1. Jan 10, 2005

### Nguyen Thanh Nam

Can I upload the images here? so that whenever you choose my topic, they're shown, no need for you to open attachments?
Any way, I am a non-native so I get difficulties solving this problem. Tell me! (it's easy but I can't use English to state some sentence)
The URL of the problem:

Thanks

2. Jan 10, 2005

### dextercioby

You can attach various types of files,provided their size in not too big...

As for the question:
HINT:Meke sure the function and the first derivative are both continuous in "1".

Daniel.

3. Jan 10, 2005

### vincentchan

let the first function is f(x) and the second function is g(x), in order for the first derivative exist @ x=1, there must meet 2 condition,
1). the function must continue at x=1, which mean f(1)=g(1) ,
2). the function must be smooth, which mean f'(x)=g'(x) @ x=1

now you have 2 equation and 2 unknown, a and b.......

4. Jan 10, 2005

### Nguyen Thanh Nam

They are continous in '1'
When you check out [f(x)-f(1)]/(x-1), will we need to let them in to ways: x->1+ and x->1-, right?
But how to write down? :-)
Andm you see, you need to download the img file, any better way so that it's shown in the post? As some mathematical functions are long and complicated

Thanks

5. Jan 10, 2005

### HallsofIvy

Staff Emeritus
No, you have to MAKE then continuous at x= 1! That's the whole point of the problem.

Since $\sqrt{2-x^2}$ is continuous from the right, its value at x= 1 is 1
Since $x^2+ bx+ c$ is continuous for all x, we must have 1+ b+ c= 1. That gives one equation for b and c.

Now differentiate both formulas:

$\frac{df}{dx}= -x(2-x^2)^{-1/2}$ if x< 1
$\frac{df}{dx}= 2x+ a$ if x> 1

When x= 1, those are -1= 2+ a.