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bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1when x = 4, f(x) must be 0

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1there's an alternative method...

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1which is kinda what i did, but i skipped steps

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.13x^211x4, when x<=4.. plug in 4 into that equation: 3*4^211*44=3*16444=48444=0

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1Thus the second equation must also be 0 when x=4

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1348118755000:dw lol btw that's not what the functions look like, but you have to make them touch each other

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1So kx^22x1=0 when x=4

across
 2 years ago
Best ResponseYou've already chosen the best response.1You want\[\lim_{x\to4}\left(kx^22x1\right)=0.\]

lizlozada
 2 years ago
Best ResponseYou've already chosen the best response.0so the answer k= 9/16, that's it?

across
 2 years ago
Best ResponseYou've already chosen the best response.1@bahrom7893 is right. That's the only value of \(k\) for which the piecewise defined function is continuous.

lizlozada
 2 years ago
Best ResponseYou've already chosen the best response.0ok, what's the process? plugging in the 4?

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1it's looking at where the function "breaks"... in this case it's at 4

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1Plug that part into the known piece of the function.. so you know that y is some number, in our case 0

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1So in order for our function not to break, the second piece must start where the first piece left off, or at 0

lizlozada
 2 years ago
Best ResponseYou've already chosen the best response.0they will always give you what you have to plug in?

across
 2 years ago
Best ResponseYou've already chosen the best response.1I'll ask an interesting question: is the function differentiable at that point? ;)

lizlozada
 2 years ago
Best ResponseYou've already chosen the best response.0we haven't learned what that means...
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