## anonymous one year ago help please ! to two decimal places, find the value of k that will make the function f(x) continuous everywhere.

1. anonymous

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2. freckles

evaluate both the left and right limit of x=4

3. freckles

$\lim_{x \rightarrow 4^-}f(x)=? \\ \lim_{x \rightarrow 4^+}f(x)=?$

4. freckles

oops -4

5. anonymous

my choices are: 11.00 -2.47 -0.47 none of these

6. freckles

ok can you evaluate both: $\lim_{x \rightarrow -4^-}f(x)=? \\ \lim_{x \rightarrow -4^+}f(x)=?$

7. freckles

hint ^- means look to the left (which is the left function of x=-4 and use it to plug in -4 into) hint ^+ means look to the right (which is the right function of x=-4 and use it to plug in -4 into) both the left and right limit need to be equal so that you can have the actual limit at x=-4 exist

8. anonymous

o.o im lost lol

9. freckles

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10. freckles

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11. anonymous

is it A?

12. freckles

I don't know. Haven't done the problem.

13. freckles

would you know how to evaluate: $\lim_{x \rightarrow -4}(3x+k) \text{ or } \lim_{x \rightarrow -4}(kx^2-5)$

14. anonymous

no. o.o

15. freckles

Both functions are continuous at x=-4 why don't you evaluate the limits by replacing x with -4?

16. freckles

and you want both (left and right limits of x=-4) sides to be equal so you have $3(-4)+k=k(-4)^2-5$

17. freckles

can you solve linear equations?

18. anonymous

Do you have to take a limit here? Can you just substitute -4 for x and set the two parts equal to each other?

19. freckles

one of the things we need for continuity at x=-4 is: $\lim_{x \rightarrow -4}f(x)=L$ we get to have this if : $\lim_{x \rightarrow -4^{-} }f(x)=\lim_{x \rightarrow -4^{+}}f(x)=L$ so formally the answer is yes to that question

20. freckles

to the limit one

21. freckles

and informally ( I would say) yes to the second question