Determining if a function is differentiable at the indicated point

dugga boy
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How do you determine this?

F(x) x^2 +1 if x<1
F(x) 2x if x >= 1 at x=1

Are there designated steps? I understand that it is the derivative, but I don't understand the differentiable at the indicated point part..
 
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For a function to be differentiable at a point, its left sided and right sided derivatives must be equal. Usually the function is the same on both sides of a point, but for this function it has two different pieces.
 
I don't understand how is it different on the left sided and right sided? Would you put 1 for the value of x and get a solution that way?
 
On the "left" of 1, the function is defined as F(x)=x^2 +1, then F'(1)=_____
On the "right" of 1, the function is defined as F(x)=2x, then F'(1)=_____

If these two values exist and are equal, then the function is differentiable at 1.
 
Ok so only if first there is a vale for f'(x) and if the two valeus match is the function "differentiable" at the indicated point...Awesome! THANK YOU
One more question if you would be so kind...how do you evaluate if F'(x) = an integer with no variable? F'(1) = 2 since F(x) =2x
 
If there's no variable then the function doesnt...vary. ie no matter what x is, the function is just constant.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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