Question

Hello! Any math tutors in the house? Calculus Q: I'm given a function where x<-2 and then another f(x) where x is greater than or equal to -2. The second f(x) (with the greater or equal to sign) has the variable "a" in it. The question then is: What value must be chosen for a in order to make this function continuous at -2?

Answer 1

A function is continuous at a point if
1. The limit exists
2. The limit as x approaches a is the same answer as f(a)

Answer 2

so would I substitute (-2) into the functions. set both functions equal to eachother and solve for a?

Answer 3

Do you have the actual functions?

Answer 4

yup. this is my first time on this site. I just don't know how to type them in here, but i'll try

Answer 5

Ah. Well, you can hit the equation button below the typing box. Feel free to get used to it.

Answer 6

oh! I never noticed that! are you a university student as well?

Answer 7

Yeah, I am. Im actually a math tutor where I go to school, lol. Im just the.....least experienced math tutor, but meh xD

Answer 8

ahhhhh. I hope you're getting paid for this.

Answer 9

Not for on this site. I just come onhere not only to hang out, but doing this helps me practice and make sure I know what I need to know or find out what I need to learn.

Answer 10

To be honest with you, a month ago I probably couldnt have given you that definition I gave. I intuitively knew it, but wouldnt have known the actual definition of continuity.

Answer 11

\(\begin{cases}
f_1(x) & \text{ if } x<-2 \\
f_2(x) & \text{ if } x\ge -2
\end{cases}\)

Right click to see the \(\LaTeX\) code used to do cases.

Answer 12

That's pretty impressive. To be honest, I hate math. I need this course to apply for graduate school. Yeah, i'm taking a first year calculus course at ubc, it's been quite the roller coaster for me.

Answer 13

k, ill post the equation on here, maybe someone can help. when i equate them to eachother, one of the denominator equals 0, which i know is no no.

Answer 14

Well, just to see what your two pieces to the function are mighthelp.

Answer 15

f(x) = 6x\[f(x)=6x^3 + 13x^2 + 12x + 20 / x + 2\]

Answer 16

the first one doesn't count lol

Answer 17

i'll write the second one now

Answer 18

Lol, alright.

Answer 19

\[f(x) -6x^2 + 6x + a\]

Answer 20

\(f(x)=\dfrac{6x^3 + 13x^2 + 12x + 20}{x + 2}\) perhaps?

Answer 21

for the first one: x is less than -2. for the second one x is greater than or equal to -2.

Answer 22

Yeah, so seems like youre looking for a value of a such thatthe two graphs intersect at x = -2

Answer 23

oooo thanks. that's fancy.

Answer 24

Yeah, theres a fraction button, too xD

Answer 25

q: What value must be chosen for a in order to make this function continuous at -2? ... why can't i just equal them to each other and sole for a

Answer 26

\(f(x)=\begin{cases}
\dfrac{6x^3 + 13x^2 + 12x + 20}{x + 2} & \text{ if } x<-2 \\ \\
-6x^2 + 6x + a & \text{ if } x\ge -2
\end{cases}\)

Answer 27

omgd. e.mccormick. you're amazing.

Answer 28

Nah, I have just use \(\LaTeX\) longer than many here and can hand type it. I have done some papers with it.

Answer 29

You can actually factor that top piecewise to get the x+2 to cancel out. Will make finding out your y-value much easier.

Answer 30

but x is cubed. and 20 doesn't have an x.

Answer 31

Yah, you basically need to find what it would be at -2 on the top if it was valid. Then you find a value of the bottom that would have that same answer.

Answer 32

yeah, just need to get rid of the denominator

Answer 33

I tested it just to see. Divide it out and check for yourself, it does divide well :3

Answer 34

kk, i'll give it a shot, thanks! xoxo

Answer 35

Yep, sure. Once you divide it out, you can plug in x = -2 for that function and find the y-value. Then your job is to fine a such that x = -2 in that function produces the same y-value.

Answer 36

thanks!!! i'll post an update if i get it. lol

Answer 37

i mean, "when" lol

Answer 38

Yeah, sure xD

Answer 39

I'd show it to you on a graph, but that would give it away.

Answer 40

wow. I can't seem to factor the x cubed. any suggestions?

Answer 41

i know i have to factor out (x+2) from each term

Answer 42

atleast i think

Answer 43

Well, I just literally divided it by x+2 to check

6x^2 + x + 10
x+2 | 6x^3 + 13x^2 + 12x + 20
-[6x^3 +12x^2]
x^2 + 12x
-[ x^2 + 2x]
10x + 20
-[10x + 20]
0

Answer 44

i want to curl up in a corner and never wake up.

Answer 45

So if you did the division, obviously (x+2) is a factor because of no remainder. What I have left is what happens after that (x+2) term cancels. So since the x+2 term in the denominator is now gone, we can plug in x = -2 to see what the y-value is:
6(-2)^2 -2 + 10 = 32

So now we need a value of a in the other function such that as x approaches -2, the answer is 32.

And no o.o No need to beat yourself up or something o.o

Answer 46

Did you go over synthetic division? There is also factor by grouping as a possibility.

Answer 47

no, I just googled it. grouping I learned a couple of years ago.

Answer 48

http://www.purplemath.com/modules/synthdiv.htm

Answer 49

thanks!

Answer 50

a=68!

Answer 51

holy shinizzle.

Answer 52

Lol.

Answer 53

thanks for your help Psymon and e.mcc (going to call you mc for short)

Answer 54

FYI: Calculus requires and uses every type of math before it. For example, take a very simple integral:
\(\int 2 x \text{d}x \)
Once you learn the rules, you know this has a value added to the exponent and the result of that is divided out of the whole thing. So:
\(\int 2 x \text{d}x \implies \int 2 x^1 \text{d}x \implies \dfrac{2x^{1+1}}{2}+c\implies x^2+c\)

Now, I just used \(1+1=2\) and \(\dfrac{2}{2}=1\) and was doing calculus.

Answer 55

And here is that graph I talked about:
https://www.desmos.com/calculator/na0jflqgdz

Very good tool to use for checking yourself!

Answer 56

Thanks you two. appreciate it MC!

Answer 57

Yep yep.

Answer 58

And a great site for brushing up on all the math you forgot:
https://www.khanacademy.org/

Answer 59

I used KA to get back up to speed and it made a HUGE difference!

Have fun!

Answer 60

oh man. that site is a life saver. esp for logs. need to brush up on some algebra.
this girl is pretty amazing too: http://www.youtube.com/user/TheIntegralCALC

Answer 61

Yah! I saw a couple of hers. Good. Ever seen PatricJMT's videos?

Answer 62

PatrickJMT that is...

Answer 63

nope on youtube?

Answer 64

Yah. He lists his by topic and is a math major. Does it on paper or white board woth a marker, so all you see is the math.

Answer 65

perfect. wrote it down on my notebook.

Answer 66

are you from north america as well?

Answer 67

California. You are a tad furthur north in Canada.

Answer 68

I've always wanted to explore California. Your summers seem amazing!

Answer 69

Well, it depends on where. The state ranges from snow country in the winter to, well, dry and sunny in the winter. In the summer that becomes sunny and breezy to roasted.

Answer 70

eeeeek. well, i've always wanted to visit there. esp after the hit song california girls. jk jk

Answer 71

Just don't visit this part in the middle of summer:
http://www.livescience.com/38054-why-death-valley-hot.html

Answer 72

NEVER. that's our future billions of years from today.