CALCU-Determine whether Rolle's Theorem can be app

Determine whether Rolle's Theorem can be applied to the function f(x) = cos(pi x) on the closed interval [negative 1 over 4, 1 over 4]. If Rolle's Theorem can be applied, find all numbers c in the open interval (negative 1 over 4, 1 over 4) such that f prime (c) = 0. Question 1 options: Rolle's Theorem applies; c = 0 Rolle's Theorem applies; c = negative 1 over 8, 1 over 8 Rolle's Theorem applies; c = 0, negative 1 over 8, 1 over 8 Rolle's Theorem applies; c = 1 over 8 Rolle's Theorem does not apply.

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Solución por Pasos

paso 1

Verificamos si se cumplen las condiciones del Teorema de Rolle para la función

f(x) = \cos(\pi x)

en el intervalo cerrado

\left[-\frac{1}{4}, \frac{1}{4}\right]

paso 2

La función

f(x) = \cos(\pi x)

es continua en

\left[-\frac{1}{4}, \frac{1}{4}\right]

y derivable en el intervalo abierto

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

paso 3

Evaluamos

f

en los extremos del intervalo:

f\left(-\frac{1}{4}\right) = \cos\left(\pi \left(-\frac{1}{4}\right)\right) = \cos\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

f\left(\frac{1}{4}\right) = \cos\left(\pi \left(\frac{1}{4}\right)\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

Como

f\left(-\frac{1}{4}\right) = f\left(\frac{1}{4}\right)

, se cumple la tercera condición del Teorema de Rolle

paso 4

Calculamos la derivada de

f(x)

:

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

Buscamos los valores de

c

en el intervalo abierto

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

donde

f'(c) = 0

:

-\pi \sin(\pi c) = 0

\sin(\pi c) = 0

Los valores de

c

que satisfacen esta ecuación en el intervalo dado son:

\pi c = n\pi \implies c = n

Donde

n

es un número entero. En el intervalo

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

, los valores posibles de

c

son

c = 0

Respuesta

Rolle's Theorem applies;

c = 0

Concepto Clave

Teorema de Rolle

Explicación

El Teorema de Rolle establece que si una función es continua en un intervalo cerrado, derivable en el intervalo abierto, y toma el mismo valor en los extremos del intervalo, entonces existe al menos un punto en el intervalo abierto donde la derivada de la función es cero. En este caso, la función

f(x) = \cos(\pi x)

cumple con todas estas condiciones en el intervalo

\left[-\frac{1}{4}, \frac{1}{4}\right]

, y el valor de

c

donde

f'(c) = 0

es

c = 0

.

Determine whether Rolle's Theorem can be applied to the function f(x) = cos(pi x) on the closed interval [negative 1 over 4, 1 over 4]. If Rolle's Theorem can be applied, find all numbers c in the open interval (negative 1 over 4, 1 over 4) such that f prime (c) = 0. Question 1 options: Rolle's Theorem applies; c = 0 Rolle's Theorem applies; c = negative 1 over 8, 1 over 8 Rolle's Theorem applies; c = 0, negative 1 over 8, 1 over 8 Rolle's Theorem applies; c = 1 over 8 Rolle's Theorem does not apply.

Solution by Steps

step 1

To determine if Rolle's Theorem can be applied, we first check if the function

f(x) = \cos(\pi x)

is continuous on the closed interval

\left[-\frac{1}{4}, \frac{1}{4}\right]

. Since

\cos(\pi x)

is a trigonometric function, it is continuous everywhere, including on

\left[-\frac{1}{4}, \frac{1}{4}\right]

step 2

Next, we check if

f(x) = \cos(\pi x)

is differentiable on the open interval

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

. Again, since

\cos(\pi x)

is differentiable everywhere, it is differentiable on

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

step 3

We now check if

f(a) = f(b)

, where

a = -\frac{1}{4}

and

b = \frac{1}{4}

. We have

f\left(-\frac{1}{4}\right) = \cos\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

and

f\left(\frac{1}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

. Since

f\left(-\frac{1}{4}\right) = f\left(\frac{1}{4}\right)

