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worty [1.4K]
4 years ago
7

The graph of g(x) is transformed from its parent function, f(x). Apply concepts involved in determining the key features of a ra

tional function to determine the domain and range of the function, .
A. What is the domain of the function, g(x)?
B. What is the range of the function, g(x)?

Mathematics
2 answers:
Alik [6]4 years ago
4 0

We are given the function:

g(x) = \frac{1}{x-6}

Domain:

The domain is the set of all possible x-values which will make the function "work", and will output real y-values.

In case of fractions, we must not have the denominator as zero , otherwise the function will become undefined.

So equating denominating equal to zero to find restriction.

x-6=0

x=6

So at x=6 , the function becomes undefined.

So domain is all real numbers except x=6.

Range:

For a fraction, we find domain of inverse function and that gives the range.

g(x) = \frac{1}{x-6}

replacing g(x) by y

y = \frac{1}{x-6}

switching y by x and x by y

x = \frac{1}{y-6}

solving for y,

y=\frac{1}{x} +6

Now here we find domain of this function.

Again for a fraction denominator cannot be zero

So range is :

g(x) >0 and g(x) <0

postnew [5]4 years ago
3 0

Answer:

A.f(x) = (one-half) Superscript x  D.The domains of both functions are the same    E.The translation from f(x) to g(x) is right 4 units and down 2 units.  

Step-by-step explanation:

A. D. E.

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Solving a Two-Step Matrix Equation<br> Solve the equation:
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Answer:

\boxed {x_{1} = 3}

\boxed {x_{2} = -4}

Step-by-step explanation:

Solve the following equation:

\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

3x_{1} + 2x_{2} + 1 = 2

-Second equation:

5x_{1} + 5x_{2} + 2 = -3

-Choose one of the two following equations, which I choose the first one, then you solve for x_{1} by isolating

3x_{1} + 2x_{2} + 1 = 2

-Subtract 1 to both sides:

3x_{1} + 2x_{2} + 1 - 1 = 2 - 1

3x_{1} + 2x_{2} = 1

-Subtract 2x_{2} to both sides:

3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1

3x_{1} = -2x_{2} + 1

-Divide both sides by 3:

3x_{1} = -2x_{2} + 1

x_{1} = \frac{1}{3} (-2x_{2} + 1)

-Multiply -2x_{2} + 1 by \frac{1}{3}:

x_{1} = \frac{1}{3} (-2x_{2} + 1)

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

-Substitute -\frac{2x_{2} + 1}{3} for x_{1} in the second equation, which is 5x_{1} + 5x_{2} + 2 = -3:

5x_{1} + 5x_{2} + 2 = -3

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

Multiply -\frac{2x_{2} + 1}{3} by 5:

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

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-Combine like terms:

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

\frac{5}{3}x_{2} + \frac{11}{3} = -3

-Subtract \frac{11}{3} to both sides:

\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}

\frac{5}{3}x_{2} = -\frac{20}{3}

-Multiply both sides by \frac{5}{3}:

\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}

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x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

-Multiply -\frac{2}{3} and -4:

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

x_{1} = \frac{8 + 1}{3}

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x_{1} = \frac{8 + 1}{3}

x_{1} = \frac{9}{3}

\boxed {x_{1} = 3}

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