Solved on Sep 20, 2023

Find the local max/min of $f(x)=6+x+x^{2}-x^{3}$ , and the intervals where it's increasing/decreasing. State answers to 2 decimal places.

STEP 1

Assumptions1. The function is $f(x)=6+x+x^{}-x^{3}$ . We need to find the local maximum and minimum values of the function and the value of $x$ at which each occurs.
3. We need to find the intervals on which the function is increasing and on which it is decreasing.

STEP 2

To find the local maximum and minimum values of the function, we first need to find the derivative of the function. The derivative will help us find the critical points of the function, which are potential local maximum and minimum points.
$f'(x) = \frac{d}{dx}(6+x+x^{2}-x^{})$

STEP 3

Apply the power rule to differentiate each term in the function.
$f'(x) =0 +1 +2x -3x^{2}$

STEP 4

implify the derivative.
$f'(x) =1 +2x -3x^{2}$

STEP 5

To find the critical points, we set the derivative equal to zero and solve for $x$ .
$1 +2x -3x^{2} =0$

STEP 6

Rearrange the equation to a standard quadratic form.
$-3x^{2} +2x +1 =0$

STEP 7

olve the quadratic equation for $x$ using the quadratic formula.
$x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

STEP 8

Substitute $a = -3$ , $b =2$ , and $c =1$ into the quadratic formula.
$x = \frac{-2 \pm \sqrt{(2)^{2} -4(-3)(1)}}{2(-3)}$

STEP 9

implify the expression under the square root.
$x = \frac{-2 \pm \sqrt{4 +12}}{-6}$

STEP 10

Calculate the value under the square root.
$x = \frac{-2 \pm \sqrt{16}}{-6}$

STEP 11

Take the square root of16.
$x = \frac{- \pm4}{-6}$

STEP 12

Calculate the two possible values for $x$ .
$x = \frac{-2 +4}{-6} = -\frac{}{}, \quad x = \frac{-2 -4}{-6} =$

STEP 13

Now that we have the critical points, we need to determine whether each point is a local maximum, a local minimum, or neither. We can do this by using the second derivative test.
First, find the second derivative of the function.
$f''(x) = \frac{d}{dx}( +2x -3x^{2})$

STEP 14

Apply the power rule to differentiate each term in the derivative.
$f''(x) =0 +2 -6x$

STEP 15

implify the second derivative.
$f''(x) =2 -x$

STEP 16

Evaluate the second derivative at each critical point. If the second derivative is positive at a critical point, then the function has a local minimum at that point. If the second derivative is negative at a critical point, then the function has a local maximum at that point.
$f''(-\frac{}{3}) =2 -6(-\frac{}{3}) =4, \quad f''() =2 -6() = -4$

STEP 17

Therefore, the function has a local minimum at $x = -\frac{}{3}$ and a local maximum at $x =$ .
To find the corresponding $y$ -values, substitute these $x$ -values into the original function.
$f(-\frac{}{3}) =6 + (-\frac{}{3}) + (-\frac{}{3})^{2} - (-\frac{}{3})^{3} = \frac{57}{10}, \quad f() =6 + +^{2} -^{3} =7$

STEP 18

So, the local maximum is $(,7)$ and the local minimum is $\left(-\frac{}{3}, \frac{57}{10}\right)$ .

STEP 19

Now, to find the intervals on which the function is increasing and decreasing, we need to consider the sign of the first derivative.
The function is increasing where the derivative is positive, and decreasing where the derivative is negative.

STEP 20

We already found the derivative to be $f'(x) = +x -3x^{}$ .

STEP 21

We also found the critical points to be $x = -\frac{1}{3}$ and $x =1$ .

STEP 22

We can use these critical points to divide the number line into intervals. Then we can test a number from each interval in the derivative to determine whether the function is increasing or decreasing on that interval.

STEP 23

The intervals are $(-\infty, -\frac{1}{3})$ , $(-\frac{1}{3},1)$ , and $(1, \infty)$ .

STEP 24

Choose a test point from each interval and substitute it into the derivative.
For the interval $(-\infty, -\frac{1}{3})$ , choose $x = -1$ .
$f'(-1) =1 +(-1) -3(-1)^{} = -$ Since $f'(-1) <0$ , the function is decreasing on the interval $(-\infty, -\frac{1}{3})$ .

