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lara31 [8.8K]
3 years ago
14

Indentify the range of the the function {(2,5), (3,8), (4,11)}

Mathematics
2 answers:
faltersainse [42]3 years ago
7 0

The domain of a function is the complete set of possible values of the independent variable (input values).

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain (output values).

For the function {(2,5), (3,8), (4,11)} you have:

  • the domain is {2,3,4};
  • the range is {5,8,11}.

In other words, the function {(2,5), (3,8), (4,11)} can be written as

  • f(2)=5;
  • f(3)=8;
  • f(5)=11.

The values 2, 3, 4  are input values (the domain) and values 5, 8, 11 are output values (the range).

Answer: the range is {5,8,11}

Lilit [14]3 years ago
4 0

Range:  {5,  8, 11}

Hope it helps

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1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

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