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Vlad [161]
3 years ago
12

I only really need the domain and range, intervals, and end behavior. My Precalculus class never covered piecewise functions so

any help is much appreciated. Thanks so much

Mathematics
1 answer:
Serjik [45]3 years ago
3 0

(a)

Domain:

we know that domain is all possible values of x for which any function ios defined

Here, curve is defined for all values of x

so, domain is

(-\infty,\infty)

Range:

Range is all possible values of y for which x is defined

minimum value of y is -4

maximium value of y is +inf

so, range is

[-4,\infty)

(b)

x-intercept:

It is the value of x where f(x)=0

so, the point where curve intersects on x-axis

and we get

2x^2-4=0

x=\sqrt{2}

and

x=-5

(c)

Increasing interval:

where curve is moving upward

we can see that

(0,\infty)

Decreasing interval:

where curve is moving downward

we get

(-\infty,-1)U(-1,0)

Constant:

constant means horizontal line

there is no horizontal line here

so, it does not exist

(D)

End behavior:

when x-->+inf

y is moving upward

so, y---->+inf

when x-->-inf

y is moving upward

so, y---->+inf

(E)

we can see that curve jupms at x=-1

so, there is discontinuity at x=-1

and this is jump type of discontinuity

(F)

For odd:

f(-x)=-f(x)

For even:

f(-x)=f(x)

we can see that none of them holds true

so, this is neither

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jekas [21]

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The second one.

Step-by-step explanation:

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3 years ago
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ser-zykov [4K]

Answer:

The answer is 2x^{3}y^{-2}\sqrt{6}

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∵ \sqrt{x^{6}}=(x^{\frac{1}{2}})^{6}=x^{3}

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2 years ago
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2 years ago
An object of mass weighing 5.24 kilograms is raised to a height of 1.63 meters. What is the potential energy of the object at th
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<span>Find force,
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Find work,
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Example:
To solve this given word problem we can first identify the given and the apt formula to use in this phenomenon: Given: Force = 4, 500 N = 4, 500 kg-m/s^2 Acceleration = 5 m/s^2   </span>
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