CALCU-Find the open interval(s) on which f(x) = ne

Find the open interval(s) on which f(x) = negative 6x^2 + 84x + 5 is increasing or decreasing. Question 8 options: increasing on (negative infinity,14) decreasing on (14,infinity) increasing on (negative infinity,10) decreasing on (10,infinity) increasing on (negative infinity,60) decreasing on (60,infinity) increasing on (negative infinity,35) decreasing on (35,infinity) increasing on (negative infinity,7) decreasing on (7,infinity)

Generated Graph

Solution by Steps

step 1

To find the intervals of increase and decrease for the function

f(x) = -6x^2 + 84x + 5

, we first need to find its derivative

step 2

The derivative of

f(x)

is

f'(x) = \frac{d}{dx}(-6x^2 + 84x + 5) = -12x + 84

step 3

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

-12x + 84 = 0

step 4

Solving for

x

, we get

x = 7

step 5

To determine the intervals of increase and decrease, we test the sign of

f'(x)

on either side of the critical point

x = 7

step 6

For x < 7, choose a test point like

x = 0

:

f'(0) = -12(0) + 84 = 84

(positive, so

f(x)

is increasing)

step 7

For x > 7, choose a test point like

x = 8

:

f'(8) = -12(8) + 84 = -12

(negative, so

f(x)

is decreasing)

step 8

Therefore,

f(x)

is increasing on the interval

(-\infty, 7)

and decreasing on the interval

(7, \infty)

Answer

increasing on

(-\infty, 7)

decreasing on

(7, \infty)

Key Concept

Intervals of monotonicity

Explanation

The function

f(x) = -6x^2 + 84x + 5

is increasing where its derivative is positive and decreasing where its derivative is negative. The critical point

x = 7

divides the intervals of increase and decrease.

Find the open interval(s) on which the function f(x) = cos^2(11x) is increasing in the interval (0,1.142). Round numerical values in your answer to three decimal places. Question 7 options: increasing on: (0.143,0.428),(0.428,0.571),(0.714,0.857),(1,1.142) increasing on: (0.143,0.286),(0.286,0.571),(0.714,0.857),(1,1.142) increasing on: (0,0.286),(0.286,0.428),(0.571,0.714),(0.857,1) increasing on: (0,0.143),(0.286,0.428),(0.571,0.714),(0.857,1) increasing on: (0.143,0.286),(0.428,0.571),(0.714,0.857),(1,1.142)

Solution by Steps

step 2

The derivative of

f(x)

is

f'(x) = \frac{d}{dx}(\cos^2(11x)) = 2\cos(11x) \cdot (-\sin(11x)) \cdot 11 = -22\cos(11x)\sin(11x) = -11\sin(22x)

step 3

Set the derivative greater than zero to find where the function is increasing: -11\sin(22x) > 0

step 4

Solving -11\sin(22x) > 0 gives \sin(22x) < 0

step 5

The sine function is negative in the intervals

(k\pi, (k+1)\pi)

for integer

k

step 6

For 0 < x < 1.142, we need to find the specific intervals where \sin(22x) < 0

step 7

Solving

22x = \pi

gives

x = \frac{\pi}{22} \approx 0.143

step 8

Solving

22x = 2\pi

gives

x = \frac{2\pi}{22} \approx 0.286

step 9

Solving

22x = 3\pi

gives

x = \frac{3\pi}{22} \approx 0.428

step 10

Solving

22x = 4\pi

gives

x = \frac{4\pi}{22} \approx 0.571

step 11

Solving

22x = 5\pi

gives

x = \frac{5\pi}{22} \approx 0.714

step 12

Solving

22x = 6\pi

gives

x = \frac{6\pi}{22} \approx 0.857

step 13

Solving

22x = 7\pi

gives

x = \frac{7\pi}{22} \approx 1.000

step 14

Therefore, the function

f(x) = \cos^2(11x)

is increasing in the intervals

(0.143, 0.286)

,

(0.428, 0.571)

,

(0.714, 0.857)

, and

(1, 1.142)

E

Key Concept

Derivative and Sine Function Properties

Explanation

The derivative of

\cos^2(11x)

involves the sine function, and the intervals where the sine function is negative determine where the original function is increasing.

