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DanielleElmas [232]
2 years ago
6

Help needed on this composition math problem

Mathematics
1 answer:
Marrrta [24]2 years ago
3 0

Given that f(x) = x/(x - 3) and g(x) = 1/x and the application of <em>function</em> operators, f ° g (x) = 1/(1 - 3 · x) and the domain of the <em>resulting</em> function is any <em>real</em> number except x = 1/3.

<h3>How to analyze a composed function</h3>

Let be f and g functions. Composition is a <em>binary function</em> operation where the <em>variable</em> of the <em>former</em> function (f) is substituted by the <em>latter</em> function (g). If we know that f(x) = x/(x - 3) and g(x) = 1/x, then the <em>composed</em> function is:

f\,\circ\,g \,(x) = \frac{\frac{1}{x} }{\frac{1}{x}-3}

f\,\circ\,g\,(x) = \frac{\frac{1}{x} }{\frac{1-3\cdot x}{x} }

f\,\circ\,g\,(x) = \frac{1}{1-3\cdot x}

The domain of the function is the set of x-values such that f ° g (x) exists. In the case of <em>rational</em> functions of the form p(x)/q(x), the domain is the set of x-values such that q(x) ≠ 0. Thus, the domain of f ° g (x) is \mathbb{R} - \{\frac{1}{3} \}.

To learn more on composed functions: brainly.com/question/12158468

#SPJ1

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Answer:

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  • range: y ∉ {1}
  • horizontal asymptote: y=1
  • vertical asymptote: x=3

Step-by-step explanation:

The expression reduces to ...

  \dfrac{x^2+9x+20}{x^2+x-12}=\dfrac{(x+4)(x+5)}{(x+4)(x-3)}=\dfrac{x+5}{x-3}\quad (x\ne-4)

The domain is limited to values of x where the expression is defined. It is undefined where the denominator is zero, at x=-4 and x=3. The graph of the expression has a "hole" at x=4, where the numerator and denominator factors cancel.

  • the domain is all real numbers except -4 and +3

The function approaches the value of 1 as x gets large in magnitude, but it cannot take on the value of 1.

  • the range is all real numbers except 1

As discussed in 'range', there is a horizontal asymptote at y=1. That is the value you would get if you were to determine the quotient of the division:*

  (x+5)/(x-3) = 1 + (8/(x-3)) . . . . quotient is 1

There is a vertical asymptote at the place where the denominator is zero in the simplified expression: x = 3.

  • vertical asymptote at x=3; horizontal asymptote at y=1

_____

* For some rational functions, the numerator has a higher degree than the denominator. In those cases, the quotient may be some function of x. The "end behavior" of the expression will match that function. (Sometimes it is a "slant asymptote", sometimes a higher-degree polynomial.)

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