, the conditions for Rolle's Theorem are satisfied

step 4

According to Rolle's Theorem, there exists at least one

c

in the open interval

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

such that

f'(c) = 0

. We find

f'(x)

by differentiating

f(x) = \cos(\pi x)

:

f'(x) = -\pi \sin(\pi x)

step 5

We set

f'(c) = 0

to find the values of

c

:

-\pi \sin(\pi c) = 0

. This simplifies to

\sin(\pi c) = 0

. The solutions to

\sin(\pi c) = 0

in the interval

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

are

c = 0

Answer

Rolle's Theorem applies;

c = 0

Key Concept

Rolle's Theorem

Explanation

Rolle's Theorem states that if a function is continuous on a closed interval, differentiable on the open interval, and the function values at the endpoints of the interval are equal, then there exists at least one point in the open interval where the derivative of the function is zero.

why is not -1/8,1/8

Solution by Steps

step 1

To apply Rolle's Theorem, we need to ensure that the function

f(x) = \cos(\pi x)

is continuous on the closed interval

[-1/4, 1/4]

and differentiable on the open interval

(-1/4, 1/4)

step 2

We also need to check that

f(-1/4) = f(1/4)

. Since

f(x) = \cos(\pi x)

, we have

f(-1/4) = \cos(-\pi/4) = \cos(\pi/4) = f(1/4)

step 3

Since

f(x)

meets the conditions of Rolle's Theorem, there must be at least one

c

in the open interval

(-1/4, 1/4)

such that

f'(c) = 0

step 4

The derivative of

f(x)

is

f'(x) = -\pi \sin(\pi x)

. Setting this equal to zero, we get

-\pi \sin(\pi c) = 0

, which simplifies to

\sin(\pi c) = 0

step 5

The solutions to

\sin(\pi c) = 0

are

c = k

where

k

is an integer. In the interval

(-1/4, 1/4)

, the only integer is

0

. Therefore,

c = 0

is the only solution

step 6

The values

-1/8

and

1/8

are not solutions because

\sin(\pi \cdot -1/8) \neq 0

and

\sin(\pi \cdot 1/8) \neq 0

Answer

The only valid solution for

c

in the interval

(-1/4, 1/4)

is

c = 0

.

Key Concept

Rolle's Theorem

Explanation

Rolle's Theorem guarantees that if a function is continuous on a closed interval, differentiable on an open interval, and the function values at the endpoints of the interval are equal, then there is at least one point in the open interval where the derivative is zero. In this case,

c = 0

is the only such point.

Determine whether the Mean Value Theorem can be applied to the function f(x) = x^3 on the closed interval [0, 2]. If the Mean Value Theorem can be applied, find all numbers c in the open interval (0, 2) such that f prime (c) = (f(2) minus f(0)) over (2 minus 0). Question 2 options: MVT applies; negative 2sqrt(3) over 3 MVT applies; 0.5 MVT applies; 2sqrt(3) over 3 MVT applies; 1 MVT does not apply

Solution by Steps

step 1

To determine if the Mean Value Theorem (MVT) can be applied to the function

f(x) = x^3

on the closed interval

[0, 2]

, we first need to check if

f(x)

is continuous on

[0, 2]

and differentiable on

(0, 2)

step 2

The function

f(x) = x^3

is a polynomial, which is continuous and differentiable everywhere. Therefore,

f(x)

is continuous on

[0, 2]

and differentiable on

(0, 2)

step 3

Since

f(x)

meets the conditions of the MVT, we can apply the theorem. According to the MVT, there exists at least one number

c

in the open interval

(0, 2)

such that

f'(c) = \frac{f(2) - f(0)}{2 - 0}

step 4

Calculate

f(2)

and

f(0)

:

f(2) = 2^3 = 8

and

f(0) = 0^3 = 0

step 5

Compute the average rate of change:

\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2 - 0} = 4

step 6

Find the derivative of

f(x)

:

f'(x) = 3x^2

step 7

Set

f'(c) = 4

:

3c^2 = 4

step 8

Solve for

c

:

c^2 = \frac{4}{3}

, so

c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2\sqrt{3}}{3}

step 9

Since

c

must be in the open interval

(0, 2)

, we discard the negative value. Thus,

c = \frac{2\sqrt{3}}{3}

Answer

MVT applies;

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

Key Concept

Mean Value Theorem (MVT)

Explanation

The MVT states that for a function continuous on a closed interval and differentiable on the open interval, there exists at least one point where the instantaneous rate of change (derivative) equals the average rate of change over the interval.