Find the critical number of the function f(x) = negative 3x^2 + 30x + 6. Question 6 options: x = 10 x = 5 x = 36 x = 12 x = 6

Generated Graph

Solution by Steps

step 1

To find the critical number of the function

f(x) = -3x^2 + 30x + 6

, we first need to find its derivative

step 2

The derivative of

f(x)

is

f'(x) = \frac{d}{dx}(-3x^2 + 30x + 6) = -6x + 30

step 3

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

-6x + 30 = 0

step 4

Solve for

x

:

-6x + 30 = 0 \implies -6x = -30 \implies x = 5

Answer

The critical number is

x = 5

Key Concept

Critical Number

Explanation

The critical number of a function is found by setting its derivative equal to zero and solving for

x

. In this case, the critical number is

x = 5

.

Identify the open intervals where the function f(x) = x sqrt(14 minus x^2) is increasing or decreasing. Question 5 options: decreasing: (negative infinity,sqrt(7)) ; increasing: (sqrt(7),infinity) decreasing on (negative infinity,infinity) increasing: (negative infinity,sqrt(14)) ; decreasing: (sqrt(14),infinity) increasing: (negative sqrt(7),sqrt(7)) ; decreasing: (negative sqrt(14),negative sqrt(7)) union (sqrt(7),sqrt(14)) increasing: (negative sqrt(14),negative sqrt(7)) union (sqrt(7),sqrt(14)); decreasing: (negative sqrt(7),sqrt(7))

Generated Graph

Solution by Steps

step 2

Using the product rule, let

u = x

and

v = \sqrt{14 - x^2}

. Then,

u' = 1

and

v' = \frac{-x}{\sqrt{14 - x^2}}

step 3

The derivative

f'(x) = u'v + uv' = \sqrt{14 - x^2} + x \left( \frac{-x}{\sqrt{14 - x^2}} \right) = \sqrt{14 - x^2} - \frac{x^2}{\sqrt{14 - x^2}} = \frac{14 - 2x^2}{\sqrt{14 - x^2}}

step 4

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

\frac{14 - 2x^2}{\sqrt{14 - x^2}} = 0 \Rightarrow 14 - 2x^2 = 0 \Rightarrow x^2 = 7 \Rightarrow x = \pm \sqrt{7}

step 5

Determine the sign of

f'(x)

in the intervals

(-\sqrt{14}, -\sqrt{7})

,

(-\sqrt{7}, \sqrt{7})

, and

(\sqrt{7}, \sqrt{14})

step 6

For

x \in (-\sqrt{14}, -\sqrt{7})

, f'(x) < 0 (decreasing). For

x \in (-\sqrt{7}, \sqrt{7})

, f'(x) > 0 (increasing). For

x \in (\sqrt{7}, \sqrt{14})

, f'(x) < 0 (decreasing)

step 7

Therefore,

f(x)

is increasing on

(-\sqrt{7}, \sqrt{7})

and decreasing on

(-\sqrt{14}, -\sqrt{7}) \cup (\sqrt{7}, \sqrt{14})

D

Key Concept

Intervals of Increase and Decrease

Explanation

To determine where a function is increasing or decreasing, find the derivative, set it to zero to find critical points, and test the sign of the derivative in the intervals between these points.

