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Sholpan [36]
3 years ago
6

Let f(x) be a function with domain [−3,∞) and range (−∞,2). If g(x)= f^−1(x) what is the domain of g(x+4)−7?

Mathematics
1 answer:
Crank3 years ago
7 0

Answer:

  • the domain of the function g(x+4)-7 is (-\infty,-2);
  • the range of the function g(x+4)-7 is [-10,\infty).

Step-by-step explanation:

If the function f(x) has the domain [-3,\infty) and the range (-\infty,2), then the function g(x)=f^{-1}(x) has the domain that id the function's f(x) range and the range that is function's f(x) domain. Therefore,

  • the domain of the function g(x) is (-\infty,2);
  • the range of the function g(x) is [-3,\infty).

The function g(x+4)-7 is the translation of the function g(x) 4 units to the left and 7 units down, hence

  • the domain of the function g(x+4)-7 is (-\infty,-2);
  • the range of the function g(x+4)-7 is [-10,\infty).
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azamat

Answer:

\large \boxed{y = (x - 5)^{2} + 2 }

Step-by-step explanation:

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We must convert it to the vertex form

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We can do this by completing the square.

\begin{array}{rcll}y & = & x^{2} - 10x + 27 & \\y - 27 & = & x^{2} - 10x & \text{Subtracted 27 from each side}\\y - 27&= & x^{2} - 10x + 25 - 25 & \text{Added and subtracted (b/2)}^{2}\\y - 27&= & (x - 5)^{2} - 25 & \text{Wrote the first three terms as the square of a binomial}\\y& = & \mathbf{(x - 5)^{2} + 2} & \text{Added 27 to each side}\\\end{array}\\\text{The vertex form of the parabola is $\large \boxed{\mathbf{y = (x - 5)^{2} + 2 }}$}The figure below shows that your parabola has its vertex at (5,2).

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