Determine whether Rolle's Theorem can be applied to f(x) = negative x^2 + 22x on the closed interval [0, 22]. If Rolle's Theorem can be applied, find all values of c in the open interval (0, 22) such that f prime (c) = 0. Question 3 options: Rolle's Theorem applies; c = 10, c = 11 Rolle's Theorem applies; c = 4, c = 10 Rolle's Theorem applies; c = 11 Rolle's Theorem applies; c = 4 Rolle's Theorem does not apply

Solution by Steps

step 1

To determine if Rolle's Theorem can be applied, we first check if the function

f(x) = -x^2 + 22x

is continuous on the closed interval

[0, 22]

and differentiable on the open interval

(0, 22)

step 2

The function

f(x) = -x^2 + 22x

is a polynomial, which is continuous and differentiable everywhere. Therefore, it is continuous on

[0, 22]

and differentiable on

(0, 22)

step 3

Next, we check if

f(0) = f(22)

step 4

Calculate

f(0)

:

f(0) = -(0)^2 + 22(0) = 0

step 5

Calculate

f(22)

:

f(22) = -(22)^2 + 22(22) = -484 + 484 = 0

step 6

Since

f(0) = f(22) = 0

, all conditions for Rolle's Theorem are satisfied

step 7

To find the values of

c

in the open interval

(0, 22)

such that

f'(c) = 0

, we first find the derivative

f'(x)

step 8

The derivative of

f(x) = -x^2 + 22x

is

f'(x) = -2x + 22

step 9

Set

f'(x) = 0

and solve for

x

:

-2x + 22 = 0 \implies 2x = 22 \implies x = 11

step 10

Therefore, the value of

c

in the open interval

(0, 22)

such that

f'(c) = 0

is

c = 11

Answer

Rolle's Theorem applies;

c = 11

Key Concept

Rolle's Theorem

Explanation

Rolle's Theorem states that if a function is continuous on a closed interval, differentiable on the open interval, and the function values at the endpoints of the interval are equal, then there exists at least one point in the open interval where the derivative is zero.

Determine whether Rolle's Theorem can be applied to f(x) = (x^2 minus 16) over x on the closed interval [negative 16, 16]. If Rolle's Theorem can be applied, find all values of c in the open interval (negative 16, 16) such that f prime (c) = 0. Question 4 options: c = 2 c = 3, c = 4 c = 4, c = 2 c = 3 Rolle's Theorem does not apply

Solution by Steps

step 1

To apply Rolle's Theorem, we need to check if the function

f(x) = \frac{x^2 - 16}{x}

is continuous on the closed interval

[-16, 16]

and differentiable on the open interval

(-16, 16)

step 2

The function

f(x) = \frac{x^2 - 16}{x}

can be simplified to

f(x) = x - \frac{16}{x}

for

x \neq 0

step 3

Check continuity:

f(x)

is continuous on

[-16, 16]

except at

x = 0

. Since

0

is within the interval,

f(x)

is not continuous on the entire interval

step 4

Since

f(x)

is not continuous on the closed interval

[-16, 16]

, Rolle's Theorem does not apply

Answer

Rolle's Theorem does not apply

Key Concept

Continuity and differentiability conditions for Rolle's Theorem

Explanation

Rolle's Theorem requires the function to be continuous on the closed interval and differentiable on the open interval. Since

f(x)

is not continuous at

x = 0

, the theorem does not apply.

