1. ## Continuity mx problem (Edited!)

Hi Community,

I've done this problem 3 times and am obviously forgetting something. Maybe I'm supposed to take the limit? Anyway here's the problem and my work...
Code:
find M so that f is continuous...

piecewise:
f(x) = mx - 11         if x< -8
x^2 + 3x - 3  if x >= -8
here's my work

Code:
f(-8-) = f(-8+)
-8x - 11 = 8^2 - (-8(3)) - 3
-8x - 11 = 85
0 = -8x - 96
96 / -8 != 12?
As always any help is appreciated!

2. Yes, you're supossed to take the limit. And for f to be continuous on -4, the must be the same (definition of continuity)

3. Originally Posted by mant1s
Hi Community,

I've done this problem 3 times and am obviously forgetting something. Maybe I'm supposed to take the limit? Anyway here's the problem and my work...
Code:
find M so that f is continuous...

piecewise:
f(x) = mx - 9         if x< -4
x^2 + 7x - 5  if x >= -4
here's my work

Code:
f(-8-) = f(-8+)
-8x - 11 = 8^2 - (-8(3)) - 3
-8x - 11 = 85
0 = -8x - 96
96 / -8 != 12?
As always any help is appreciated!
why you take -8 you want to study the continuity at x=-4 not -8 so

$\displaystyle -4m-9=16-28-5$

$\displaystyle 4m+9=17$

$\displaystyle 4m=8\rightarrow m=2$ right

4. Originally Posted by mant1s
Hi Community,

I've done this problem 3 times and am obviously forgetting something. Maybe I'm supposed to take the limit? Anyway here's the problem and my work...
Code:
find M so that f is continuous...

piecewise:
f(x) = mx - 9         if x< -4
x^2 + 7x - 5  if x >= -4
here's my work

Code:
f(-8-) = f(-8+)
-8x - 11 = 8^2 - (-8(3)) - 3
-8x - 11 = 85
0 = -8x - 96
96 / -8 != 12?
As always any help is appreciated!

sorry guys I typed the problem out wrong... it is supposed to be -8 instead of -4. I accidently typed the value from a sample problem.

5. Originally Posted by Ruun
Yes, you're supossed to take the limit. And for f to be continuous on -4, the must be the same (definition of continuity)
Do I take the limit of each side? Sorry I'm a little confused :/

6. Yes the one without m, will gave some real number $\displaystyle k$, and the second one will give you one real number depending of m. One equation, one unknowkn, solve for m, and you're ready

7. I hate to be a bother but could I please have an example?

8. Check this out

We say that a function is continuos a $\displaystyle x=c$ if and only if $\displaystyle \lim_{x\to{c^-}}f(x)=\lim_{x\to{c^+}}f(x)$

So by this definition

$\displaystyle \lim_{x\to{-8^-}}(mx-11)=\lim_{x\to{-8^+}}(x^2+3x-3)$$\displaystyle \Longleftrightarrow{m(-8)-11}=(-8)^2+3(-8)-3$

Now solve for m

9. Thanks Von,

I solved for

$\displaystyle \Longleftrightarrow{m(-8)-11}=(-8)^2+3(-8)-3$

by

Code:
-8m - 11 = -91
-8m = -80
m = -80/-8 ---> 10
However the answer is wrong. Did I do my algebra right?

10. Originally Posted by mant1s
Code:
-8m - 11 = -91
-8m = -80
m = -80/-8 ---> 10
However the answer is wrong. Did I do my algebra right?
No you did not.
Have you ever considered that may be a great deal of your trouble?
$\displaystyle (-8)^2+3(-8)-3=37$

11. Thanks Plato. I dono what my deal was on that one. I think it was just a +/- human error.