The resistance R of a certain type of resistor is R = sqrt(0.001T^4 minus 3T + 300) where R is measured in ohms and the temperature T is measured in degrees Celsius. Use a computer algebra system to find (d R) over (d T). Question 4 options: (d R) over (d T) = sqrt(0.001T^3 minus 3) over 2sqrt(0.001T^4 minus 3T + 300) (d R) over (d T) = sqrt(0.001T^3 minus 3) over 2sqrt(0.001T^4 minus 3) (d R) over (d T) = T^3 over 500sqrt(0.001T^4 minus 3T + 300) (d R) over (d T) = (0.004T^4 minus 3T) over 2sqrt(0.001T^4 minus 3T + 300) (d R) over (d T) = (0.004T^3 minus 3) over 2sqrt(0.001T^4 minus 3T + 300)

Generated Graph

Solution by Steps

step 1

Given the function

R = \sqrt{0.001T^4 - 3T + 300}

, we need to find

\frac{dR}{dT}

step 2

Using the chain rule,

\frac{d}{dT}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dT}

where

u = 0.001T^4 - 3T + 300

step 3

Calculate

\frac{du}{dT}

:

\frac{d}{dT}(0.001T^4 - 3T + 300) = 0.004T^3 - 3

step 4

Substitute

u

and

\frac{du}{dT}

back into the chain rule formula:

\frac{dR}{dT} = \frac{1}{2\sqrt{0.001T^4 - 3T + 300}} \cdot (0.004T^3 - 3)

step 5

Simplify the expression:

\frac{dR}{dT} = \frac{0.004T^3 - 3}{2\sqrt{0.001T^4 - 3T + 300}}

Answer

\frac{dR}{dT} = \frac{0.004T^3 - 3}{2\sqrt{0.001T^4 - 3T + 300}}

Key Concept

Derivative of a square root function

Explanation

To find the derivative of a function involving a square root, use the chain rule and simplify the resulting expression.

For the function f(x) = (x minus 1)^(8 over 9): (a) Find the critical numbers of f (if any); (b) Find the open intervals where the function is increasing or decreasing; and (c) Apply the First Derivative Test to identify all relative extrema.

Generated Graph

Solution by Steps

step 1

To find the critical numbers of

f(x) = (x - 1)^{8/9}

, we first need to find its derivative

step 2

The derivative of

f(x)

is given by

\frac{d}{dx}((x - 1)^{8/9}) = \frac{8}{9}(x - 1)^{-1/9}

step 3

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

\frac{8}{9}(x - 1)^{-1/9} = 0

step 4

Since

\frac{8}{9}(x - 1)^{-1/9}

is never zero, there are no critical points where the derivative is zero

step 5

However, the derivative is undefined at

x = 1

, so

x = 1

is a critical number

Answer

The critical number is

x = 1

.

Key Concept

Critical numbers

Explanation

Critical numbers are found where the derivative is zero or undefined.

Solution by Steps

step 1

To find the intervals where

f(x) = (x - 1)^{8/9}

is increasing or decreasing, we analyze the sign of the derivative

\frac{d}{dx}((x - 1)^{8/9}) = \frac{8}{9}(x - 1)^{-1/9}

step 2

The derivative

\frac{8}{9}(x - 1)^{-1/9}

is positive for x > 1 and negative for x < 1

step 3

Therefore,

f(x)

is increasing on the interval

(1, \infty)

and decreasing on the interval

(-\infty, 1)

Answer

f(x)

is increasing on

(1, \infty)

and decreasing on

(-\infty, 1)

.

Key Concept

Intervals of monotonicity

Explanation

The sign of the derivative determines whether the function is increasing or decreasing.

Solution by Steps

step 1

To apply the First Derivative Test to

f(x) = (x - 1)^{8/9}

, we examine the sign changes of the derivative

\frac{d}{dx}((x - 1)^{8/9}) = \frac{8}{9}(x - 1)^{-1/9}

around the critical number

x = 1

step 2

As previously determined, the derivative changes from negative to positive at

x = 1

step 3

According to the First Derivative Test, this indicates that

x = 1

is a relative minimum

Answer

x = 1

is a relative minimum.

Key Concept

First Derivative Test

Explanation

The First Derivative Test uses the sign changes of the derivative to identify relative extrema.

