Question

Need confirmation on my question. will post pic and will post my solution.

Answer 1

what i did was:
I got the lim x->2- f(x) = lim x-> 2+ f(x)

so...
4 = 4a - 2b + 3
0 = 4a - 2b - 1 <--- i used this as the lim x-> 3-

lim x-> 3- (f(x)) = lim x-> 3+ (f(x))
4a - 2b - 1 = 12 - a + b
a = (13 + 3b)/5


i used the value of a in:
0 = 4a - 2b - 1
b = -47/2

and i used b in:
a = (13+ 3b)/5
a = (-115/10)

Is that correct?

Answer 2

I think you're wrong, and anyway you're answer for a isn't simplified enough, considering that both -115 and 10 have a factor of five.

Answer 3

ok how do you know im wrong? and the simplified a is -23/2. ill fix it later

Answer 4

\[10a-4b=9\]Substitute your a and b into the limit as x approaches 3
10 times a is -115 and negative 4 times -47/2 is 94\[-115+94=-21\]

Answer 5

Above I set \[9a-3b+3=12-a+b\]

Answer 6

but for this one i used:
4a - 2b - 1 = 12 - a + b

Answer 7

I substituted into your original equations...?
Unless I'm miscomprehending, I think you accidentally set 12-a+b=0, which isn't stated anywhere. In order for the function to be continous at x=3,
\[ax ^{2}-bx+3=4x-a+b\] and then you substitute 3 for x

Answer 8

ok so if i substitute 3 for x, i would get
either
a = (9 + 4b)/10

or
b = (10a - 9)/4

say i got one of those, where would I then use it?

Answer 9

In your original equations which have to be equal to one another, the quadratic and the line. That eliminates either a or b and allows you to solve for a single variable.

Answer 10

hmm..
so..
I'll use a = (9 + 4b)/10:
on the 1st equation and the 2nd equation.

1st equation is:
x + 2 || x<2

2nd equation:
ax^2 - bx + 3 || 2<= x < 3

so they both exist at 2.

lim x-> 2- (x+2) = lim x->2+ (ax^2 - bx + 3)
2 + 2 = 4a - 2b + 3
4 = 4((9 + 4b)/10) - 2b + 3
40 = 36 + 14b + 3
b = 1/14


so i use this value in the original ax2−bx+3=4x−a+b:
lim x->3- (ax^2 - bx + 3) = lim x->3+ (4x - a + b)
9a - 3b + 3 = 12 - a + b
9a - 3(1/14) + 3 = 12 - a + (1/14)
63a = 63 - 7a + 2
a = 13/14

is that how you plug them in?

Answer 11

@L.T. Are you still there?

Answer 12

I got it it. haha.
I took

lim x-->2- f[1] = lim x-->2+ f[2]

so...

4 = 4a - 2b + 3 [Eqn 1]

leave that for awhile.



then

lim x-->3- f[2] = lim x-->3+ f[3]

so...

9a - 3b + 3 = 12 - a + b

then either get a or b



a = (9 + 4b)/10



use the a in [Eqn 1]:

4 = 4((9 + 4b)/10) - 2b + 3

then solve for b



b = 13/2

then use this value in a

a = (9 + 4(13/2))/10

a = 7/2

thats it.