Identify the domain and the range of the function y-3x-6-2 for x as even integers between 4-5 to 10-5-

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Domain and Range of a Function

Identify the ...

Question

Identify the domain and the range of the function $y = \frac{3 x + 6}{2}$ for $x$ as even integers between $4.5$ to $10.5.$

Domain

= {5, 6, 7, 8, 9, 10}

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Range

= {12, 15, 18}

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Domain

= {6, 8, 10}

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Range is all real values.

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Solution

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y = [[x - 3]]

- 3 < x < 3

X

Y

x - y = 2 n; where n is a integer

X : {4, 5, 6, 7} Y : {2, 6, 7, 9} x \in X & y \in Y

\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}

[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0

f (x) = \sqrt{([x] - 1)} + \sqrt{(4 - [x])}

Identify the domain and range of the function f(x), defined in N, as below:

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$6$	$8$	$10$
$\frac{3 (6) + 6}{2} = \frac{24}{2} = 12$	$\frac{3 (8) + 6}{2} = \frac{30}{2} = 15$	$\frac{3 (10) + 6}{2} = \frac{36}{2} = 18$