Find a function f that has derivative f prime (x) = 6x + 2 and with graph passing through the point (negative 5, 2). Question 5 options: f(x) = negative (2 over 25)x^2 f(x) = 6x^2 + 2x minus 61 f(x) = 3x^2 + 2x minus 60 f(x) = 3x^2 + 2x minus 63 f(x) = negative (2 over 5)x

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Solution by Steps

step 1

To find a function

f(x)

that has a derivative

f'(x) = 6x + 2

and passes through the point

(-5, 2)

, we first integrate the derivative

step 2

The integral of

6x + 2

is

3x^2 + 2x + C

, where

C

is the constant of integration

step 3

We now use the point

(-5, 2)

to find the value of

C

. Substituting

x = -5

and

f(x) = 2

into the equation

f(x) = 3x^2 + 2x + C

, we get:

2 = 3(-5)^2 + 2(-5) + C

step 4

Simplifying the equation:

2 = 75 - 10 + C \implies 2 = 65 + C \implies C = 2 - 65 \implies C = -63

step 5

Therefore, the function

f(x)

is

f(x) = 3x^2 + 2x - 63

Answer

f(x) = 3x^2 + 2x - 63

Key Concept

Integration to find the original function

Explanation

To find a function with a given derivative, we integrate the derivative and use any given points to solve for the constant of integration.

Determine whether Rolle's Theorem can be applied to the function f(x) = x^2 minus 14x + 40 on the closed interval [4, 10]. If Rolle's Theorem can be applied, find all values of c in the open interval (4, 10) such that f prime (c) = 0. Question 6 options: Rolle's Theorem applies; c = 7 Rolle's Theorem applies; c = 8.5 Rolle's Theorem applies; c = 5.5 Rolle's Theorem applies; c = negative 7 Rolle's Theorem does not apply

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Solution by Steps

step 1

To determine if Rolle's Theorem can be applied to the function

f(x) = x^2 - 14x + 40

on the closed interval

[4, 10]

, we first need to check if

f(x)

is continuous on

[4, 10]

and differentiable on

(4, 10)

step 2

The function

f(x) = x^2 - 14x + 40

is a polynomial, which is continuous and differentiable everywhere. Therefore,

f(x)

is continuous on

[4, 10]

and differentiable on

(4, 10)

step 3

Next, we need to check if

f(4) = f(10)

step 4

Calculate

f(4)

:

f(4) = 4^2 - 14 \cdot 4 + 40 = 16 - 56 + 40 = 0

step 5

Calculate

f(10)

:

f(10) = 10^2 - 14 \cdot 10 + 40 = 100 - 140 + 40 = 0

step 6

Since

f(4) = f(10) = 0

, all conditions for Rolle's Theorem are satisfied

step 7

According to Rolle's Theorem, there exists at least one

c

in the open interval

(4, 10)

such that

f'(c) = 0

step 8

Find

f'(x)

:

f'(x) = 2x - 14

step 9

Set

f'(x) = 0

and solve for

x

:

2x - 14 = 0 \implies 2x = 14 \implies x = 7

step 10

Therefore,

c = 7

is the value in the open interval

(4, 10)

where

f'(c) = 0

Answer

Rolle's Theorem applies;

c = 7

Key Concept

Rolle's Theorem

Explanation

Rolle's Theorem states that if a function is continuous on a closed interval, differentiable on the open interval, and the function values at the endpoints of the interval are equal, then there exists at least one point in the open interval where the derivative of the function is zero.

Determine whether the Mean Value Theorem can be applied to the function f(x) = 2sin(x) + sin(2x) on the closed interval [8pi, 9pi]. If the Mean Value Theorem can be applied, find all numbers c in the open interval (8pi, 9pi) such that f prime (c) = (f(9pi) minus f(8pi)) over (9pi minus 8pi). Question 7 options: MVT applies; 17pi over 2 MVT applies; 25pi over 4 MVT applies; 25pi over 6 MVT applies; 25pi over 3 MVT does not apply

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Solution by Steps

step 1

To determine if the Mean Value Theorem (MVT) can be applied to the function

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

on the closed interval

[8\pi, 9\pi]

, we first need to check if

f(x)

is continuous on

[8\pi, 9\pi]

and differentiable on

(8\pi, 9\pi]

