Making Piecewise Function Continuous w/ First & Second Derivatives

Dollydaggerxo
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I have to make a piecewise function continuous with first and second derivatives at x=0 by finding a value for a and b.

I have made the function continuous by equating the RHS and LHS limits and then solving for the variable a.

The limit of the first derivatives of the LHS and RHS are the same, both 0, so that makes the first derivative continuous...

However, the limits of the second derivates for the RHS and LHS are different. One is infinty when the other is 0.

My question is what does this mean exactly? Is the question not possible if my second derivatives are not continuous?
 
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First, please tell us what the problem really is! You say "by finding a value of a and b" but tell us nothing about how "a" and "b" are related to the function. Are you given a formula for the function involving a and b? If so, that formula is important information.
 
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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