is not within the domain

step 7

Since

x = -28

is not within the domain, there are no points of inflection for the function

f(x) = x \sqrt{x + 21}

Answer

There are no points of inflection for the function

f(x) = x \sqrt{x + 21}

.

Key Concept

Points of Inflection

Explanation

Points of inflection occur where the second derivative changes sign. In this case, the critical point found is not within the domain of the function, so no points of inflection exist.

Determine the open intervals on which the graph of y = negative 4x^3 + 5x^2 + 4x minus 7 is concave downward or concave upward.

Generated Graph

Solution by Steps

step 1

To determine the concavity of the function

y = -4x^3 + 5x^2 + 4x - 7

, we first need to find the second derivative

step 2

The first derivative of

y

is

y' = \frac{d}{dx}(-4x^3 + 5x^2 + 4x - 7) = -12x^2 + 10x + 4

step 3

The second derivative of

y

is

y'' = \frac{d^2}{dx^2}(-4x^3 + 5x^2 + 4x - 7) = -24x + 10

step 4

To find the points of inflection, we set the second derivative equal to zero:

-24x + 10 = 0

step 5

Solving for

x

, we get

x = \frac{10}{24} = \frac{5}{12}

step 6

To determine the concavity, we test the intervals around

x = \frac{5}{12}

step 7

For x < \frac{5}{12}, choose

x = 0

: y''(0) = -24(0) + 10 = 10 > 0, so the graph is concave upward

step 8

For x > \frac{5}{12}, choose

x = 1

: y''(1) = -24(1) + 10 = -14 < 0, so the graph is concave downward

Answer

The graph of

y = -4x^3 + 5x^2 + 4x - 7

is concave upward on the interval

(-\infty, \frac{5}{12})

and concave downward on the interval

(\frac{5}{12}, \infty)

.

Key Concept

Concavity of a function

Explanation

The concavity of a function is determined by the sign of its second derivative. If the second derivative is positive, the function is concave upward; if negative, the function is concave downward.

Find all relative extrema of the function f(x) = x^2 + 4x plus 1. Use the Second Derivative Test where applicable.

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Solution by Steps

step 1

To find the relative extrema of the function

f(x) = x^2 + 4x + 1

, we first need to find the first derivative of the function

step 2

The first derivative of

f(x)

is

f'(x) = 2x + 4

step 3

Set the first derivative equal to zero to find the critical points:

2x + 4 = 0

step 4

Solving for

x

, we get

x = -2

step 5

To determine the nature of the critical point, we use the second derivative test. The second derivative of

f(x)

is

f''(x) = 2

step 6

Since

f''(x) = 2

is positive, the function is concave up at

x = -2

, indicating a local minimum

step 7

The value of the function at

x = -2

is

f(-2) = (-2)^2 + 4(-2) + 1 = 4 - 8 + 1 = -3

Answer

The function

f(x) = x^2 + 4x + 1

has a local minimum at

x = -2

with a value of

-3

.

Key Concept

Relative Extrema

Explanation

The relative extrema of a function can be found by determining the critical points using the first derivative and then applying the second derivative test to classify the nature of these points.

so there is not a maximum?

No, there is no mention of a maximum in the provided chat history.

Determine the open intervals on which the graph of the function y = 4x minus tan(2x), (negative pi over 4, pi over 4) is concave upward or concave downward.

Generated Graph

Solution by Steps

step 1

To determine the concavity of the function

y = 4x - \tan(2x)

on the interval

(-\frac{\pi}{4}, \frac{\pi}{4})

, we first need to find the second derivative of the function

step 2

The first derivative of

y = 4x - \tan(2x)

is

y' = 4 - 2\sec^2(2x)

step 3

The second derivative of

y = 4x - \tan(2x)

is

y'' = -8\tan(2x)\sec^2(2x)

step 4

To determine the concavity, we need to analyze the sign of

y''

on the interval

(-\frac{\pi}{4}, \frac{\pi}{4})

step 5

The function

y'' = -8\tan(2x)\sec^2(2x)

is negative when

\tan(2x)

is positive and positive when

\tan(2x)

is negative

step 6

On the interval

(-\frac{\pi}{4}, \frac{\pi}{4})

,

\tan(2x)

is positive for

x \in (0, \frac{\pi}{4})

and negative for

x \in (-\frac{\pi}{4}, 0)

step 7

Therefore, the function

y = 4x - \tan(2x)

is concave downward on

(0, \frac{\pi}{4})

and concave upward on

(-\frac{\pi}{4}, 0)

Answer

The function

y = 4x - \tan(2x)

is concave downward on

(0, \frac{\pi}{4})

and concave upward on

(-\frac{\pi}{4}, 0)

.

Key Concept

Concavity of a function

Explanation

The concavity of a function is determined by the sign of its second derivative. If the second derivative is positive, the function is concave upward; if it is negative, the function is concave downward.

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