Question

Find the values of a and b that make f continuous everywhere.
f(x) =


(x2 − 4) / (x − 2)
if x < 2

ax2 − bx + 1 if 2 ≤ x < 3
4x − a + b if x ≥ 3

Answer 1

could you make the question a bit clearer?

Answer 2

not too sure how..... another one like it tho

Answer 3

Find the values of a and b that make f continuous everywhere.
f(x) =


(x2 − 4) / (x − 2) if x < 2

ax2 − bx + 3 if 2 ≤ x < 3

4x − a + b if x ≥ 3


a =
b =

Answer 4

So you know the left limit needs to equal the right limit for both x=2 and x=3?

Answer 5

You have these equations:
\[\lim_{x \rightarrow 2^-}f(x)=\lim_{x \rightarrow 2^+}f(x)\]
\[\lim_{x \rightarrow 3^-}f(x)=\lim_{x \rightarrow 3^+}f(x)\]

Answer 6

Let me know if you need further help. :)

Answer 7

this is not a limit problem though, it is problem where i need to evaluate both a and b

Answer 8

You needs limits here.

Answer 9

If the function is continuous at x=2, then you the following things
f(2)=limx->2(f(x))
and of course for both sides to exist.

In order for limx->2(f(x)) to exist, the left limit needs to equal the right limit.

Answer 10

Using what I wrote above, you will get a system of two linear equations involving the unknowns a and b.

Answer 11

IM SO CONFUSED

Answer 12

On the limit part?

Answer 13

no how to evaluate

Answer 14

like are my values 2 and 3

Answer 15

I looked at where the function was broken up at.
I'm trying to make the function exist at those two numbers since those are the only 2 numbers that could possibly make our function discontinuous. If we find the right a and b, we can ensure the function will be continuous at x=2 and x=3.

So looking at the limits I asked you to evaluate...we have:
\[\lim_{x \rightarrow 2^-}f(x)=\lim_{x \rightarrow 2^+}f(x) \]
\[\lim_{x \rightarrow 2^-} \frac{x^2-4}{x-2}=\lim_{x \rightarrow 2^+} (ax^2-bx+1)\]

The left hand side can be reduced
The right hand side is a polynomial and polynomails are continuous everywhere for any real input. You can just plug in x=2 on the right hand side of the equation.

Now the other equation is from
\[\lim_{x \rightarrow 3^-}f(x)=\lim_{x \rightarrow 3^+}f(x) \]
\[\lim_{x \rightarrow 3^-}(ax^2-bx+1)=\lim_{x \rightarrow 3^+}(4x-a+b) \]
Both the left and right expressions are polynomials. Just plug in 3 into both.

Answer 16

so a(3)^2-b(3)+1 ???

Answer 17

So you are looking at the last equation I wrote I assume:

a(3)^2-b(3)+1=4(3)-a+b

Answer 18

So you have that equation (which you can simplify and combine like terms)
Now go back to the first one and evaluate both limits (simplify and combine like terms)
You will see we have a system of two linear equations with two unknowns.

Answer 19

Did you reduce that fraction yet?

Answer 20

no i am so confused im sorry

Answer 21

\[\frac{x^2-4}{x-2}\]

Do you know how to factor x^2-4?

Answer 22

yes
they cancel

Answer 23

Yep you reduced the fraction now plug in what x is approaching

Answer 24

and you were talking about the (x+2)'s canceling, right?

Answer 25

oops the (x-2)'s

Answer 26

yes
so...3? so 3-2=1

Answer 27

:( no no...
what...where did 3' come from?

Answer 28

We are looking at the first equation I wrote.

Answer 29

The one where x is approaching 2 not 3.

Answer 30

We are done with the approaching 3 equation for now.

Answer 31

We will return to it in a bit.

Answer 32

So you said (x^2-4)/(x-2)=x+2 right?

Answer 33

since the (x-2)'s canceled.

Answer 34

Now since x is approaching 2, just plug it in. 2+2=4

Answer 35

So that was the left hand side of our equation. Not pluggin 2 on the right hand side of the equation gives you?

Answer 36

i dont know where to look

Answer 37

At the equations I gave you.

Answer 38

I gave you two.

Answer 39

We already looked at the second.
We are looking at the first now.

Answer 40

equation 1:
\[\lim_{x \rightarrow 2^-} \frac{x^2-4}{x-2}=\lim_{x \rightarrow 2^+} (ax^2-bx+1) \]

equation 2:
\[\lim_{x \rightarrow 3^-}(ax^2-bx+1)=\lim_{x \rightarrow 3^+}(4x-a+b) \]


We already evaluated both sides of equation 2. You were also suppose to simplify and combine like terms for equation 2.

Now we are looking back equation 1.
We reduced the fraction on the left hand side giving us:
\[\lim_{x \rightarrow 2^-} (x+2)=\lim_{x \rightarrow 2^+} (ax^2-bx+1) \]
Now since both of these functions are continuous at x=2, plug in x=2.