Can I Upload Images Here? Problem Solving for Non-Native English Speakers

[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

 

[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.

 

[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...

 

[Nguyen Thanh Nam]

They are continuous in '1'
When you check out [f(x)-f(1)]/(x-1), will we need to let them into 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

 

[HallsofIvy]

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.