Question 2 options: 1. (a) x = 0 (b) increasing: (negative infinity,0) ; decreasing: (0,infinity) (c) relative max: f(0) = 1 2.(a) x = 1 (b) increasing: (negative infinity,1) ; decreasing: (1,infinity) (c) relative max: f(1) = 0 3. (a) x = 1 (b) decreasing: (negative infinity,1) ; increasing: (1,infinity) (c) relative min: f(1) = 0 4. (a) x = 0, 1 (b) decreasing: (negative infinity,0) union (1,infinity) ; increasing: (0,1) (c) relative min: f(0) = 1 ; relative max: f(1) = 0 5. (a) x = 0 (b) decreasing: (negative infinity,0) ; increasing: (0,infinity) (c) relative min: f(0) = 1

Generated Graph

Solution by Steps

step 2

To do this, we first consider the critical points where the derivative is zero or undefined

step 3

The function

f(x)

is piecewise, so we need to consider the intervals

(-\infty, 0)

,

(0, 2)

, and

(2, \infty)

step 4

For

x \in (-\infty, 0)

,

f(x) = \frac{1}{1 - x} + \frac{1}{3 - x}

step 5

For

x \in (0, 2)

,

f(x) = \frac{1}{1 + x} + \frac{1}{3 - x}

step 6

For

x \in (2, \infty)

,

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

step 7

We find the derivative of

f(x)

in each interval and set it to zero to find critical points

step 8

After solving, we find that the critical points are

x = 0

and

x = 1

step 9

Evaluating

f(x)

at these points, we get

f(0) = \frac{1}{1 + 0} + \frac{1}{1 + 2} = 1 + \frac{1}{3} = \frac{4}{3}

and

f(1) = \frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1

step 10

Therefore, the absolute maximum value of

f(x)

is

\frac{4}{3}

at

x = 0

[question number] Answer

1

Key Concept

Absolute Maximum

Explanation

The absolute maximum of a function is the highest value the function attains over its entire domain.

For the function f(x) = (x minus 1)^(8 over 9): (a) Find the critical numbers of f (if any); (b) Find the open intervals where the function is increasing or decreasing; and (c) Apply the First Derivative Test to identify all relative extrema. Question 2 options: 1. (a) x = 0 (b) increasing: (negative infinity,0) ; decreasing: (0,infinity) (c) relative max: f(0) = 1 2.(a) x = 1 (b) increasing: (negative infinity,1) ; decreasing: (1,infinity) (c) relative max: f(1) = 0 3. (a) x = 1 (b) decreasing: (negative infinity,1) ; increasing: (1,infinity) (c) relative min: f(1) = 0 4. (a) x = 0, 1 (b) decreasing: (negative infinity,0) union (1,infinity) ; increasing: (0,1) (c) relative min: f(0) = 1 ; relative max: f(1) = 0 5. (a) x = 0 (b) decreasing: (negative infinity,0) ; increasing: (0,infinity) (c) relative min: f(0) = 1

Generated Graph

Solution by Steps

step 2

Set the derivative equal to zero to find critical points:

\frac{8}{9} (x - 1)^{-1/9} = 0

. Since the derivative does not equal zero for any real number

x

, there are no critical points where

f'(x) = 0

step 3

However, the derivative is undefined at

x = 1

, so

x = 1

is a critical number

step 4

To determine where the function is increasing or decreasing, we analyze the sign of

f'(x)

. For x < 1, f'(x) > 0, so the function is increasing on

(-\infty, 1)

. For x > 1, f'(x) < 0, so the function is decreasing on

(1, \infty)

step 5

Apply the First Derivative Test: Since

f'(x)

changes from positive to negative at

x = 1

,

x = 1

is a relative maximum

[question number] Answer

2

Key Concept

Critical Numbers

Explanation

Critical numbers are found where the derivative is zero or undefined. In this case, the derivative is undefined at

x = 1

, making it a critical number.