. Since

f(x)

is a trigonometric function, it is continuous and differentiable everywhere

step 2

Next, we calculate

f(9\pi)

and

f(8\pi)

:

f(9\pi) = 2\sin(9\pi) + \sin(18\pi) = 0

f(8\pi) = 2\sin(8\pi) + \sin(16\pi) = 0

Thus,

f(9\pi) - f(8\pi) = 0

step 3

The Mean Value Theorem states that there exists at least one

c

in

(8\pi, 9\pi)

such that:

f'(c) = \frac{f(9\pi) - f(8\pi)}{9\pi - 8\pi} = \frac{0}{\pi} = 0

We need to find

c

such that

f'(c) = 0

step 4

Differentiate

f(x)

:

f'(x) = \frac{d}{dx}(2\sin(x) + \sin(2x)) = 2\cos(x) + 2\cos(2x)

Set

f'(c) = 0

:

2\cos(c) + 2\cos(2c) = 0

\cos(c) + \cos(2c) = 0

Using the double-angle identity

\cos(2c) = 2\cos^2(c) - 1

, we get:

\cos(c) + 2\cos^2(c) - 1 = 0

2\cos^2(c) + \cos(c) - 1 = 0

Solve the quadratic equation for

\cos(c)

:

\cos(c) = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}

\cos(c) = \frac{1}{2} \quad \text{or} \quad \cos(c) = -1

Since

c

is in

(8\pi, 9\pi)

,

\cos(c) = \frac{1}{2}

is the valid solution

step 5

Find

c

such that

\cos(c) = \frac{1}{2}

in

(8\pi, 9\pi)

:

c = 2k\pi \pm \frac{\pi}{3}

For

c

in

(8\pi, 9\pi)

, we get:

c = 8\pi + \frac{\pi}{3} = \frac{25\pi}{3}

Thus,

c = \frac{25\pi}{3}

is the solution

Answer

MVT applies;

\frac{25\pi}{3}

Key Concept

Mean Value Theorem (MVT)

Explanation

The MVT states that for a continuous and differentiable function on a closed interval, there exists at least one point where the derivative equals the average rate of change over that

Determine whether the Mean Value Theorem can be applied to the function f(x) = x^2on the closed interval [negative 6, 2]. If the Mean Value Theorem can be applied, find all numbers c in the open interval (negative 6, 2) such that f prime (c) = (f(2) minus f(negative 6)) over (2 minus (negative 6)). Question 8 options: MVT applies; c = negative 2 MVT applies; c = negative 1 MVT applies; c = negative 4 MVT applies; c = negative 3 MVT applies; c = 0

Solution by Steps

step 1

To determine if the Mean Value Theorem (MVT) can be applied, we need to check if the function

f(x) = x^2

is continuous on the closed interval

[-6, 2]

and differentiable on the open interval

(-6, 2)

step 2

The function

f(x) = x^2

is a polynomial, which is continuous and differentiable everywhere. Therefore,

f(x)

is continuous on

[-6, 2]

and differentiable on

(-6, 2)

step 3

Since

f(x)

meets the conditions of the MVT, we can apply the theorem. According to the MVT, there exists at least one number

c

in the open interval

(-6, 2)

such that

f'(c) = \frac{f(2) - f(-6)}{2 - (-6)}

step 4

Calculate

f(2)

and

f(-6)

:

f(2) = 2^2 = 4

and

f(-6) = (-6)^2 = 36

step 5

Compute the difference quotient:

\frac{f(2) - f(-6)}{2 - (-6)} = \frac{4 - 36}{2 + 6} = \frac{-32}{8} = -4

step 6

Find

f'(x)

:

f'(x) = 2x

step 7

Set

f'(c) = -4

:

2c = -4 \implies c = -2

step 8

Verify that

c = -2

is in the open interval

(-6, 2)

. Since

-2

is within

(-6, 2)

, it satisfies the conditions of the MVT

Answer

MVT applies;

c = -2

Key Concept

Mean Value Theorem (MVT)

Explanation

The Mean Value Theorem states that for a function continuous on a closed interval and differentiable on the open interval, there exists at least one point where the derivative equals the average rate of change over the interval.

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