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hichkok12 [17]
3 years ago
14

Piece function:

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
5 0

First, make the table so you can sketch the graph:

    x   |    y

    5  |   20        

    4  |    9              x² - 5   ; x > 3

 <u>   3  |    </u><u>4 </u><u>  </u><u><em>(open dot)                      </em></u><em>   ⇒ </em>from the right

  <u>  3  |    </u><u>5</u><u>                  5     ; x = 3       </u>   ⇒ at x = 3

    3  |    2  <em>(open dot)                          ⇒ </em>from the left

    2  |    4              -2x + 8  ; x < 3          

<u>     1   |    6                                           </u>


Next, look at the graph (or table) to find the limits:

lim 3⁺ = 4  <em>as x approaches 3 from the right, y approaches 4</em>

lim 3⁻ = 2  <em>as x approaches 3 from the left, y approaches 2</em>

lim 3 = DNE <em>lim 3⁺ ≠ lim 3⁻ so the limit does not exist</em>

f(3) = 5  <em>when x = 3, y = 5</em>

f(x) is NOT continuous at x = 3 <em>because lim 3 ≠ f(3)</em>

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\qquad \: D = \sf \red{\sqrt{ {(x_2 -x_1)}^{2}  +   {(y_2 -y_1)}^{2} }}

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: \implies \: D  =   \sf \red{\sqrt{ {(x_2 -x_1)}^{2}  +   {(y_2 -y_1)}^{2} }} \\  \\ : \implies \: D  = \sf \sqrt{  {(5-( - 3))}^{2}  +   {(2 -( - 2))}^{2} }\\  \\ : \implies \: D  = \sf \sqrt{  {8}^{2}  +   {4}^{2} } \\  \\ : \implies \: D  = \sqrt{64 + 16}  \\  \\ : \implies \: D  = \sqrt{80}

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