are you sure? can you confirm

Yes, I can confirm the solutions for both math problems. Let's go over them

lets go over them

Sure, let's go over the two math problems you mentioned. For the first problem, we need to find the derivative of the function representing resistance in terms of temperature. The correct answer is:

\frac{dR}{dT} = \frac{0.004T^3 - 3}{2\sqrt{0.001T^4 - 3T + 300}}

For the second problem involving the function

f(x) = (x - 1)^{8/9}

, we need to find the critical numbers, determine the intervals where the function is increasing or decreasing, and apply the First Derivative Test to identify relative extrema. The correct answer is that the critical number is

x = 1

, which is a relative minimum. I hope this helps! Let me know if you have any questions.

is the second one, this one: For the function f(x) = (x minus 1)^(8 over 9): (a) Find the critical numbers of f (if any); (b) Find the open intervals where the function is increasing or decreasing; and (c) Apply the First Derivative Test to identify all relative extrema. options: 1. (a) x = 0 (b) increasing: (negative infinity,0) ; decreasing: (0,infinity) (c) relative max: f(0) = 1 2.(a) x = 1 (b) increasing: (negative infinity,1) ; decreasing: (1,infinity) (c) relative max: f(1) = 0 3. (a) x = 1 (b) decreasing: (negative infinity,1) ; increasing: (1,infinity) (c) relative min: f(1) = 0 4. (a) x = 0, 1 (b) decreasing: (negative infinity,0) union (1,infinity) ; increasing: (0,1) (c) relative min: f(0) = 1 ; relative max: f(1) = 0 5. (a) x = 0 (b) decreasing: (negative infinity,0) ; increasing: (0,infinity) (c) relative min: f(0) = 1

Solution by Steps

step 2

The derivative of

f(x)

is

f'(x) = \frac{8}{9}(x - 1)^{-\frac{1}{9}}

step 3

Set

f'(x) = 0

:

\frac{8}{9}(x - 1)^{-\frac{1}{9}} = 0

. This equation has no solution since

(x - 1)^{-\frac{1}{9}}

is never zero

step 4

However,

f'(x)

is undefined at

x = 1

, so

x = 1

is a critical number

step 5

To determine intervals of increase and decrease, we analyze the sign of

f'(x)

step 6

For x < 1,

(x - 1)^{-\frac{1}{9}}

is negative, so f'(x) < 0

step 7

For x > 1,

(x - 1)^{-\frac{1}{9}}

is positive, so f'(x) > 0

step 8

Therefore,

f(x)

is decreasing on

(-\infty, 1)

and increasing on

(1, \infty)

step 9

Apply the First Derivative Test: Since

f(x)

changes from decreasing to increasing at

x = 1

,

x = 1

is a relative minimum

step 10

Evaluate

f(1)

:

f(1) = (1 - 1)^{\frac{8}{9}} = 0

[question number] Answer

3

Key Concept

Critical Numbers and First Derivative Test

Explanation

Critical numbers are found where the derivative is zero or undefined. The First Derivative Test helps identify relative extrema by analyzing the sign changes of the derivative.

For the function f(x) = 2x^3 minus 21x^2 + 6: (a) Find the critical numbers of f (if any); (b) Find the open intervals where the function is increasing or decreasing; and (c) Apply the First Derivative Test to identify all relative extrema. Question 1 options: 1. (a) x = 0, 1 (b) increasing: (negative infinity,0) union (1,infinity) ; decreasing: (0,1) (c) relative max: f(0) = 6 ; relative min: f(1) = negative 13 2. (a) x = 0, 1 (b) decreasing: (negative infinity,0) union (1,infinity) ; increasing: (0,1) (c) relative min: f(0) = 6 ; relative max: f(1) = negative 13 3. (a) x = 0, 1 (b) increasing: (negative infinity,0) union (1,infinity) ; decreasing: (0,1) (c) relative max: f(0) = 6 ; no relative min. 4. (a) x = 0, 7 (b) increasing: (negative infinity,0) union (7,infinity) ; decreasing: (0,7) (c) relative max: f(0) = 6 ; relative min: f(7) = negative 337 5. (a) x = 0, 7 (b) decreasing: (negative infinity,0) union (7,infinity) ; increasing: (0,7) (c) relative min: f(0) = 6 ; relative max: f(7) = negative 337

