## lizlozada Group Title For what values of k is f(x) continuous? 3x^2-11x-4 x<=4 kx^2-2x-1 x>4 one year ago one year ago

1. bahrom7893 Group Title

when x = 4, f(x) must be 0

2. bahrom7893 Group Title

k*4^2-2*4-1=0

3. bahrom7893 Group Title

there's an alternative method...

4. bahrom7893 Group Title

which is kinda what i did, but i skipped steps

5. bahrom7893 Group Title

3x^2-11x-4, when x<=4.. plug in 4 into that equation: 3*4^2-11*4-4=3*16-44-4=48-44-4=0

6. bahrom7893 Group Title

Thus the second equation must also be 0 when x=4

7. bahrom7893 Group Title

|dw:1348118755000:dw| lol btw that's not what the functions look like, but you have to make them touch each other

8. bahrom7893 Group Title

So kx^2-2x-1=0 when x=4

9. across Group Title

You want$\lim_{x\to4}\left(kx^2-2x-1\right)=0.$

10. bahrom7893 Group Title

16k-8-1=0 k=9/16

so the answer k= 9/16, that's it?

12. across Group Title

@bahrom7893 is right. That's the only value of $$k$$ for which the piece-wise defined function is continuous.

ok, what's the process? plugging in the 4?

then solving for k?

15. bahrom7893 Group Title

it's looking at where the function "breaks"... in this case it's at 4

16. bahrom7893 Group Title

Plug that part into the known piece of the function.. so you know that y is some number, in our case 0

17. bahrom7893 Group Title

So in order for our function not to break, the second piece must start where the first piece left off, or at 0

18. bahrom7893 Group Title

when x = 4

they will always give you what you have to plug in?

20. across Group Title

I'll ask an interesting question: is the function differentiable at that point? ;)