How to Find a and b for a Derivative at 1?

[Nguyen Thanh Nam]

Ok, they tell me tofind a and b so that the function:
f(x)= *Root(2-x^2) if -root(2)<=x<=1
*x^2 + ax + b if x>1
has derivative at 1
I got that the condition for this graph to be continuous at 1 is a+b=0
And I moved to check out the derivative stuff:
When x->1+, the derivative is 0
But when 1->1-, igot stuck:
lim (x->1-) of [f(x)-f(1)]/(x-1) = lim (x->1-) of (x^2+ax+b-1)/(x-1). So how should I move on?

 

[dextercioby]

Equate the lateral limits for the derivative and get another relation between "a" and "b".
Then use the 2 equations found to determine "a" and "b".

Daniel.

 

[dextercioby]

Ok, they tell me tofind a and b so that the function:
f(x)= *Root(2-x^2) if -root(2)<=x<=1
*x^2 + ax + b if x>1
has derivative at 1
I got that the condition for this graph to be continuous at 1 is a+b=0
And I moved to check out the derivative stuff:
When x->1+, the derivative is 0
But when 1->1-, igot stuck:
lim (x->1-) of [f(x)-f(1)]/(x-1) = lim (x->1-) of (x^2+ax+b-1)/(x-1). So how should I move on?

How did u compute those derivatives??Post your work.My advice from the previous post remains valid as soon as u dig those derivatives rightly.

Daniel.

 

[Nguyen Thanh Nam]

OK.
We'll check when x>1:
lim (x->1+) of [f(x)-f(1)]/(x-1) = lim (x->1+) of [x^2+ax+b-1]/(x-1)] (from the function at my first post, I got that when x>1, f(x)=x^2+ax+b)

 

[Nguyen Thanh Nam]

Come on, anybody help me out!

 

[HallsofIvy]

Do you really need to use the basic formula for a derivative? You should know, by the time you are doing a problem like this, that the derivative of x<sup>2+ax+b is
2x+ b.
That tells you that the derivative, as you approach 1 from the right, is 2+ b, not 0.

The derivative of $\sqrt{2- x^2}= (2- x^2)^{\frac{1}{2}}$ is $\frac{1}{2}(2-x^2)^{-\frac{1}{2}}(-2x)$ which is -1 at x= 1. That is, the derivative, as you approach from the left is -1. In order that f be differentiable at x= 1, you must have 2+ b= -1.</sup>

 

[Nguyen Thanh Nam]

Yup, my class is at that lession, tis is an apllication. So, I didn't really understand you.:-) But I'll try to read it more.