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

1. bahrom7893

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

2. bahrom7893

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

3. bahrom7893

there's an alternative method...

4. bahrom7893

which is kinda what i did, but i skipped steps

5. bahrom7893

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

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

7. bahrom7893

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

8. bahrom7893

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

9. across

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

10. bahrom7893

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

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

12. across

@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

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

16. bahrom7893

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

17. bahrom7893

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

when x = 4

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

20. across

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