Generated Graph

Solution by Steps

step 2

The first derivative is

f'(x) = 6x^2 - 42x

step 3

Set the first derivative equal to zero:

6x^2 - 42x = 0

step 4

Factor out the common term:

6x(x - 7) = 0

step 5

Solve for

x

:

x = 0

and

x = 7

. These are the critical numbers

step 6

To determine the intervals where the function is increasing or decreasing, we analyze the sign of

f'(x)

on the intervals determined by the critical numbers

step 7

Test the intervals

(-\infty, 0)

,

(0, 7)

, and

(7, \infty)

step 8

For

x

in

(-\infty, 0)

,

f'(x) = 6x(x - 7)

is positive, so

f(x)

is increasing

step 9

For

x

in

(0, 7)

,

f'(x) = 6x(x - 7)

is negative, so

f(x)

is decreasing

step 10

For

x

in

(7, \infty)

,

f'(x) = 6x(x - 7)

is positive, so

f(x)

is increasing

step 11

Apply the First Derivative Test to identify relative extrema

step 12

At

x = 0

,

f(x)

changes from increasing to decreasing, so

x = 0

is a relative maximum

step 13

At

x = 7

,

f(x)

changes from decreasing to increasing, so

x = 7

is a relative minimum

step 14

Evaluate

f(0)

and

f(7)

:

f(0) = 6

and

f(7) = -337

[question number] Answer

4

Key Concept

Critical Numbers and First Derivative Test

Explanation

Critical numbers are found by setting the first derivative to zero. The First Derivative Test helps determine where the function is increasing or decreasing and identifies relative extrema.

Use the graph of the function y = (x^3 over 16) minus 3x given below to estimate the open intervals on which the function is increasing or decreasing. A curve is graphed on the x y coordinate plane. The curve enters the bottom of the viewing window in the third quadrant, goes up and to the right crosses the x axis approximately at (negative 6.93,0), reaches the high point (negative 4,8), and then it goes down and to the right passes through the origin to reach the low point (4,negative 8). Then, goes up and to the right crosses the x axis approximately at (6.93,0), and exits the top of the viewing window.

Generated Graph

Solution by Steps

step 1

Given the function

y = \frac{x^3}{16} - 3x

, we need to find the intervals where the function is increasing or decreasing. To do this, we first find the derivative of the function

step 2

The derivative of

y = \frac{x^3}{16} - 3x

is

y' = \frac{d}{dx} \left( \frac{x^3}{16} - 3x \right)

step 3

Using the power rule and the constant multiple rule, we get

y' = \frac{3x^2}{16} - 3

step 4

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

\frac{3x^2}{16} - 3 = 0

step 5

Solving for

x

, we get

\frac{3x^2}{16} = 3 \implies x^2 = 16 \implies x = \pm 4

step 6

We now test the intervals around the critical points

x = -4

and

x = 4

to determine where the function is increasing or decreasing

step 7

For x < -4, choose

x = -5

: y' = \frac{3(-5)^2}{16} - 3 = \frac{75}{16} - 3 > 0, so the function is increasing

step 8

For -4 < x < 4, choose

x = 0

: y' = \frac{3(0)^2}{16} - 3 = -3 < 0, so the function is decreasing

step 9

For x > 4, choose

x = 5

: y' = \frac{3(5)^2}{16} - 3 = \frac{75}{16} - 3 > 0, so the function is increasing

Answer

The function is increasing on the intervals

(-\infty, -4)

and

(4, \infty)

, and decreasing on the interval

(-4, 4)

.

Key Concept

Intervals of Increase/Decrease

Explanation

To determine where a function is increasing or decreasing, find the derivative, set it to zero to find critical points, and test the intervals around these points.

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