Write the piecewise defined function for the total cost of parking in the garage. That is, state the function C(x), where x is the number of hours a car is parked in the garage.

Respuesta :

Answer:

[tex]C(x) = \left[\begin{array}{ccc}4x &0 \le x \le 2& \\4 +2x &2 < x \le 6& \\16 &6<x\le 8& \end{array}\right[/tex]

Step-by-step explanation:

Given

See attachment for question

Required

The piece-wise function

From the attachment, we have:

(1) $4/hr for first 2 hours

This is represented as:

[tex]C(x) = 4x[/tex]

The domain is: [tex]0 \le x \le 2[/tex]

(2) $2/hr for next 4 hours

Here, we have:

[tex]Rate = 2[/tex]

The total cost in the first 2 hours is:

[tex]C(x) = 4x[/tex]

[tex]C(2) = 4*2 = 8[/tex]

So, this function is represented as:

[tex]C(x) = C(2) + Rate * (Time - 2)[/tex] ----- 2 represents the first 2 hours

So, we have:

[tex]C(x) = C(2) + Rate * (Time - 2)[/tex]

[tex]C(x) =8 + 2(x - 2)[/tex]

Open brackets

[tex]C(x) =8 + 2x - 4[/tex]

Collect like terms

[tex]C(x) =8 - 4+ 2x[/tex]

[tex]C(x) =4+ 2x[/tex]

The domain is:

[tex]2 < x \le 2 + 4[/tex]

[tex]2 <x \le 6[/tex]

(3) 0 charges for the last 2 hours

The maximum charge from (2) is:

[tex]C(x) =4+ 2x[/tex]

[tex]C(6) = 4 + 2*6[/tex]

[tex]C(6) = 4 + 12[/tex]

[tex]C(6) = 16[/tex]

Since there will be no additional charges, then:

[tex]C(x) = 16[/tex]

And the domain is:

[tex]6 < x \le 8[/tex] --- 8 represents the limit

So, we have:

[tex]C(x) = \left[\begin{array}{ccc}4x &0 \le x \le 2& \\4 +2x &2 < x \le 6& \\16 &6<x\le 8& \end{array}\right[/tex]

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