The inverse notation f^-1 used in a pure mathematics problem is not always used when finding inverses of applied problems. rather, the universe of a function such as c =c (q) will be q=q(c). the following problem illustrates this idea.
in a certain country the following function represents the income tax t (in dollars) do for a person who is adjusted gross income is g dollars where 30,500 ? g ? 74,100.
t(g)=4250+0.25(g-30,500)
a.) what is the domain of the function t?
given that the income tax do t is an increase in linear function of adjusted gross income g find the range of the function t
c.) find adjusted gross income g as a function of income tax t . what are the domain and the range of this function?

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  A)  30,500 < G ≤ 74,100

  B)  4,250 < T ≤ 15,150

  C) G = 4(T -4,250) +30,500; domain: 4,250 < T ≤ 15,150; range: 30,500 < G ≤ 74,100

Step-by-step explanation:

A) The domain of T is given in the problem statement:

  30,500 < G ≤ 74,100

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B) The range of T is the set of values of T that correspond to the domain values. The minimum will be T(30,500) = 4,250. The maximum will be T(74,100) = 4,250 +0.25(74,100 -30,500) = 15,150.

  4,250 < T ≤ 15,150

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C) The inverse function can be found by solving the given equation for G.

  T = 4250 +0.25(G -30500)

  T -4250 = 0.25(G -30500)

  4(T -4250) = G -30500

  G = 30,500 +4(T -4,250)

The domain and range of this function are the range and domain of the function T(g), respectively.

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step-by-step explanation:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.