STEP 25

For the interval $(-\frac{1}{3},1)$ , choose $x =0$ .
$f'(0) =1 +(0) -3(0)^{} =1$ Since $f'(0) >0$ , the function is increasing on the interval $(-\frac{1}{3},1)$ .

STEP 26

For the interval $(1, \infty)$ , choose $x =$ .
$f'() =1 +() -3()^{} = -5$ Since $f'() <0$ , the function is decreasing on the interval $(1, \infty)$ .

STEP 27

So, the function is increasing on the interval $\left(-\frac{1}{3},1\right)$ and decreasing on the intervals $(-\infty, -\frac{1}{3})$ and $(1, \infty)$ .

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Find the local max/min of f(x)=6+x+x2−x3f(x)=6+x+x^{2}-x^{3}f(x)=6+x+x2−x3, and the intervals where it's increasing/decreasing. State answers to 2 decimal places.

STEP 1

Assumptions1. The function is f(x)=6+x+x−x3f(x)=6+x+x^{}-x^{3}f(x)=6+x+x−x3 . We need to find the local maximum and minimum values of the function and the value of xxx at which each occurs.3. We need to find the intervals on which the function is increasing and on which it is decreasing.

STEP 2

STEP 3

Apply the power rule to differentiate each term in the function.f′(x)=0+1+2x−3x2f'(x) =0 +1 +2x -3x^{2}f′(x)=0+1+2x−3x2

STEP 4

implify the derivative.f′(x)=1+2x−3x2f'(x) =1 +2x -3x^{2}f′(x)=1+2x−3x2

STEP 5

To find the critical points, we set the derivative equal to zero and solve for xxx.1+2x−3x2=01 +2x -3x^{2} =01+2x−3x2=0

STEP 6

Rearrange the equation to a standard quadratic form.−3x2+2x+1=0-3x^{2} +2x +1 =0−3x2+2x+1=0

STEP 7

olve the quadratic equation for xxx using the quadratic formula.x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}x=2a−b±b2−4ac​​

STEP 8

Substitute a=−3a = -3a=−3, b=2b =2b=2, and c=1c =1c=1 into the quadratic formula.x=−2±(2)2−4(−3)(1)2(−3)x = \frac{-2 \pm \sqrt{(2)^{2} -4(-3)(1)}}{2(-3)}x=2(−3)−2±(2)2−4(−3)(1)​​

STEP 9

implify the expression under the square root.x=−2±4+12−6x = \frac{-2 \pm \sqrt{4 +12}}{-6}x=−6−2±4+12​​

STEP 10

Calculate the value under the square root.x=−2±16−6x = \frac{-2 \pm \sqrt{16}}{-6}x=−6−2±16​​

STEP 11

Take the square root of16.x=−±4−6x = \frac{- \pm4}{-6}x=−6−±4​

STEP 12

Calculate the two possible values for xxx.x=−2+4−6=−,x=−2−4−6=x = \frac{-2 +4}{-6} = -\frac{}{}, \quad x = \frac{-2 -4}{-6} =x=−6−2+4​=−​,x=−6−2−4​=

STEP 13

STEP 14

Apply the power rule to differentiate each term in the derivative.f′′(x)=0+2−6xf''(x) =0 +2 -6xf′′(x)=0+2−6x

STEP 15

implify the second derivative.f′′(x)=2−xf''(x) =2 -xf′′(x)=2−x

STEP 16

STEP 17

STEP 18

So, the local maximum is (,7)(,7)(,7) and the local minimum is (−3,5710)\left(-\frac{}{3}, \frac{57}{10}\right)(−3​,1057​).

STEP 19

Now, to find the intervals on which the function is increasing and decreasing, we need to consider the sign of the first derivative.The function is increasing where the derivative is positive, and decreasing where the derivative is negative.

STEP 20

We already found the derivative to be f′(x)=+x−3xf'(x) = +x -3x^{}f′(x)=+x−3x.

STEP 21

We also found the critical points to be x=−13x = -\frac{1}{3}x=−31​ and x=1x =1x=1.

STEP 22

We can use these critical points to divide the number line into intervals. Then we can test a number from each interval in the derivative to determine whether the function is increasing or decreasing on that interval.

