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1 year ago

Given f(x)=e^-x^3 find the vertical and horizontal asymptotes

Mathematics

1 answer:

Aleonysh [2.5K]1 year ago

5 0

Given:

$f\mleft(x\mright)=e^{-x^3}$

To find the vertical and horizontal asymptotes:

The line x=L is a vertical asymptote of the function f(x) if the limit of the function at this point is infinite.

But, here there is no such point.

Thus, the function f(x) doesn't have a vertical asymptote.

The line y=L is a vertical asymptote of the function f(x) if the limit of the function (either left or right side) at this point is finite.

$\begin{gathered} y=\lim _{x\rightarrow\infty}e^{-x^3} \\ =e^{-\infty} \\ y=0 \\ y=\lim _{x\rightarrow-\infty}e^{-x^3} \\ y=e^{\infty} \\ =\infty \end{gathered}$

Thus, y = 0 is the horizontal asymptote for the given function.

You might be interested in

Which expressions are equivalent to (5⋅x)⋅3 ? Drag and drop the equivalent expressions into the box.

marysya [2.9K]

<h2>Greetings!</h2>

Answer:

3⋅(5⋅x)

5⋅(x⋅3)

15x

Step-by-step explanation:

As the values are inside the brackets, it does not matter what side the (x3) is on, so 3⋅(5⋅x) is equivalent.

Multiplying the contents of the brackets in the third one (x * 3) by 5 gives the same value as 3 * (x * 5) so 5⋅(x⋅3) is also equivalent.

On multiplying the brackets out:

5 * x = 5x

5x * 3 = 15x

So 15x is also equivalent.

<h2>Hope this helps!</h2>

3 0

3 years ago

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$