If f(x) is continuous and differentiable and f(x) = ... then b?
f(x) =
{ax^4 + 5x, x< or = 2
{bx^2 - 3x, x> 2,
then b = ?
Help pls

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- myininaya

1st set these two expressions equal to each other: \[\lim_{x \rightarrow 2^-}f(x)=\lim_{x \rightarrow 2^+}f(x)\]

- myininaya

2nd set these two expressions equal to each other:
\[\lim_{x \rightarrow 2^-}f'(x)=\lim_{x \rightarrow 2^+}f'(x)\]

- anonymous

What should I do after that? Add those two equations together?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- myininaya

You will have a system of equations to solve just like from algebra class.

- anonymous

Well I got, 4a - 10 = 4b - 6 and then -32 + 5 = 4b - 3. Is that correct?

- myininaya

The first equation on the left hand side you have 4a-10
I think it should be 16a+10.
What happen to a in the second expression?

- myininaya

I mean equation err

- anonymous

oops you are right.

- myininaya

\[a(2)^4+5(2)=b(2)^2-3(2) \text{ first equation }\]

- myininaya

\[4a(2)^3+5=2b(2)-3 \text{ second equation }\]

- anonymous

Wait isn't it a(-2)^4 + 5 (-2) since the limit is coming from the left side?

- myininaya

It isn't approaching -2. It is approaching 2.

- myininaya

You have the left is just approaching 2 from the left.
You have the right is just approaching 2 from the right.
Either way they are both still approaching 2 and not some other number.

- anonymous

Ahh ok I got it :) After I get these 2 equations then what should I do afterward?

- myininaya

Have you ever solved a system of two linear equations?

- anonymous

Yes, like 4 years ago haha.

- anonymous

16a + 10
4b - 6
32a + 5
4b - 3 ( Can't I just substitute it and then solve it) ?

- myininaya

You can put them both in ax+by=c form if you want.
Like if we had something like 2x+3y=1
-2x+3y=1
This system is already set up for elimination. Add the equations together and solve for y first would be my strategy.
6y=2
y=2/6=1/3
Then plug into one of the other equations and solve for x.

- myininaya

Or this system:
3x-6y=2
x+y=2
This one isn't set up for elimination. But you could multiply the bottom equation by -3 so we would have it setup for elmination.
3x-6y=2
-3x-3y=-6
----------------add
-9y=-4
y=4/9
and then so on to find x.

- myininaya

\[a(2)^4+5(2)=b(2)^2-3(2) \text{ first equation }\]
\[16a+10=4b-6\]
\[16a-4b=-16 \text{ rewrote the equation 1} \]

- myininaya

Do you think you can rewrite equation 2 in that form?

- anonymous

32a - 4b = -8 ( 2nd equation right) right?

- myininaya

looks good

- myininaya

So we have
16a-4b=-16
32a-4b=-8

- myininaya

If we multiply the second or even the first (just not both of them) by -1, then we can set this up for elimination method.

- myininaya

I'm trying to eliminate b. (you could go for a but I prefer working with smaller numbers)

- myininaya

So
-16a+4b=16
32a-4b=-8
----------------add the equations together and what do we get?

- anonymous

I got the final answer of 6 for b. Is that right?

- anonymous

and .5 for a

- myininaya

yep. good job.

- anonymous

yayyy, you should just be my tutor haha. Thank you so much :D

- myininaya

np

Looking for something else?

Not the answer you are looking for? Search for more explanations.