STEP 23

The intervals are (−∞,−13)(-\infty, -\frac{1}{3})(−∞,−31​), (−13,1)(-\frac{1}{3},1)(−31​,1), and (1,∞)(1, \infty)(1,∞).

STEP 24

STEP 25

For the interval (−13,1)(-\frac{1}{3},1)(−31​,1), choose x=0x =0x=0.f′(0)=1+(0)−3(0)=1f'(0) =1 +(0) -3(0)^{} =1f′(0)=1+(0)−3(0)=1Since f′(0)>0f'(0) >0f′(0)>0, the function is increasing on the interval (−13,1)(-\frac{1}{3},1)(−31​,1).

STEP 26

For the interval (1,∞)(1, \infty)(1,∞), choose x=x =x=.f′()=1+()−3()=−5f'() =1 +() -3()^{} = -5f′()=1+()−3()=−5Since f′()<0f'() <0f′()<0, the function is decreasing on the interval (1,∞)(1, \infty)(1,∞).

STEP 27

So, the function is increasing on the interval (−13,1)\left(-\frac{1}{3},1\right)(−31​,1) and decreasing on the intervals (−∞,−13)(-\infty, -\frac{1}{3})(−∞,−31​) and (1,∞)(1, \infty)(1,∞).

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Find the local max/min of $f(x)=6+x+x^{2}-x^{3}$ , and the intervals where it's increasing/decreasing. State answers to 2 decimal places.

Assumptions1. The function is $f(x)=6+x+x^{}-x^{3}$ . We need to find the local maximum and minimum values of the function and the value of $x$ at which each occurs.
3. We need to find the intervals on which the function is increasing and on which it is decreasing.

Apply the power rule to differentiate each term in the function.
$f'(x) =0 +1 +2x -3x^{2}$

implify the derivative.
$f'(x) =1 +2x -3x^{2}$

To find the critical points, we set the derivative equal to zero and solve for $x$ .
$1 +2x -3x^{2} =0$

Rearrange the equation to a standard quadratic form.
$-3x^{2} +2x +1 =0$

olve the quadratic equation for $x$ using the quadratic formula.
$x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

Substitute $a = -3$ , $b =2$ , and $c =1$ into the quadratic formula.
$x = \frac{-2 \pm \sqrt{(2)^{2} -4(-3)(1)}}{2(-3)}$

implify the expression under the square root.
$x = \frac{-2 \pm \sqrt{4 +12}}{-6}$

Calculate the value under the square root.
$x = \frac{-2 \pm \sqrt{16}}{-6}$

Take the square root of16.
$x = \frac{- \pm4}{-6}$

Calculate the two possible values for $x$ .
$x = \frac{-2 +4}{-6} = -\frac{}{}, \quad x = \frac{-2 -4}{-6} =$

Apply the power rule to differentiate each term in the derivative.
$f''(x) =0 +2 -6x$

implify the second derivative.
$f''(x) =2 -x$

So, the local maximum is $(,7)$ and the local minimum is $\left(-\frac{}{3}, \frac{57}{10}\right)$ .

Now, to find the intervals on which the function is increasing and decreasing, we need to consider the sign of the first derivative.
The function is increasing where the derivative is positive, and decreasing where the derivative is negative.

We already found the derivative to be $f'(x) = +x -3x^{}$ .

We also found the critical points to be $x = -\frac{1}{3}$ and $x =1$ .

The intervals are $(-\infty, -\frac{1}{3})$ , $(-\frac{1}{3},1)$ , and $(1, \infty)$ .

For the interval $(-\frac{1}{3},1)$ , choose $x =0$ .
$f'(0) =1 +(0) -3(0)^{} =1$ Since $f'(0) >0$ , the function is increasing on the interval $(-\frac{1}{3},1)$ .

For the interval $(1, \infty)$ , choose $x =$ .
$f'() =1 +() -3()^{} = -5$ Since $f'() <0$ , the function is decreasing on the interval $(1, \infty)$ .

So, the function is increasing on the interval $\left(-\frac{1}{3},1\right)$ and decreasing on the intervals $(-\infty, -\frac{1}{3})$ and $(1, \infty)$ .