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3 years ago

For each question below, use the function f (x) = x3

Mathematics

2 answers:

Tomtit [17]3 years ago

6 0

Answer:

A. The function is negative on the interval (-∞, 0)

B. The only root is x = 0

C. The function approaches ∞ when x gets really large (goes towards ∞). The function approaches -∞ when x gets really small (goes towards -∞).

Step-by-step explanation:

The function given is $f(x)=x^3$

The graph is attached. Looking at the graph will make understanding the solutions easier.

We need to find the intervals that is divided by the roots of the function.

To find x-intercepts, we set $f(x)=0$

$f(x)=x^3\\0=x^3\\x=0$

*Looking at the graph, we can also see that x=0 is the only x-intercept*

Thus the intervals are from -∞ to 0 and from 0 to ∞

We take value in each interval to determine whether the function is negative or positive. If the answer is positive, function is increasing, if negative, the function is decreasing.

From -∞ to 0, we take x = -1

$f(-1)=(-1)^3\\=-1$

Hence it is decreasing

From 0 to ∞, we take x = 1

$f(1)=(1)^3\\=1$

Hence it is increasing

The function is negative on the interval (-∞, 0)

To find the roots, x-intercepts, we set $f(x)=0$

$f(x)=x^3\\0=x^3\\x=0$

The only root is x = 0

To find end behavior, we look at the graph.

The end behavior means what happens to the function when x goes to ∞ and -∞

Looking at the graph, the function approaches ∞ when x gets really large (goes towards ∞). The function approaches -∞ when x gets really small (goes towards -∞).

*Also, if you plug in large x values, f(x) goes towards infinity. If you plug in small x values, f(x) goes towards negative infinity*

givi [52]3 years ago

5 0

Answer:

A. (-∞ , 0)

B. x = 0 triple root

C. The function tends to infinity

Step-by-step explanation:

For the function f(x) = x^3 we have:

Negative intervals:

(-∞ , 0)

The roots of this function is:

x = 0

The final behavior of this function tends toward infinity

A graphic is attached below

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A parallelogram has sides of length 7.7 millimeters and 17.2 millimeters. What is the perimeter?

Arlecino [84]

Solutions

We know that a rectangle has four sides and the perimeter of a rectangle is the distance around the outside of the rectangle.The opposite sides of a rectangle are congruent To find the perimeter you have to use the formula 2 × (Side A + Side B).

We have the side lengths 7.7 mm and 17.2 mm.

Solve

2 × (Side A + Side B)
2 × ( Side A (7.7) + Side B (17.2) )
= (7.7 x 2) + (17.2 x 2)
= 49.6

The perimeter is 49.6 mm².

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3 years ago

Read 2 more answers

What is the simplified form of 162

SashulF [63]

Answer:

9 √ 2

Step-by-step explanation:

1. Rewrite 162 as $9^{2}$ ⋅2

2. factor 81 out of 162

$\sqrt{81 (2)}$

3. Rewrite 81 as $9^{2}$

$\sqrt{9^{2}.2 }$

4. Pull the terms from under the radical

$\sqrt[9]{2}$

I hope this helps :)

5 0

3 years ago

Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

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

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

8 0

3 years ago

Beth got on the bus at 8:06 a.m. thirth-five minutes later what time she got to school

Tresset [83]

8:41 a.m.
Thirty-five minutes after 8:06 a.m. means
8:06
0:35 +
_____
8:41 a.m.

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4 years ago

F(x)=x2+8x+6. What is the value of f(4)?

dezoksy [38]

46

When it says f(4) that means x=4

So you then just fill in all the x’s with 4’s and solve as normal

4 0

3 years ago