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Savatey [412]
4 years ago
14

Identify the key characteristics of the parent fifth-root function f (x) = ^5sqrtx. Include the following: domain, range, interv

als where the function is increasing, decreasing, or constant, intercepts, where the function is even, odd, both or neither, whether the function is continuous or discontinuous, asymptotes, and end behavior.
sqrt=square root.
The square root is supposed to be to the fifth power.

Mathematics
1 answer:
Andre45 [30]4 years ago
7 0

Given functin is :

f\left(x\right)=\sqrt[5]{x}

We know that the domain of the expression is all real numbers except where the expression is undefined. In given function, there is no real number that makes the expression undefined. Hence domain is all real numbers.

Domain: (-∞,∞)

Range is the set of y-values obtained by plugging values from domain so the range will also same.

Range: (-∞,∞)

If we increase value of x then y-value will also increase so that means it is an INCREASING function. You can also verify that from graph.

It crosses x and y-axes both at the origin

Hence x-intercept=0 and y-intercept=0

Graph is not symmetric about y-axis hence it can't be EVEN

Graph is not symmetric about origin so it is ODD.

There is  no breaking point in the graph so that means it is a Continuous function.

There is no hoirzontal or vertical or slant line which seems to be appearing to touch the graph at infinity so there is NO asymptote.

END behaviour means how y-changes when x approaches infinity.

From graph we can see that when x-approaches -∞ then y also approaches ∞.

when x-approaches +∞ then y also approaches +∞.

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3 0
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Consider the system of differential equations dx/dt=−2y dy/dt=−2x. . Convert this system to a second order differential equation
Musya8 [376]

Answer:

Step-by-step explanation:

we have the following differential equations

\frac{dx}{dt}=-2y\\\frac{dy}{dt}=-2x\\

by differentiating the second equation we have

\frac{d}{dt}(\frac{dy}{dt})=-2\frac{dx}{dt}\\\frac{d^{2}y}{dt^{2}}=-2\frac{dx}{dt}\\\frac{dx}{dt}=\frac{-1}{2}\frac{d^{2}y}{dt^{2}}

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\frac{-1}{2}\frac{d^{2}y}{dt^{2}}=-2y\\\frac{d^{2}y}{dt^{2}}-4y=0

and by using the characteristic polynomial

m^{2}+4=0\\m=\±2i

the solution is

y(t)=Acos(2t)+Bsin(2t)

and to compute x(t) we have

\frac{dx}{dt}=-2Acos(2t)-2Bsin(2t)\\\\\int dx = \int[-2Acos(2t)-2Bsin(2t)]dt\\\\x(t)=-Asin(2t)+Bcos(2t)

and if we use x(0)=4 and y(0)=3, we can calculate the constants A and B

x(0)=B=4\\y(0)=A=3

I hope this is useful for you

regards

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Marie has a small copy of rene magritte's famous painting, the schoolmaster. Her copy has dimensions 2 inches by 1.5 inches. The
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First, lets convert our dimensions into cm. 

2*2.54 = 5.08 cm.
1.5*2.54 = 3.81 cm.

Now, lets multiply by for the get the dimensions of the original painting in inches. 

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Finally, lets divide our new dimensions.

203.2/2.54 = 80 in.
152.4/2.54 =  60 in.

The dimensions of the original painting is 80 in. by 60 in.
8 0
3 years ago
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