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vagabundo [1.1K]
4 years ago
6

Match each function to its domain and range.

Mathematics
1 answer:
Vlad1618 [11]4 years ago
4 0
The domain of a function is the set of the possible values of x, while the range of a function is the set of the resulting values of the functions from the domain.

By inspection it can be seen that the function
f(x) = 4 - 4x has the domain and range as follows:
<span>domain: {0, 1, 3, 5, 6}
range: {-20, -16, -8, 0, 4}

i.e. 4 - 4(0) = 4 - 0 = 4
4 - 4(1) = 4 - 4 = 0
4 - 4(3) = 4 - 12 = -8
4 - 4(5) = 4 - 20 = -16
4 - 4(6) = 4 - 24 = -20


Also, the function
f(x) = 5x - 3 has the domain and range as follows:
</span><span>domain: {-2, -1, 0, 3, 4}
range: {-13, -8, -3, 12, 17}

i.e. 5(-2) - 3 = -10 - 3 = -13
5(-1) - 3 = -5 - 3 = -8
5(0) - 3 = 0 - 3 = -3
5(3) - 3 = 15 - 3 = 12
5(4) - 3 = 20 - 3 = 17


Also, the function
f(x) = -10x has the domain and range as follows:
</span><span>domain: {-4, -2, 0, 2, 4}
range: {-40, -20, 0, 20, 40}

i.e. -10(-4) = 40
-10(-2) = 20
-10(0) = 0
-10(2) = -20
-10(4) = -40

Also, the function
f(x)=\frac{3}{x}+1.5 has the domain and range as follows:
</span><span>domain: {-3, -2, -1, 2, 6}
range: {0.5, 0, -1.5, 3, 2}

i.e.
\frac{3}{-3}+1.5=-1+1.5=0.5 \\  \\ \frac{3}{-2}+1.5=-1.5+1.5=0 \\  \\ \frac{3}{-1}+1.5=-3+1.5=-1.5 \\  \\ \frac{3}{2}+1.5=1.5+1.5=3 \\  \\ \frac{3}{6}+1.5=0.5+1.5=2</span>
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Answer:

Solving the expression \frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3} we get \mathbf{\frac{1}{y^{\frac{5}{2} }x^3z^{\frac{5}{4} }}}

Step-by-step explanation:

We need to solve the expression:

\frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3}

We know the exponent rule: (a^n)^m = a^{nm}

\frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3}\\\\=\frac{y^2z^{\frac{1}{4}} }{z^{\frac{3}{2}}.x^3y^{\frac{9}{2} }}

Now, another exponent rule says that: \frac{a^m}{a^n}=a^{m-n}

=\frac{y^{2-\frac{9}{2}} z^{\frac{1}{4}-\frac{3}{2} } }{x^3}\\=\frac{y^{\frac{4-9}{2}} z^{\frac{1-3*2}{4} } }{x^3}\\=\frac{y^{\frac{-5}{2}}z^{\frac{-5}{4} } }{x^3}

We also know that: a^{-m}=\frac{1}{a^m}

=\frac{1}{y^{\frac{5}{2} }x^3z^{\frac{5}{4} }}

So, solving the expression \frac{y^2z^{\frac{1}{4}} }{(z^{\frac{1}{2}}.xy^{\frac{3}{2}})^3} we get \mathbf{\frac{1}{y^{\frac{5}{2} }x^3z^{\frac{5}{4} }}}

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Step-by-step explanation:

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