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Calculus continuity question

Does anyone know how I solve this? o.o

http://i943.photobucket.com/albums/ad275/BlackyoshiCP/Untitled2_zps64aef08e.png

October 9, 2013

7 Comments • Newest first

Blackyoshi

Ohhh ok wow, I wish she could've explained it that clearly. xD Thanks everyone! And thank you Arthador for the more philosophical interpretation haha.

Reply October 9, 2013
Arthador

I believe your prof is more focused on the concept of continuity which is better than just graphing everything. What does it mean to be continuous? To be continuous at a point, the graph must approach the same number from the left and right of that point. You should set both equations equal to each other, plug in x=0, and solve for d. Note: It has been some time since I've done problems like this but i think this is the right way to do it. Good Luck

Reply October 9, 2013
momozzz

lt wants it to be continuous so d-x must be equal to -2.5x^2 + 3

d-x = -2.5x^2 + 3
d = -1.5x^2 + 3
d = -1.5(0)^2 + 3
d = -1.5(0) + 3
d = +3

3 - x = -2.5x^2 + 3
3 - 0 = -2.5(0) + 3
3 = 3

Reply October 9, 2013
vaelietta

limit as x approaches f(x) from the right must be EQUAL to limit as x approaches f(x) from the left. Figure it out from there.

Reply October 9, 2013
bowcreaters

Set first equation to 0, find x. Plug that x into 2nd equation, get D.

Looks more like PreCalc than actual Calc

[Edit]: My bad, I meant to say set x=0 for first equation. Then find the value. Find the same value for D that would make 1st equation equal to 2nd equation.

Reply October 9, 2013 - edited
Blackyoshi

[quote=LegendaryNee]didn't your teacher teach you to graph it and see it in picture? It's easier that way.[/quote]

No, our prof never mentioned anything like that. I'll try it, thanks.

Reply October 9, 2013 - edited
LegendaryNee

didn't your teacher teach you to graph it and see it in picture? It's easier that way.

Reply October 9, 2013 - edited