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MariettaO [177]
2 years ago
5

What is the end behavior of the function f of x equals negative 4 times the cube root of x?

Mathematics
1 answer:
vampirchik [111]2 years ago
8 0

The function f(x) = 4\sqrt[3]{x} is a cube root function

The function end behavior is:

\quad \mathrm{as}\:x\to \:+\infty \:,\:f\left(x\right)\to \:+\infty \:,\:\:\mathrm{and\:as}\:x\to \:-\infty \:,\:f\left(x\right)\to \:+\infty

<h3>How to determine the end behavior?</h3>

The equation of the function is given as:

f(x) = 4\sqrt[3]{x}

To determine the end behavior, we plot the graph of the function f(x).

From the attached graph of the function, we can see that:

As x approaches infinity, the function f(x) approaches infinity, and vice versa

Hence, the function end behavior is:

\quad \mathrm{as}\:x\to \:+\infty \:,\:f\left(x\right)\to \:+\infty \:,\:\:\mathrm{and\:as}\:x\to \:-\infty \:,\:f\left(x\right)\to \:+\infty

Read more about function end behavior at:

brainly.com/question/23968442

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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

3 0
3 years ago
Jack crawled 3 ft and Christine crawled 3⁄4 the distance Jack crawled. If you add the distance they each crawled together, what
LUCKY_DIMON [66]
Christine crawled:
3 ft* (3/4)= 2.25 ft.

3 ft+ 2.25 ft= 5.25 ft

We already know that 1 ft= 12 in.

5.25 ft* (12in/ 1 ft)= 63 in.

The total distance the two children crawled together is 63 in~
5 0
3 years ago
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