A split function with 2 parameters: (x^2+ab+b)/(x^2+2x-15), x<3; -(x^2)/3+7/2, x>= 3

[fthku]

A split function with 2 parameters: (x^2+ab+b)/(x^2+2x-15), x<3; -(x^2)/3+7/2, x>= 3

Hi guys, I'm doing old exams practicing for one coming up, and I've been stuck at this particular problem:

Find a and b if the function is continuous at 3

. . . . .\(\displaystyle f(x)\, =\, \begin{cases} \dfrac{x^2\, +\, ax\, +\, b}{x^2\, +\, 2x\, -\, 15}&\mbox{for }\, x\, <\, 3 \\ -\dfrac{x^2}{3}\, +\, \dfrac{7}{2}&\mbox{for }\, x\, \ge\, 3 \end{cases}\)

I don't have much to show, I've calculated the right hand side and the limit and value of the function are 0.5 at that point, meaning the limit of the left hand side should also be 0.5, but I'm stuck at that point.

Thanks in advance!

 

[Deleted member 4993]

Hi guys, I'm doing old exams practicing for one coming up, and I've been stuck at this particular problem:

Find a and b if the function is continuous at 3

I don't have much to show, I've calculated the right hand side and the limit and value of the function are 0.5 at that point, meaning the limit of the left hand side should also be 0.5, but I'm stuck at that point.
Thanks in advance!

Why? How? - show work.

 

[fthku]

Why? How? - show work.

Supposedly it is minus infinity. I didn't see a way to factor anything and l'hopital doesn't seem applicable here. I'd appreciate any help

 

[pka]

particular problem:
Find a and b if the function is continuous at 3

Here is a hint: We need \(\displaystyle (x-3)~\&~(x+1)\) to be factors of \(\displaystyle x^2+ax+b\).
You Must Explain WHY!

 

[Ishuda]

Supposedly it is minus infinity. I didn't see a way to factor anything and l'hopital doesn't seem applicable here. I'd appreciate any help

To explain it slightly different than pka, we need x-3 to be a factor of the numerator in order to get rid of that infinity. That is the denominator is
d = (x-3)(x+A)
and the numererator is
n = (x-3)(x+B)
so that the expression is
E = \(\displaystyle \dfrac{x^2\, +\, a\, x\, +\, b}{x^2\, +\, 2\, x\, -\, 15}\)
=\(\displaystyle \dfrac{(x-3)(x+A)}{(x-3)(x+B)}\, =\, \dfrac{x+A}{x+B}\)

Now do some division (synthetic or long) to find A and B and thus a and b.

 

[Ishuda]

Hi guys, I'm doing old exams practicing for one coming up, and I've been stuck at this particular problem:

Find a and b if the function is continuous at 3

I don't have much to show, I've calculated the right hand side and the limit and value of the function are 0.5 at that point, meaning the limit of the left hand side should also be 0.5, but I'm stuck at that point.
Thanks in advance!

Another way to find b (in terms of a) is to note that since f has a removable singularity at x=3 and the denominator is 0 at x=3, the numerator must be zero at x=3. So
9 + 3 a + b = 0
and
b = - 3 (a + 3)

 

[fthku]

Thanks a lot guys, appreciate the help!