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>
Answer:
Quadrant 2
Step-by-step explanation:
Answer: x=-1,5.
Step-by-step explanation:
Answer:
last one
Step-by-step explanation:
The answer is whatever the other guy said
