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1 year ago

Which function below has the end behavior f(x) →-infinity as x → infinity and f (x) → infinity as x→-infinity

Mathematics

1 answer:

Sholpan [36]1 year ago

6 0

To find the function that has the following end behavior:

$\begin{gathered} f(x)\rightarrow-\infty\text{ as }x\rightarrow\infty \\ f(x)\rightarrow\infty\text{ as }x\rightarrow-\infty \end{gathered}$

Considering the function which is given in option C.

When x tends to infinity,

$\begin{gathered} f\mleft(x\mright)=-x^3-4x^2+x \\ \lim _{x\to\infty}(-x^3-4x^2+x)=-\infty \\ \lim _{x\to-\infty}(-x^3-4x^2+x)=\infty \end{gathered}$

In other words, the degree of the given function is 3.

That is, odd.

The leading coefficient is -1.

That is, negative.

Hence, the end behavior is,

$\begin{gathered} f(x)\rightarrow-\infty\text{ as }x\rightarrow\infty \\ f(x)\rightarrow\infty\text{ as }x\rightarrow-\infty \end{gathered}$

Hence, the correct option is C.

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Find the area of the rectangle below

algol13

30 square cm that’s the answer

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3 years ago

Use the method of lagrange multipliers to find

Yanka [14]

Answer:

a) The function is: f(x, y) = x + y.

The constraint is: x*y = 196.

Remember that we must write the constraint as:

g(x, y) = x*y - 196 = 0

Then we have:

L(x, y, λ) = f(x, y) + λ*g(x, y)

L(x, y, λ) = x + y + λ*(x*y - 196)

Now, let's compute the partial derivations, those must be zero.

dL/dx = λ*y + 1

dL/dy = λ*x + 1

dL/dλ = (x*y - 196)

Those must be equal to zero, then we have a system of equations:

λ*y + 1 = 0

λ*x + 1 = 0

(x*y - 196) = 0

Let's solve this, in the first equation we can isolate λ to get:

λ = -1/y

Now we can replace this in the second equation and get;

-x/y + 1 = 0

Now let's isolate x.

x = y

Now we can replace this in the last equation, and we will get:

(x*x - 196) = 0

x^2 = 196

x = √196 = 14

then the minimum will be:

x + y = x + x = 14 + 14 = 28.

b) Now we have:

f(x) = x*y

g(x) = x + y - 196

Let's do the same as before:

L(x, y, λ) = f(x, y) + λ*g(x, y)

L(x, y, λ) = x*y + λ*(x + y - 196)

Now let's do the derivations:

dL/dx = y + λ

dL/dy = x + λ

dL/dλ = x + y - 196

Now we have the system of equations:

y + λ = 0

x + λ = 0

x + y - 196 = 0

To solve it, we can isolate lambda in the first equation to get:

λ = -y

Now we can replace this in the second equation:

x - y = 0

Now we can isolate x:

x = y

now we can replace that in the last equation

y + y - 196 = 0

2*y - 196 = 0

2*y = 196

y = 196/2 = 98

The maximum will be:

x*y = y*y = 98*98 = 9,604

6 0

3 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra

Nana76 [90]

The length of BC is $4 \sqrt{5}$ .

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

Leg rule of geometric mean theorem:

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

$\Rightarrow 16\times 5 = x\times x$

$\Rightarrow 80= x^2$

$\Rightarrow 16\times 5= x^2$

Taking square root on both sides.

$\Rightarrow \sqrt{16\times 5} = \sqrt{x^2}$

$\Rightarrow \sqrt{4^2\times 5} = \sqrt{x^2}$

square and square roots get canceled, we get

$\Rightarrow 4\sqrt{ 5} = x$

The length of BC is $4 \sqrt{5}$ .

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3 years ago

How do you break apart the factor 42 using place values

VLD [36.1K]

Let's break apart 42.

The 4 is in the tens. The 2 is in the ones.

Place Value starting from Thousands

Thousands Hundreds Tens Ones

_________ ________ ____ _____

4 2

So this is how we break up numbers using place value.

Hope this helps!

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3 years ago

Solve for x. 3.rº 80° 20° A. 10 B. 20 C. 50 D. 80

ohaa [14]

Answer:

x = 20°

Step-by-step explanation:

The angle measures should add up to 180 degrees. So that leaves us with 100 degrees left to find if we take out the 80 given.

3x + 2x simplifies to 5x. 5x must equal 100.

100/5 = 20.

Therefore x = 20.

6 0

3 years ago

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