Question

Find the value that of a that makes the following function continuous (-infinity,infinity)

http://webwork.math.ttu.edu/wwtmp/equations/e8/efb7a5bbdb022e204e3e5552a1c8a01.png

Answer 1

find the limit as x goes to -5 of the first expression. set the second one equal to that limit, and solve for "a"

Answer 2

you got this?

Answer 3

factor the numerator and cancel with the denominator to find the limit of the first expression as x goes to -5

Answer 4

ok one second.

Answer 5

you know it has to factor as
\[(x+5)\times \text{something}\] because if you replace x by -5 you get 0, so don't think too hard about factoring

Answer 6

ok

Answer 7

well i know the other term has to be (6x^2 + something but the only thing i can think to put there is -1x but that doesn't work because that will leave me with a -5x rather than a +5x

Answer 8

and that also won't give me 50

Answer 9

you know the constant has to be 10

Answer 10

right because it need to get 50 for the last term

Answer 11

you can always divide if it is too hard to figure out this way

Answer 12

really

Answer 13

I did not know that.

Answer 14

but also if the constant is ten we also have 50x as well as 50 as the last to terms in the numerator

Answer 15

\[6x^3+29x^2+5x+50)(x+5)(6x^2+cx+10)\]

Answer 16

\[(6x^3+29x^2+5x+50)=(x+5)(6x^2+cx+10)\]

we need c now you will get 10x when you multiply out, and you have 5x so you know c = -1

Answer 17

oh ok I see c=-1

Answer 18

the x terms will come from 10x + cx and you have 5x making c = -1

Answer 19

if this is too annoying you can just divide and see what you get

Answer 20

now i can cancel ok

Answer 21

yes of course otherwise there is no limit.

Answer 22

so now do i one again factor?

Answer 23

leaving you with
\[6x^2-x+10\] replace x by -5 and see what you get
then put
\[-x^2+x+a= \text { your limit}\] and solve for a

Answer 24

no you factor and cancel. then you want the limit so replace x by -5 to find it

Answer 25

ok that makes sense I really appreciate it I do have one more trig problem I need help with I think I've posted it but i will repost in a second if not

Answer 26

the one with tangent?

Answer 27

same idea as with cosine. look up the double angle formula, then replace theta by arcsine or arccos whichever you have

Answer 28

yes thats the one is tan(2tan^-1(x)) i believe or something like that

Answer 29

yeah that is the idea. i forget the formulas (obviously) so look them up before computing

Answer 30

ok thanks man